New Transition Maths [2 ed.] 9781909417359

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New

2nd Edition

A Structured Work Plan for Transition Year including Student Workbook

Michael O’Keeffe Editor: Horst Punzet

MENTOR BOOKS 43 Furze Road Sandyford Industrial Estate Dublin 18 Tel: 01-2952112 Fax: 01-2952114 Website: www.mentorbooks.ie Email: [email protected] All Rights Reserved.

Text:

Michael O’Keeffe

Editor:

Horst Punzet

Edited by:

Daniel McCarthy

Design, layout & cover:

Kathryn O’Sullivan

Acknowledgements: The publishers would like to thank Getty Images and Alamy Images for permission to use images in this book. The author would like to acknowledge the creative skills of Kathryn O’Sullivan, the dedication of Treasa O’Mahony, the advice of Horst Punzet and the enthusiasm and energy of Danny McCarthy – all of which were essential to the publication of this book. ISBN: 978-1-909417-35-9

© Michael O’Keeffe 2015

1 3 5 7 9 10 8 6 4 2

Dedication To Chrissie, Anna and Eamon and my parents Jim and Abby

Contents Calculator Use Reference Guide . . . . . . . . . . . . . . . . . . . . . . . 5 1.

Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.

Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.

Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.

Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.

Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

6.

Pythagoras’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

7.

Ratio and Fibonacci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

8.

Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

9.

Pi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

10. Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 11. Combinations and Permutations . . . . . . . . . . . . . . . . . . . . 105 12. Fermat’s Last Theorem (and other Conjectures) . . . . . 114 13. Chaos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 14. Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 15. Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 16. Interesting Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 17. Infinity and Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

Calculator Use Reference Guide

Calculator Use Reference Guide Here are some useful calculator buttons. The operation varies slightly, but most scientific calculators have the following functions: (a) Square

X2

To find 42

4

X2

= . . . . . . . . . . . . . . . . . . 16

(b) Cube

X3

To find 43

4

X3

= . . . . . . . . . . . . . . . . . . 64

^

To find 45

4

^

5

4

= ...................2

6

4

(c) Higher Power

(d) Square Root

To find 4

3

(e) Cube Root

3

To find

(f) Higher Roots

x

To find 5 7776

5

3

64

7

x

7

7

=

=

. . . . . . . . . . . 1,024

..............4

6 .................6

NOTE Some of these functions may not have a button for themselves and are written above a different function. If this is the case, simply use the Shift button or 2nd function Higher Roots

5

(g) Fraction

d c

button. To find 5 7776

x

Shift

7

x

7

7

6

= .....6

To convert 5·2 into a fraction

5

·

2

d c

...........................

5

26 5

Transition Year Maths

NOTE On some calculators the number must be entered before the calculation is done.

5

·

2

d c

=

26 5

...............

(h)*Factorial x! To find 5! (5! means 5 × 4 × 3 × 2 × 1)

5

X!

=

. . . . . . . . . . . . . . . . . . . . . . . . . . . 120

(i)** Combinations nCr How many combinations of two things can be made from five things. 5 × 4 20 5 nCr 2 means = = 10 2 ×1 2

5

nCr

2

. . . . . . . . . . . . . . . . . . . . . . . . . . . 10

(j)** Permutations nPr How many permutations of two things can be made from five things. 5 nPr 2 means 5 × 4 = 20

5

nPr

2

. . . . . . . . . . . . . . . . . . . . . . . . . . . 20

* See Chapter 16 (Interesting Numbers). ** See Chapter 10 (Probability) and Chapter 11 (Combinations and Permutations).

(k) Brackets

(

) The computation in the bracket will always be done first. 3+4×5

3

+

4

×

5

=

(3 + 4) × 5

(

3

+

4

)

×

5

=

. . . . . . . . 35

3 + (4 × 5)

3

+

(

4

×

5

)

=

. . . . . . . 23

×

5

%

=

................4

. . . . . . . . . . . . . . . 23

(l) Percentage % To find 5% of €80

8

0

6

Calculator Use Reference Guide

(m) Scientific notation EXP Enter the number 300,000,000,000 into your calculator. Note the number of zeros.

3 EXP 1

1

=

300,000,000,000 = 3 × 100,000,000,000 = 3 × 1011

Scientific Notation in more detail: Scientific notation is used to deal with large and small numbers. It is useful because instead of writing down all the zeros, the number of zeros is written down.

RULE Move decimal once to right, power goes down one. Move decimal once to left, power goes up one.

NOTE 100 is defined as being equal to 1.

Write the following numbers in scientific notation: (a) 2 . . . . . . . . . . 2 × 100 (g) 2 . . . . . . . . . . . . . . . 2 × 100 (b) 20 . . . . . . . . 2 × 101 (h) 0·2 . . . . . . . . . . . . . . 2 × 10-1 2 (c) 200 . . . . . . . . 2 × 10 (i) 0·02 . . . . . . . . . . . . . 2 × 10-2 (d) 2,000 . . . . . . 2 × 103 (j) 0·002 . . . . . . . . . . . . 2 × 10-3 4 (e) 20,000 . . . . . 2 × 10 (k) 0·0002 . . . . . . . . . . . 2 × 10-4 (f) 200,000 . . . . 2 × 105 (l) 0·00002 . . . . . . . . . . 2 × 10-5

Example 1:

PROJECT

Seconds of Age

How many seconds old will you be when your school finishes today? Assume you were born at 12 noon on your birthday.

NOTE There are 365 days in a year. Leap years have one extra day i.e. 366 days. The following are the leap years between 1990 and 2020: 1992, 1996, 2000, 2004, 2008, 2012, 2016 and 2020.

Do we need these numbers? 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000 = 1 × 1091 This is the number of atoms in the observable universe according to a recent estimation. 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,0 00,000,000,000,000,000,000,000,000,000,000,000,000,000. One with a hundred noughts, named by Edward Kasner in 1938, is a ‘googol’. This was the largest number with a name, at that time. There are much larger named numbers now, like a ‘googolplex’ and a ‘googolplexian’. Look them up on the internet.

7

Transition Year Maths

Example 2:

Write the following numbers without scientific notation: (a) 4·4 × 100 = . . . . . 4·4 (g) 4·4 × 100 = . . . . . . . . . . 4·4 (b) 4·4 × 101 = . . . . . . 44 (h) 4·4 × 10-1 = . . . . . . . . . 0·44 2 (c) 4·4 × 10 = . . . . . 440 (i) 4·4 × 10-2 = . . . . . . . . 0·044 (d) 4·4 × 103 = . . . 4,400 (j) 4·4 × 10-3 = . . . . . . . 0·0044 4 (e) 4·4 × 10 = . . 44,000 (k) 4·4 × 10-4 = . . . . . . 0·00044 (f) 4·4 × 105 = . 440,000 (l) 4·4 × 10-5 = . . . . . 0·000044 In this book calculations can be done with and without a calculator. However, a calculator makes some calculations very easy.

Adding: Add 3·2 × 103 and 1·4 × 104 Using the rule above, bring both numbers to the same power and add as normal. 3·2 × 103 = ·32 × 104 4 +1·4 × 10 +1·4 × 104 1·72 × 104 . . . . . . . . . . or using a calculator

3

·

2 EXP 3

+

1

·

4 EXP 4

=

. . . . . . . . . . . . 17,200

Multiplying: Multiply 3·2 × 103 by 1·4 × 104 3·2 by 1·4 = 4·48 103 by 104 = 107 (add power) ∴ 3·2 × 103 × 1·4 × 104 = 4·48 × 107 or using a calculator

3

·

2 EXP 3

×

1

·

4 EXP 4

=

. . . . . . . . . . . . 44,800,000

Subtraction: Take 1·4 × 104 from 3·2 × 105 Using the rule above, bring both powers to same power and subtract as normal. 3·2 × 105 = 3·2 × 105 –1·4 × 104 = –0·14 × 105 (move decimal once to left, power goes up 1) 3·06 × 105 or using a calculator

3

·

2 EXP 5

–

1

·

4 EXP 4

=

8

. . . . . . . . . . . . 306,000

Calculator Use Reference Guide

Division Divide 3·2 × 105 by 1·6 × 104 3·2 =2 1·6 105 Divide = 105− 4 = 101 4 10 3·2 × 105 = 2 × 101 1·6 × 104 or using a calculator Divide

3

·

2 EXP 5

÷

1

·

6 EXP 4

=

. . . . . . . . . . . . 20

Calculator Use Activities 1.

Evaluate with your calculator: (a) 52

(b) 202

(d) (3 × 107)2

(c) 6142

(a) ______________ (b) ______________ (c) ______________ (d) ______________ 2.

Evaluate (a) 53

(b) 203

(d) (3 × 107)3

(c) 6143

(a) ______________ (b) ______________ (c) ______________ (d) ______________ 3.

Evaluate (a) 55

(b) 205

(d) (3 × 107)5

(c) 6145

(a) ______________ (b) ______________ (c) ______________ (d) ______________ 4.

Evaluate (a) 81

(b)

1000

(c)

(d) 3 × 107

10000 0

(a) ______________ (b) ______________ (c) ______________ (d) ______________ 5.

Evaluate (a) 1000

(b)

3

81

(c) 3 1000

(d)

3

10000

(a) ______________ (b) ______________ (c) ______________ (d) ______________

9

Transition Year Maths

6.

Evaluate (a) 3!

(b) 10!

(c) 20!

(d) (3 × 101)!

(a) ______________ (b) ______________ (c) ______________ (d) ______________ 7.

Evaluate (a) 5C1

(b) 5C 2

(c) 5C 3

(e) 5C 5

(d) 5C 4

(a) ______________ (b) ______________ (c) ______________ (d) ______________ 8.

Evaluate (a) 5 P1

(b) 5 P2

(d) 5 P4

(c) 5 P3

(e) 5 P5

(a) ______________ (b) ______________ (c) ______________ (d) ______________ 9.

Write as a fraction: (a) 4·4

(b) 5·1

(c) 6·3

(d) 9·2

(e) 102·84

(a) ___________ (b) ___________ (c) ___________ (d) ___________ (e) ___________ 10. Evaluate 3 + (4 + (5 × 6) – 5) × 7 + 2 + (8 ÷ (2 + 2)) ____________________________________________________________________ 11. There are 60 seconds in 1 minute 60 minutes in 1 hour 24 hours in 1 day 365 days in 1 year 100 years in 1 century Note: 3·6 billion years is the age of the earth. Ignoring leap years write each of the following answers in scientific notation: (a) How many seconds are in 1 minute? _______________________________________________________________ (b) How many seconds are in 1 hour? _______________________________________________________________ (c) How many seconds are in 1 day? _______________________________________________________________ (d) How many seconds are in 1 year? _______________________________________________________________ (e) How many seconds are in 1 century? _______________________________________________________________

10

Calculator Use Reference Guide

(f)

How old is the Earth in (a) seconds, (b) minutes, (c) hours, (d) days, (e) years, (f) centuries? (a) _____________ (b) _____________ (c) _____________ (d) _____________ (e) _____________ (f) _____________

12. (i)

Think of a steel ball the size of our universe. Imagine a fly lands on the ball once every million years. How many seconds would it take to wear away by friction? Estimate an answer to a power of 10. Assume the fly erodes 1 x 10–30 kg every time it lands and assume the universe has a radius of 1080m.

(ii) Think of the same steel ball and the same fly landing on it every million years. How many tonnes of steel would erode away every second? Estimate an answer to a power of 10. • Volume of a sphere =

4 3

πr 3

• Take π = 3·14 • Mass = density × volume • Density of steel = 5000 kg/m3 _________________________________ _________________________________ 13. Add

(i) 3 × 106 + 2 × 106 + 2 × 107 + 4 × 108 (ii) 3 × 1014 + 4·2 × 1011 + 9 × 1012 _________________________________ _________________________________

14. Multiply (i) 3 × 106 × 2 × 107 (ii) 3 × 108 × 4 × 109 × 3·2 × 104 _________________________________ _________________________________ 15. Evaluate (i) (3 × 10 )

(ii) (4·22 −− 10 )

8 3

2 5

⎛ 2·9 × 106 + 3·1 × 105 ⎞ (iii) ⎜ ⎟⎠ 2 × 104 ⎝

7

______________________ ______________________ ______________________ 16. Find

(i) 22% of 100 (ii) 22% of 4800 (iii) 22% of

1 8

(iv) 22% of 886·25 __________________ __________________ __________________ __________________

11

Transition Year Maths

1. Number Systems We use a base 10 system with 10 digits, they are : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. This is the decimal place-value system. 437 means 7 × 100 (7) plus 3 × 101 (30) plus 4 × 102 (400). You are all familiar with this system. The second system you are likely to encounter is the Roman Numeral system. It does not have place-value. The letters have fixed values and are ordered from largest to smallest. If a letter representing a smaller value appears before a larger one (e.g. IV), the smaller value is subtracted from the larger value (e.g. 5 – 1 = 4, IV). I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000 Example 1:

Write out the first 10 numbers in Roman Numerals. I, II, III, IV, V, VI, VII, VIII, IX, X Note: 4 is IV, one before five

Example 2:

Write out the Roman Numerals for the following (a) 21 = XXI (b) 900 = CM (c) 1954 = MCMLIV (d) 3592 = MMMDXCII

It is difficult to do calculations using this system without converting back to our base 10 system first. The third system you might encounter is the Base 2 or Binary System. It uses just two digits: 0 and 1. This is a place value number system. The number 1101112 means (looking at the digits from right to left) 1 × 20 (110) plus 1 × 21 (210) plus 1 × 22 (410) plus 0 × 23(010) plus 1 × 24 (1610) plus 1 × 25 (3210) 1 + 2 + 4 + 0 + 16 + 32 = 5510 answer: 1101112 = 5510 All digital information is carried using this system. The bases for all the technology in your mobile phones, computers and digital TVs is dependent on the binary system. The two digits can be 1, 0 or on, off, or +, – or 6 volts, 0 volts etc. Electronically it is easy to represent the binary system. A subscript is used to identify the base number. The subscript (e.g. 10) is only used when discussing bases. 1410 is normally just written 14. Example 3: Write the first 20 numbers in binary 110 = 12 810 = 10002 1510 = 210 = 102 910 = 10012 1610 = 310 = 112 1010 = 10102 1710 = 410 = 1002 1110 = 10112 1810 = 510 = 1012 1210 = 11002 1910 = 610 = 1102 1310 = 11012 2010 = 710 = 1112 1410 = 11102

12

11112 100002 100012 100102 100112 101002

1. Number Systems

NOTE The numbers just go up in order. The next number that can be written after 1 using just 1 and 0 must be 10, the next 11 and so on. Example 4: 0 units 1 × 21 1 × 22 0 × 23 1 × 24

Convert 101102 to base 10 = 0 = 0 = 1×2 = 2 = 1×4 = 4 = 0×8 = 0 = 1 × 16 = 16 2210 answer: 101102 = 2210

To convert from base 10 to binary, divide the decimal by 2 and note the remainder. Keep dividing the quotient by 2 until it equals 0. The remainders are the binary number. Example 5: 22 11 5 2 1

Convert 2210 to binary Quotient Remainder ÷ 2 = 11 . . . . . . . . . 0 ÷ 2 = 5.........1 ÷ 2 = 2.........1 ÷ 2 = 1.........0 ÷ 2 = 0.........1

2210 = 101102

Adding, multiplying and other functions are the same as base 10. Example 6: Addition (i) Add 1010 + 1010 in base 2 1010 1010 10100 Multiply (ii) Multiply 11 × 11 in base 2 (iii) Multiply 101 × 10 in base 2

PROJECT 1.1

Add the columns from right to left. 0+0=0 1 + 1 = 10 Write the 0 and carry the 1 1+0+0=1 1 + 1 = 10 (ii)

11 × 11 11 110 1001

(iii)

101 × 10 1010

Base 4

Write out the first 20 numbers in base 4 (using only digits 0, 1, 2, 3). Then add 134 and 224 and multiply 24 by 134. Check your answers by converting back to base 10. Hexadecimal arithmetic is done using base 16. The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. 410 = 416 2410 = 1816 1410 = E16 3410 = 2216

13

4410 = 2C16

Transition Year Maths

PROJECT 1.2

Base 16

Write out the first 40 numbers in base 16, then add 216 to 1316 and multiply 216 × 1316. Check your answers – there are many conversion programmes available on the Internet.

Fractions can be confusing in bases other than 10. Consider the binary system: 1 is 0·12 2 10

1 4 10 is 0·012

Base 10 has become the dominant system, most likely because we have 10 fingers on which to count. Societies with base 10 as their counting system have advanced faster scientifically, probably because it is easier to use. It is also speculated that the Roman Numeral system contributed to the fall of the Roman Empire, as place-value systems of base 5 from China and base 10 from the Middle East became more widely used.

PROJECT 1.3

Converting to Base 10

Consider the numbers 11111111112 22222222223 and 11111111113. Use the method from Example 4 to convert these numbers to base 10. Work in small groups to investigate if you can find an easy or a short-cut way of doing any of these calculations. See the hint a the end of the chapter for the short-cut.

Questions

The following Questions pages are FREE to download from mentorbooks.ie/resources

NOTE A subscript indicates the base. If no subscript is given, assume a base of 10.

1.

Write the numbers from 10 to 30 in Roman Numerals. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

2.

Write the dates from 1999 to 2012 in Roman Numerals. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

14

1. Number Systems 3.

What number is MMDCCLXVII? ____________________________________________________________________ ____________________________________________________________________

4.

Write in Roman Numerals: (a) 194

(b) 1492

(c) 2222

(d) 2837

(a) ______________ (b) ______________ (c) ______________ (d) ______________

5.

Write the numbers 2010 to 3010 inclusive in base 2. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

6.

Add 1012 + 1012 + 1012 ________________________________________________ ____________________________________________________________________ ____________________________________________________________________

7.

Add 1012 + 10012 + 11102 ______________________________________________ ____________________________________________________________________ ____________________________________________________________________

8.

Add 11112 + 11112 + 100002 + 100102 ____________________________________ ____________________________________________________________________ ____________________________________________________________________

9.

Multiply 1012 × 102 ___________________________________________________ ____________________________________________________________________ ____________________________________________________________________

15

Transition Year Maths 10. Multiply 1012 × 112 ___________________________________________________ ____________________________________________________________________ ____________________________________________________________________

11. Multiply 112 × 102 × 1112 ______________________________________________ ____________________________________________________________________ ____________________________________________________________________

12. Write out the first 30 numbers in base 3. _________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

13. Write the number 14 in base 2, base 4, base 6, base 8, base 10, base 12, base 14 and base 20. __________________ __________________ __________________ __________________ __________________ __________________ __________________ __________________

14. Assuming 0·0000012 =

1 what is 1 , 1 , 3 , 5 64 32 16 16 16

__________________ __________________ __________________ __________________

15. Write

1 1 1 1 in binary code. , , , 2 10 4 10 8 10 16 10

____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

16

1. Number Systems

16. Convert (a) 3410 to binary by dividing method. ___________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ (b) 5410 to binary by dividing method. ___________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

17. Write the numbers 1 to 40 in base 16.____________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

18. Add 102 + B16 + 114. Give the answer in base 2, base 16 and base 4. __________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ Project 1.3 Hint: 1112 = 10002 – 12

17

Transition Year Maths

2. Statistics “Statistical thinking will one day be as necessary for efficient citizenship as the ability to read or write” Statistician S.S. Wilks paraphrasing from H.G. Wells’s 1903 book “Mankind in the Making” Statistics is the collection, presentation and analysis of data. The vast majority of decisions made today are taken after considering research. Governments, companies, institutions of all sorts need ‘facts’ before they make their decisions. The ‘facts’ are usually presented graphically by statisticians. Lists of numbers are difficult to read, so statisticians take the raw data, the numbers, and produce easy-to-read, bar charts, trend graphs, pie charts and a whole variety of different visual ways of presenting the data, to make it more useful.

Use of Statistics You will already have used charts to show facts clearly. Example 1:

Draw a Pie Chart of the Men’s All Ireland Hurling winners for the 24 years between 1991 and 2014 inclusive:

Data Kilkenny Cork Clare Tipperary Offaly Wexford

Convert to Degrees 12 3 3 3 2 1

12 × 360 = 180° 24 3 × 360 = 45° 24 3 × 360 = 45° 24 3 × 360 = 45° 24 2 × 360 = 30° 24 1 × 360 = 15° 24

Men’s Hurling All Ireland Winners 1991 – 2014

18

2. Statistics

PROJECT 2.1

Draw Pie Charts

Look up the winners of the Football All Ireland Championships for both the men and the women for the 24 years between 1991 and 2014. Draw a Pie Chart for each to compare which counties are strongest in the men’s Football Championship with those of the women’s Football Championship for this period. Statistics have obvious uses like this but advanced statistics can be used to do amazing things.

Language Translation An interesting example of a possible use of statistics is a project Google is reported to be currently working on. They hope to be able to translate from one language to another in real time. They hope to use statistics derived from the billions of pages of text used on their sites every day to create this service. They do not intend to use rules or grammar to translate between the 47 languages. They are using the statistical probability of one word being beside another word to do this, e.g. you speak in English on the phone and the person at the other end hears you in Chinese, she speaks back in Chinese but you hear her in English. Even if you think you have little interest in maths, everyone is affected by statistics during their life; e.g. if you get 40% in your Geography exam you might think that this is not a very good mark. However if the class average is only 34%, then it is actually quite a good mark.

Averages The average of a set of numbers is the most useful way of analysing them. We will look at three ways of finding the average and investigate which is the most suitable in different situations: sum of all numbers (the mean is what most people call the average) Mean = amount of numbers Mode = the mode is the number or value which occurs most often Median = when all the values are listed in order of size, the median is the middle one (or the average of the two middle ones)

The marks for 28 pupils in a history exam are: 35, 38, 43, 46, 46, 49, 50, 55, 56, 56, 58, 60, 60, 61, 61, 62, 64, 65, 65, 67, 68, 70, 71, 72, 82, 82, 82, 84. Find the mean, mode, median and comment on suitability of each average: Example 2:

35 + 38 + … + 84 1708 = = 61 28 28 61 + 61 Median = = 61 2 Mode = 82 Mean

=

For a pupil wishing to compare her mark to the average student in the class, both the mean and the median give a good estimation in this example. The mode in this case is poor, because, by chance, there were three pupils scoring 82.

19

Transition Year Maths

An American professional basketball team publishes its players’ wages. They have one exceptional star player. The wages per annum are: $360,000, $380,000, $400,000, $420,000, $460,000, $490,000, $500,000, $500,000, $520,000, $530,000, $600,000, $5,640,000. Find the mean, mode, median and comment on the suitability of each average

Example 3:

350, 000 + 380, 000 + … + 5, 640, 000 10, 800, 000 = = $900, 000 12 12 $490, 000 + $500, 000 = = $495, 000 2 = $500,000 (only figure which occurs twice) =

Mean Median Mode

In this example, for an ordinary player considering signing to play for the team, the median or the mode gives him a better idea of what he may earn. The mean is affected by the very high wages of one exceptional player. This is called outlier.

Normal Distribution 100 pupils were shown a 20 digit number for 30 seconds and then asked to recall the number. The number of digits they could remember and repeat before they made a mistake was recorded:

Example 4:

Digits 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 remembered 2 0 2 3 7 5 8 10 15 13 11 9 5 4 2 2 0 0 1 1 0

Frequency

Plot the data on a bar chart, the frequency goes on the y-axis. Find the three averages, for this data, the mean, the mode and the median. Mean

=

total number of digits remembered total number of pupils

2 × 0 + 0 × 1 + 2 × 2 + 3 × 3 + 7 × 4 …1 × 19 + 0 × 20 = 8·37 2 + 0 + 2 + 3… + 1 + 0 Mode = the most common number of numbers remembered = 8 Median = line all the numbers remembered in a row and the median is the middle one, or the average of the two middle ones if it is an even number i.e. = 8 + 8 = 8 2 =

20

2. Statistics

The data in this example should produce a bar chart which is close to the classical bell shape of a normal distribution. The data should be roughly symmetrical about the mean value. 16

Number of Pupils Frequency

14 12 10 8 6 4 2 0 1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

Digits Remembered

Standard Deviation The Standard Deviation of a set of numbers gives an indication of the spread of the data. The mean value gives an indication of the average. The Standard Deviation gives more information. It tells how far the figures are, on average, from the mean. It can be calculated from the formula: ∑(x − x )2 n standard deviation σ= ∑ = the sum of x = each value in the data set σ=

x = the mean of all values in the data set n = the number of values in the data set Example 5:

Eight pupils from a Transition Year class recorded their pulse rate per minute when at rest. They then recorded the pulse rate for eight family members of different ages. The results were as follows:

Transition Year Pupils

72

70

74

80

74

75

76

79

Family Members

64

58

88

73

89

68

92

68

Find the mean and standard deviation of both sets of data and comment on the results. 72 + 70 + 74 + 80 + 74 + 75 + 76 + 79 600 = 75 (Mean) = 8 8 64 + 58 + 88 + 73 + 89 + 68 + 92 + 68 600 Mean (Family members) = = = 75 (Mean) 8 8 Mean (Pupils) =

21

Transition Year Maths

SET 2 (Family Members)

SET 1 (Pupils) x1 72

x 75

(x1 - x ) -3

(x1 - x )2 9

x2 64

x 75

(x2 - x ) -11

(x2 - x )2 121

70

75

-5

25

58

75

-17

289

74

75

-1

1

88

75

13

169

80

75

5

25

73

75

-2

4

74

75

-1

1

89

75

14

196

75

75

0

0

68

75

-7

49

76

75

1

1

92

75

17

289

79

75

4

16

68

75

-7

49 1166

78

78 1166 = 3·12 Standard Deviation = = 12·07 8 8 By chance both groups had the same mean pulse rate. The Transition Year pupils all had a similar rate, causing the standard deviation to be low, 3·12. The Family Members, some possibly very young or very old had a much wider spread, resulting in a relatively high standard deviation of 12·07.

Standard Deviation = Comment:

The following project is based on Example 4; it should take approximately two class periods to complete:

PROJECT 2.2

Recall the Digits

Do this as a class group. The teacher writes a 20 digit random number on the board. The class look at it for 30 seconds; the number is then covered. The pupils then write down (independently) as much of the random number as they can remember. Reveal the random number again and each pupil records how many of the digits they got correct before their first error. The exercise is repeated a number of times with a different 20 digit number each time. Record all the data on the board. Plot a Bar Chart of the data and find the average number of digits remembered. Discuss the shape of the Bar Chart.

Example 6:

A class of 30 pupils get a short mid-term exam in English and one in French. The teacher grades the exam to the nearest 5%. The results are as follow:

French Marks English Marks 20 25 30 35 35 40 40 45 45 50 40 45 45 50 50 55 55 55 55 55 50 55 55 55 60 60 60 60 60 65 60 60 60 60 60 60 60 60 65 65 70 70 75 75 80 85 85 90 95 100 65 65 70 70 70 70 75 75 75 80

22

2. Statistics

Compare the results in French and English by (i) finding the mean (an average) of each and (ii) by plotting a bar chart for each using the same scale.

FRENCH

French Marks

ENGLISH

English Marks

23

Transition Year Maths

Compare results: (i) the average mark (the mean) in both classes is similar, one is 59 the other is 61. Therefore by looking only at the mean, you might conclude that classes have a similar ability in French and English. (ii) Comparing the results using the Bar Charts, give us extra information. Yes, the charts confirm that the average mark is approximately 60% for both subjects. However, the charts clearly show that the spread of the marks is much greater in French than in English. You can now conclude that the grades are significantly different for English and French even though the average is similar. Most pupils in the English class got 60% or within 10% of 60%. While most pupils in the French class got more than 10% above or below the 60% mark.

PROJECT 2.3

Calculate Standard Deviation

Using the data from Example 6, quantify the spread by calculating the Standard Deviation for the French and English marks

NOTE Note on describing the shapes of charts: A perfect normal distribution is symmetrical, the highest point will be the mean. Positive skew means that there is a long tail to the right. Negative skew means that there is a long tail to the left.

% of Population

No. of Workers

Example A: marks in the exam in Example 5 are normally distributed about the mean.

Negative Skew

Positive Skew

Normal Distribution

Example B: A plot of the income of the population of a small town will be skewed positively with a small number of people earning a lot.

24

Example C: the retirement age of populations in most developed countries will look like this, peaking at around 62 years.

2. Statistics

NOTE A full class period will be required to collect the data for the following project. Over the course of the class each pupil should be able to test every other pupil and get a full set of data for themselves. (50 pieces of data will give a good graph, so if you have 20 readings yourself, (Primary Data), take 30 readings from other students in the class, (Secondary Data).

PROJECT 2.4

Reaction Time

Use a 50 cm half metre stick or a 30 cm ruler. Work in pairs to collect data for this project. Ask another pupil to hold their hand out with their thumb and forefinger separated by 1 cm. Then place the 0 cm mark on the ruler between their thumb and forefinger, tell them you are going to drop the ruler sometime in the next 10 seconds and she has to catch it as quickly as possible. At a random time drop the ruler, record the distance it fell (to the nearest cm.) by reading the number of centimetres immediately above their fingers. Repeat this exercise for everyone in the class and record all the data on a bar chart, with “distance the ruler fell” recorded on the x-axis, and frequency on the y-axis. Use some secondary data from other pupils to ensure you have enough data. What shape is the graph? Normal, Positive Skew or Negative Skew? Can you explain the shape?

25

Transition Year Maths

Scatter Graphs (Correlation) Scatter Graphs show the relationship between two sets of data. 20 Transition Year boys were weighed and their heights measured. Plot the results on a Scatter Graph putting weight on the x-axis. Comment on the result:

Example 7:

Height 151 154 155 156 156 158 158 163 165 169 169 170 171 172 174 175 177 179 180 182 cm Weight kg

45

47

60

51

54

50

62

59

60

61

56

63

60

58

72

63

68

66

Scatter Graph Plot of Weight vs. Height (for Transition Year Boys).

Comment:

The mean weight can be calculated (like Example 2) to be 60·1 kg, and the mean height to be 166·7 cm. But when we draw a Scatter Graph, we are interested in the connection or relationship between the two sets of data. It is clear from the graph that there is a diagonal band of points across the graph, from the bottom left to the top right. This is called a Positive Correlation. Generally speaking, the taller boys are heavier, and the shorter boys are lighter. This is what we would have expected.

26

70

77

2. Statistics

Correlation: This is the relationship between the two sets of data. We will distinguish between five different relationships. Most sets of data will fit into one of these five categories. Weak Positive Correlation:

Variable 1

Variable 1

Strong Positive Correlation:

Line of best-fit Variable 2

Variable 2

Weak Negative Correlation:

Variable 1

Variable 1

No Correlation: no line of best fit: scatter is too dispersed

Variable 2

Strong Negative Correlation:

Variable 2

The correlation between weight and height in Example 6 is a weak positive correlation.

Variable 1

PROJECT 2.5

Variable 2

Height and Shoe Size

Work in pairs with a measuring tape or two metre sticks to collect the data. Measure the height of everyone in the class and ask them for their shoe size. Record the pairs of data and then plot them on a Scatter Graph. Put shoe size on the x-axis and height on the y-axis. When you have completed the plot, comment on the correlation.

The data you collect in this project is called Primary Data, because you collected it yourself. If you run out of time, you may need to get some sets of data from other pupils. This is known as Secondary Data as you did not collect it yourself.

27

Transition Year Maths

The Lie Factor (use of statistics to distort or exaggerate the truth) The Lie Factor =

size of the effect shown in graph size of the effect in the data

Statistics and charts have been used to present information falsely or to exaggerate the information since they were invented. The most common ways are to change the scale to exaggerate differences or to omit some data. We have already noticed how the average can be distorted depending on whether we use the mean, mode or median. Example 8:

A car company notices that the fuel efficiency of one of its car models has improved from 10 kilometres per litre in 2010 to 12 kilometres per litre in 2015. Plot (i) a Fair Chart to advertise this improvement and (ii) a Distorted Chart to exaggerate the fuel efficiency improvement. (iii) Calculate the Lie Factor of the exaggeration and comment on it.

12

12

8

Kilometre/Litre

Kilometre/Litre

10

6 4

11

10

2 9

0 2010

2010

2015

Car Fuel Efficiency Improvement 2010 to 2015

(i)

2015

Car Fuel Efficiency Improvement 2010 to 2015

The first chart gives a realistic representation of the improvement in fuel efficiency from 10 km/l to 12 km/l. It is a 20% increase in efficiency (ii) The second chart distorts the improvement by changing the vertical scale. It appears as though the cars are now 200% more efficient the % change in the size of the chart (iii) The Lie Factor = the % change in the size of the data

28

2. Statistics

⎛ 6cm − 5cm ⎞

× 100 = 20% ⎜⎝ % change in chart 5cm ⎟⎠ Chart 1: Lie Factor = % change in data = = 1 (no distortion) ⎛ 12 − 10km / 1 ⎞ ⎜⎝

× 100 = 20% 10km / 1 ⎟⎠

⎛ 6cm − 2cm ⎞

× 100 = 200% ⎜⎝ % change in chart 2cm ⎟⎠ Chart 2: Lie Factor = % change in data = = 10 (large distortion) ⎛ 12 − 10km / 1 ⎞ ⎜⎝

× 100 = 20% 10km / 1 ⎟⎠

The true improvement in efficiency is 20%; the second chart gives the impression that the improvement is 200%. The Lie Factor of 1 implies the first chart gives an accurate impression. The Lie Factor of 10 in chart 2 indicates a 10-fold distortion. There is nothing inaccurate with Chart 2; it just exaggerates visually the improvement in efficiency.

Florence Nightingale: English nurse and mathematician, (1820–1910) Florence Nightingale is best remembered for her work as a nurse, particularly during the Crimean War, and for setting up a school of nursing. However, she also had a gift for mathematics and was a pioneer in the visual presentation of information and statistical graphs. She was an early user of pie charts and circular histograms.

29

Transition Year Maths

Questions

The following Questions pages are FREE to download from mentorbooks.ie/resources

1.

Find (a) the mean, (b) mode and (c) median for the following numbers. Then find the spread of the data for each set by calculating (d) the standard deviation (correct to two places of decimal): (i) 4, 8, 6, 3, 9, 5, 6, 2, 6, 5, 4, 2 (ii) 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5 (iii) 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 4, 4 (i) (a) ____________ (b) ____________ (c) ____________ (d) ____________ (ii) (a) ____________ (b) ____________ (c) ____________ (d) ____________ (iii) (a) ____________ (b) ____________ (c) ____________ (d) ____________

2.

67 pupils from a school do a 10 kilometre run for charity. The table shows the number of pupils finishing the run in each time range. (When calculating the mean, assume the time is the midpoint of range. For example 30 – 35, use 32·5 minutes etc.) (i) Plot the data on a bar chart with time on the x-axis (ii) What is the mean time, to the nearest minute? (iii) What is the modal time range? (iv) How would you describe the shape of the chart? (i)

Time (mins) No. of Pupils 30 – 35 2 35 – 40 12 40 – 45 21 45 – 50 13 50 – 55 6 55 – 60 3 60 – 65 2 65 – 70 2 70 – 75 1 75 – 80 2 80 – 85 1 85 – 90 1 90 – 95 0 95 – 100 1

(ii) ___________________________ (iii) ___________________________ (iv) ___________________________ ___________________________ ___________________________ ___________________________

30

2. Statistics

3.

Six coins were tossed 500 times and the number of times they landed facing heads up were counted. The results were:

Number of Heads

0

1

Frequency

8

45

(i)

2

3

(i) 4

115 152 118

5

6

56

6

Plot the results on a Bar Chart, putting ‘number of heads’ on the x-axis.

(ii) Describe the shape of the chart. __________________________________________ __________________________________________

4.

(a)

Draw a Scatter Graph of the following data. These are mean annual temperatures taken at points around the Alps at different elevations:

(a)

(b) Describe the correlation. ________________________________________ ________________________________________ Elevation (metres)

Mean temperature (°C)

500

25

1000

22

2000

13

1500

20

2500

8

3000

6

800

26

1400

19

3500

4

2700

5

1100

19

700

24

31

Transition Year Maths

5.

A garage has 11 cars of the same model of different ages for sale. These are the asking prices for the 11 cars. (a) Draw a scatter graph of the data. (b) Describe the correlation of the distribution. _______________________________________________________________ _______________________________________________________________

Age (years)

1

2

10

8

3

5

1

9

8

4

6

Price (Euro) 20,000 7,000 4,000 5,000 17,000 15,000 21,000 2,000 6,000 15,000 14,000 (a)

6.

A model of mobile phone is made in a factory in China. The retail price at various distances from the factory is given below: Currency is given in the Chinese Yuan Renminbi, abbreviated to ‘CNY’

Distance to shop (km)

500

2500 1000 3000 700

1800

400

3500 1800 2000

Price (CNY)

150

350

500

450

200

400

(a) Draw a scatter graph of the data.

100

250 (a)

(b) Discuss the correlation. ____________________________________ ____________________________________ ____________________________________ ____________________________________ ____________________________________ ____________________________________

32

150

250

2. Statistics

7.

The FIFA soccer World Cup was played 20 times between 1930 and 2014. The following chart shows the winning countries. Calculate the angle for each country and draw a pie chart of the data: Pie Chart

8.

Country

No. of Wins

Brazil

5

Italy

4

Germany

4

Argentina

2

Uruguay

2

England

1

France

1

Spain

1

Angle

The twelve month moving average temperature in Dublin Airport in 1958 was 9·40C and in 2008 it was 10·00C. (i) Plot this data on a chart to show accurately the rise in temperature. (ii) Plot it on a Distorted Chart to exaggerate the rise in temperature, by choosing an appropriate scale which makes the temperature rise look larger than it really is. (i)

(ii)

(iii) Calculate the Lie Factor of your second chart. ____________________________________________________________________ ____________________________________________________________________ (iv) Discuss the second chart. Is it deliberately misleading, or can you justify your chart because you believe global warming is an important issue and you want to highlight temperature rise? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

33

Transition Year Maths

3.

Graphs

2-Dimensional (c) x2 + y2 = 9 . . . . . . . circle

(a) y = x . . . . straight line

...............

2 (d) x2 + y = 1 . . . . . ellipse 9

(b) y = x2 . . . curve with one bend

34

3. Graphs

3-Dimensional x 2 y2 z2 + + =1 4 4 4

. . . . . sphere of radius 2 Shape intersects the axis at (0, 0, 2) (0, 0, -2) (0, 2, 0) (0, -2, 0) (2, 0, 0) (-2, 0, 0)

x2 +

y2 + z 2 = 1 . . . . . . cigar shape 9

Shape intersects the axis at (0, 0, 1) (0, 0, -1) (0, 3, 0) (0, -3, 0) (1, 0, 0) (-1, 0, 0)

x2 z2 + y2 + = 1 . . . . . flying saucer 9 9

Shape intersects the axis at (3, 0, 0) (-3, 0, 0) (0, 1, 0) (0, -1, 0) (0, 0, 3) (0, 0, -3)

35

Transition Year Maths

2-Dimensional Through observation much can be deduced about a 2-dimensional graph. If both powers of x and y are one, it is always a straight line: Graph of y = x + 1

If the power of either x or y is one, and the other is two, the graph is a curve with one bend: Graph of y + 1 = x2

If the power of either x or y is one, and the other is three, the graph is a curve with two bends: Graph of y = x3 – 3x2 – x + 3

If the power of either x or y is one, and the other is four, the graph is a curve with three bends: Graph of y = x4 – 5x3 + 5x2 + 4x – 4

36

3. Graphs

Example 1

Describe the shape of the following lines: (a) y = x . . . . . . . . . . . . . . . . . . . . straight line (b) y = x2 . . . . . . . . . . . . . . . . . . . . curve, one bend (c) y = 3x . . . . . . . . . . . . . . . . . . . straight line (d) y2 = x . . . . . . . . . . . . . . . . . . . . curve, one bend (e) y = x3 – 3x2 – x + 3 . . . . . . . . . curve, two bends (f) y = 10x – 2 . . . . . . . . . . . . . . . straight line (g) y = 4x2 + x + 3 . . . . . . . . . . . . . curve, one bend (h) y = x4 – 5x3 + 5x2 + 4x – 4 . . . . curve, three bends (i) y = x4 + x3 + x2 + x + 3 . . . . . . . curve, three bends (j) y = x3 + x2 + x + 2 . . . . . . . . . . curve, two bends

RULE Rule for drawing graphs: First we need to collect a number of points to plot, to see what shape the graph has: 1. Write down the value of x from –3 to +3 on top of the chart. 2. Write down the equation on the side of the chart. 3. Values of y can be calculated by putting values of x into the equation and adding. 4. Each value of x now has a corresponding value for y which may be plotted.

Example 2

Draw the line y = 2x + 3. Use the value of x from –3 to +3.

x

–3

–2

–1

0

1

2

3

2x

–6

–4

–2

0

2

4

6

3

3

3

3

3

3

3

3

y

–3

–1

1

3

5

7

9

y

x

37

Transition Year Maths

Example 3

x x3 –3x2 –4x +12 y

Draw a graph of the curve y = x3 – 3x2 – 4x + 12 from x = –3 to x = +3 –3 –27 –27 12 12 –30

–2 –8 –12 8 12 0

0 0 0 0 12 12

–1 –1 –3 4 12 12

2 8 –12 –8 12 0

1 1 –3 –4 12 6

3 27 –27 –12 12 0

y

x

Closed Shapes When both the x and the y are squared, new shapes are formed. Square roots are estimated to one place of decimals. 1. Circles: x2 + y2 = 9. Find points on the graph. Let x = 0

⇒ y2 = 9

⇒ y = +3 or y = –3

⇒ (0,3) and (0,–3) are on the graph.

Let y = 0

⇒ x2 = 9

⇒ x = +3 or x = –3

⇒ (3, 0) and (–3, 0) are on the graph.

Let x = 1

⇒ 1 + y2 = 9

⇒ y2 = 8 ⇒ y = ± 8 ⇒ (1, 2·8) and (1, –2·8) are on the graph.

Let y = 1

⇒ x2 + 1 = 9

⇒ x2 = 8 ⇒ x = ± 8 ⇒ (2·8,1) and (–2·8,1)

Let x = –1 ⇒ 1 + y2 = 9 Let y = –1

⇒ x2 +1 = 9

⇒ y2 = 8 ⇒, 2·8) and (–1,–2·8) are on the graph. ⇒ x2 = 8 ⇒ (2·8,–1) and (–2·8,–1) are on the graph.

38

3. Graphs

Draw a graph of y = x4 – 5x3 + 5x2 + 5x – 6 on the domain x = –2 to +3 and use the graph to find the factors and roots of the function. Note: When the function = 0, the roots are where the graph cuts the x-axis. The factors will be (x – roots).

Example 4

x

–2

–1

0

1

2

3

x4

16

1

0

1

16

81

–5x3

40

5

0

–5

–40

–135

+5x2

20

5

0

+5

20

45

+5x

–10

–5

0

+5

10

15

–6

–6

–6

–6

–6

–6

–6

y

60

0

–6

0

0

0

y

x

NOTE From the data above, we can plot the general shape of the graph and find all the roots! If we wished to have a more accurate graph, we would have to find the values of the function (y) for x = –½, ½, 1½, 2½, 3½ and 4½. The roots are x = –1, 1, 2 and 3. The factors are (x + 1), (x – 1), (x – 2) and (x – 3) Note: If the function does not cut the x-axis at a whole number, the roots (and hence the factors) can only be estimated.

39

Transition Year Maths

y2 x2 + =1 4 9 y2 ⇒ =1⇒ y = ± 3 9 x2 ⇒ =1⇒x = ± 2 4

Ellipse Let x = 0 Let y = 0

⇒ points (0,3) and (0,–3) are on the graph. ⇒ points (2,0) and (–2,0) are on the graph.

More points can be found to verify the shape:

PROJECT 3.1

Draw Graph and find Factors

Draw the graph of: y = x5 + 2x4 – 10x3 – 20x2 + 9x + 18 from x = +3 to x = –3. Find the five factors.

PROJECT 3.2

Drawing 3D Equations

In three dimensions there are 3 axes, the x and y are as usual and the z-axis is perpendicular to the page. Write an equation in x, y and z to give: (a) a sphere of radius 1 (b) a 3-D disc shape of radius 4 and height 1 (c) a cigar shape of radius 1 and length 4

A circle centred on the origin (0,0) has an equation x2 + y2 = r2 where r is the radius. If we wish to plot a circle centred on a different point (h,k) the equation becomes (x – h)2 + (y – k)2 = r2

Example 5:

Find the equation of a circle centre (4,0) with radius 1. h = 4 (x – h)2 + (y – k)2 = r2 k = 0 (x – 4)2 + (y – 0)2 = 12 r = 1 x2 – 8x + 16 + y2 = 1 x2 + y2 – 8x + 15 = 0

40

3. Graphs

PROJECT 3.3

Equation for 3D Sphere

Write an equation for a 3-Dimensional sphere of radius 1 centred on the point (2, 2, 2).

PROJECT 3.4

Equation for Ellipse

Write an equation for any ellipse not centred on the origin.

Example 6:

Find the centre and radius of the circle x2 + y2 – 8x + 15 = 0 This is the reverse of Example 5. So you know the answer, but how is it done? Rewrite as x2 – 8x + y2 = –15 We must complete the square. Half –8 = –4, then square – 4 = 16. The equation becomes x2 – 8x + 16 + y2 = –15 + 16 (adding 16 to both sides). ⇒ (x – 4)2 + (y – 0))2 = 1 Therefore the centre is (4, 0) and radius is 1.

41

Transition Year Maths

Questions

The following Questions pages are FREE to download from mentorbooks.ie/resources

Note: Select the x-axis and y-axis carefully and choose a suitable scale to ensure graphs fit on the grids provided.

1.

Describe the shape (straight line, curve with one bend etc) of the following lines: (a) y = x2 ____________________________________________________________ (b) 3y = 2x___________________________________________________________ (c) 3y = 2x + x2 _______________________________________________________ (d) y = 4x3 ___________________________________________________________ (e) y3 = x ____________________________________________________________ (f) y = 5x4 + 3x3 + 2x2 + x + 1____________________________________________

2.

Draw the following lines from x = –3 to x = +3. (i) y = 5x (ii) y = 3x + 4 (iii) y = 2x2 (iv) y = 2x2 + 4 (v) y = x3 + x2 – 9x – 9 (vi) y = x4 – x3 – 7x2 + x + 6

3.

What is the radius of the following circles? (a) x2 + y2 = 1 ________________________________________________________ (b) x2 + y2 = 100 ______________________________________________________ (c) x2 + y2 = 16 _______________________________________________________

4.

2 Draw the ellipse x + y 2 = 1 25

42

3. Graphs

5.

Try to draw in three dimensions the shape: y2 x2 + + z2 = 1 16 16

6.

Draw a graph of y = x2 – 2x –3 from x = –4 to x = +4. Hence find the two roots when y = 0 and find the two factors.

7.

Draw a graph of y = x3 – 2x2 – 4x + 1 from x = –2 to x = 4. Hence estimate the three roots when y = 0 and find the three factors.

8.

Draw graphs of all the following functions from x = –4 to x = +4. Hence find the roots and the factors. (a) y = x3 – 3x2 – x + 3

(b) y = x3 + 3x2 – x – 3

43

Transition Year Maths (c) y = x3 + 2x2 – x – 2

(d) y = x4 – x3 – 7x2 + x + 6

(e) y = x3 + 3x2 – 4x – 12

(f) y = x3 + x2 – 9x – 9

NOTE In all of these graphs your main interest is finding where the graphs cut the x-axis.

9.

Draw a graph of y = – 2x3 + 9x + 2 from x = –3 to x = +3. Estimate the roots of this when y = 0. Hence estimate the factors.

44

3. Graphs 10. Sketch the following circles and write their equations. (a) centre (0, 0) radius 4

(b) centre (0, 0) radius 6

(c) centre (0, 0) radius

2

45

Transition Year Maths

11. Sketch and give the centre and radius of the following circles: (a) x2 + y2 = 36

(b) x2 + y2 = 30

(c) 9x2 + 9y2 = 1

12. Sketch the following circles and write their equations: (a) centre (1, 1) radius 1

(b) centre (15, 0) radius 5

46

3. Graphs

13. Sketch and give the centre and radius of the following circles: (a) x2 + y2 + 2y – 3 = 0

(b) x2 + y2 + 2x + 4y + 1 = 0

(c) x2 + y2 + 2x + 2y = 0

14. Draw

y2 x2 =1 + 4 36

47

Transition Year Maths

4.

Formulas

Formulas are designed to make maths easier. When presented with a new formula, learn what each letter stands for. Once this is known, the formula can be used. The important thing about formulas is to use them without worrying about where the formulas come from. This is what they are produced for. Generally in school we do not prove all the formulas we use. Some formulas can be easily proven, some are very difficult to prove mathematically and some are produced simply from observation. Consider the area of a disc = πr2. Did you ever think about where this formula came from? In the third century BC Archimedes proposed cutting a sphere into thin slices to find a formula for volume. Nicholas of Cusa, a fifteenth-century cardinal and philosopher, came up with the proof for the area of a disc:

Area of disc = πr 2

πr

r Other formulas are more obvious. Consider speed.

Speed =

Distance Time

Walk 10 metres (m) across the room in 5 seconds. distance 10 m The average speed = = = 2 m/sec. time 5 sec

48

Consider a disc of radius r. Divide the disc into thin slices as shown, stack them, every second one facing each other. If the slices are small the shape produced will be close to a rectangle, the width of the rectangle will be r, the radius, and the height will be half the circumference, πr. The area of the rectangle is height by width = πr × r = πr2. Therefore the area of the disc is also = πr2.

4. Formulas

Formulas are expressions written with letters. The most famous formula ever is probably: E = mc2 E = energy measured in joules ( J ) m = mass measured in kilogrammes (kg) c = speed of light measured in metres per second (m/s)

The energy contained in a piece of matter is equal to the mass multiplied by the speed of light squared. It is part of Einstein’s special theory of relativity. The speed of light in space is fixed and can be measured accurately to be 300,000,000 m/s = 3 × 108 m/s. Example 1:

How much energy is contained in 1 litre of milk (Note: 1 litre = 1 kg). The chemical energy is about 275,000 J (this is written on the carton). This is energy the body is able to use. The nuclear energy can be worked out from E = mc2. E = mc2 = 1 kg × (300,000,000)2 = 90,000,000,000,000,000 J. There is no known way on earth of extracting this huge amount of energy from the milk!

The Bomb On earth there are only two ways of producing nuclear energy. (a) One is fission. Fission means an element with a heavy nucleus, like uranium (number 92 on the periodic table), is broken into smaller ones. When this happens some of the mass disappears. The missing mass turns into energy. This happens in atomic bombs. (b) The second way is fusion. This is the fuel of stars. Hydrogen is fused into helium. The helium has slightly less mass. All the energy coming from our sun comes from fusion. In both cases the energy released can be calculated from E = mc2.

Changing the subject of a formula E = mc2 . . . this tells us the energy contained in a certain amount of mass, m. The reverse can happen, energy can turn into mass. To make m the subject of the formula, divide both sides by c2. This gives m =

E This tells us the mass produced from a certain amount of energy, E. c2

Finally we may be able to measure E and m, and may wish to use the formula to find a value for c, the speed of light. First divide both sides by m, then get the square root of both sides c =

E m

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Transition Year Maths

Example 2:

a +d =

e2 +2f make e the subject. u

(i) Multiply both sides by u . . . . . . . . . u(a + d) = e2 + 2f (ii) Take away 2f from both sides . . . . . u(a + d) – 2f = e2 (iii) Get the square root of both sides . . u(a + d ) − 2 f = e

Example 3:

1 s = ut + at 2 Make a the subject. 2 (i) Take away ut from both sides . . . . . . . . s − ut = 1 at 2 2 (ii) Multiply both sides by 2 . . . . . . . . . . . . . 2(s – ut) = at2 (iii) Divide both sides by t 2 . . . . . . . . . . . . .

2(s − ut ) =a t2

(iv) Multiply top . . . . . . . . . . . . . . . . . . . . . .

2s − 2ut =a t2

(v) Simplify . . . . . . . . . . . . . . . . . . . . . . . . . .

2s 2u − =a t2 t

Some Selected Common Formulas Temperature 5 9 C = (F – 32) or F = C + 32 9 5 C = Degrees Celsius F = Degrees Fahrenheit

Example 4:

Convert 0ºC Celsius to Fahrenheit 9 F = (0) + 32 = 32°C 5

50

Nicholas of Cusa German philosopher and mathematician, (1401–1464). Nicholas was a Catholic priest. He worked extensively on infinity, leading to proofs such as the one on page 38. He used his work to support the concept of God. Long before Galileo was even born, Nicholas proposed that the stars could be distant suns with inhabited planets orbiting them.

4. Formulas

Absolute Temperature K = 273 + C or C = K – 273 K = Kelvins C = Degrees Celsius

Convert 20ºC to Kelvin K = 273 + 20 = 293K

Example 5:

Simple Interest I=

100 × I P × R ×T or P = T ×R 100

or R =

100 × I P ×T

or T =

100 × I P ×R

I = Simple Interest, P = Principal, R = Rate, T = Time Example 6:

How much interest will €400 make in 2 years at a rate of 3% p.a.? I=

400 × 3 × 2 = €24 100

Compound Interest r ⎞ ⎛ A = P ⎜1 + ⎝ 100 ⎟⎠

n

A = Amount (Principal + Interest), P = Principal, r = Rate, n = Time Example 7:

How much will a principal of €1,000 amount to after 5 years at 4%? 5

4 ⎞ ⎛ A = 1, 000 ⎜ 1 + = €1,216·65 ⎝ 100 ⎟⎠

Volume of a Cylinder V = πr2h or h =

V or r = πr 2

V πh

V = volume of cylinder, r = radius, h = height Example 8:

Find the radius of a cylindrical glass if its volume is 0·5 litres and its height is 30 cm (0·5l = 500 cm3). r=

500 = 2·3 cm 30π

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Transition Year Maths

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

How much nuclear energy is there in a 1 kg book? (c = 3 × 108 m/s) ____________________________________________________________________

2.

A teenager uses about 14,000 kJ (14,000,000 J) of energy a day. If the energy from the nucleus of an atom in one litre of milk could be harnessed, for how many days would it supply all of her energy needs (c = 3 × 108 m/s)? ____________________________________________________________________

3.

If an atomic bomb of 1 kg of uranium turns into 999 g of waste products, how much energy will be released in the explosion? (c = 3 × 108 m/s) ____________________________________________________________________

4.

If a star is losing one million tonnes of mass every second, how much energy is it giving off? (c = 3 × 108 m/s); (1 tonne = 1,000 kg). ____________________________________________________________________

5.

Make a the subject of the formula v = u + at. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

6.

Make x the subject of: v ________________________________________________________ x ________________________________________________________________

(a) U =

________________________________________________________________ ________________________________________________________________

52

4. Formulas

(w + y )x ____________________________________________________ w ________________________________________________________________

(b) S =

________________________________________________________________ (c)

d + 100e ________________________________________________ 2x ________________________________________________________________ C=

________________________________________________________________ b + 2c + 3d (d) a = ________________________________________________ 4ee + 5 f + x ________________________________________________________________ ________________________________________________________________ (e) ax10 = y5 _________________________________________________________ ________________________________________________________________ ________________________________________________________________ (f) a + x = b – x ______________________________________________________ ________________________________________________________________ ________________________________________________________________

7.

Convert the following temperatures to Celsius and to Kelvin: Cairo . . . . . . . 88°F

Celsius: _______________________________________ Kelvin: ________________________________________

Dublin . . . . . . 76°F

Celsius: _______________________________________ Kelvin: ________________________________________

Miami . . . . . . 101°F

Celsius: _______________________________________ Kelvin: ________________________________________

Sydney . . . . . . 60°F

Celsius: _______________________________________ Kelvin: ________________________________________

South Pole . . . . 2°F

Celsius: _______________________________________ Kelvin: ________________________________________

What season is it in the Northern hemisphere?____________________________

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Transition Year Maths

8.

Use the simple interest formula three times to find out how much interest €5,000 will make over 3 years at 4·4% p.a. (interest added annually). Now use the compound interest formula to do the same question. ________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

9.

Find the volume of the following cylinders: (i) h = 30 cm r = 5 cm ________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (ii) h = 35 cm r = 2·4 cm ______________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

10. Find the radius of the following cylinders (i) h = 30 cm, vol = 1,000 cm3 _________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (ii) h = 15 cm, vol = 2 litres ___________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

54

4. Formulas

11. Find the height of the following cylinders: (i) r = 3 cm, vol = 1,500 cm3 __________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (ii) r = 2·2 cm, vol = 1 litre _____________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

12. If the surface area of a sphere is 4πr2, what is the radius in terms of surface area? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

13. The total surface area of a solid cylinder is 4πrh + 2πr2 What is the height (h) in terms of s and r ? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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Transition Year Maths

5.

Co-Ordinate Geometry

This sheet of paper is a plane. It is a two-dimensional (2D) space. A two-dimensional space extends forever in every direction beyond the boundaries of the page. It has length and height. Experiment

Imagine a 2D person called Mr Flat, living in a 2D world on a sheet like this. Mr Flat’s house would consist of a line. The door is a gap in the line.

Mr Flat cannot imagine anything outside his world. He has no idea that there could be anything above or below him. Now, you (a human living in our familiar 3D world) arrive and stick a compass through his world, through the sheet of paper. Mr Flat will get a shock. As the compass goes through the paper, he will see a flat metal disc appear in his world. As the compass is pulled out, Mr Flat will see the metal disc disappear. What would Mr Flat think when he sees this metal disc appear and disappear?

All points on a plane like Mr Flat’s world can be associated with an ordered pair of real numbers. Draw two number lines at right angles to each other, the horizontal one is called the x-axis, the vertical one is called the y-axis. Every point on the sheet can now be associated with an ordered pair, the first number gives the x co-ordinate, the second the y co-ordinate. Always put curved brackets around the couple, and a comma between the first number and the second. It is very important to remember that the horizontal co-ordinate, the x co-ordinate comes first in the bracket. Grid reference on maps work in the same way. As in co-ordinate geometry, the most important thing to remember is that the number on the x-axis, the horizontal axis comes first.

56

5. Co-Ordinate Geometry

Example 1:

Draw a plane with an x-axis from 0 to 4 and a y-axis from 0 to 7. Plot the points: A (1,1) B (1,6) C (3,1) D (3,6) Draw the lines [AB]; [BC]; [CD] and reveal a letter.

y 7 6 5 4 3 2 1 1

PROJECT 5.1

2

3

Plot Lines to reveal a common shape

Draw a plane with an x-axis from –9 to +9 and a y-axis from –3 to +3 to reveal the name of a common shape. Plot the following points and lines. Points: A (–5,3); B (–7,3); C (–8,2); D (–8,–2); E (–7, –3); F (–5,–3); G (–3,3); H (–4,2); I (–4,–2); J (–3,–3); K (–2,–3); L (–1,–2); M (–1,2); N (–2,3); O (1,3); P (1,–3); Q (4,–3); R (4,3); S (8,3); T (5,3); U (5,0); V (8,0); W (5,–3); X (8,–3); Lines:

[AB]; [BC]; [CD]; [DE]; [EF]; [HI]; [IJ]; [JK]; [KL]; [LM]; [MN];[NG]; [GH]; [PO]; [OQ]; [QR]; [TW];[TS]; [UV]; [WX];

Do the project on this grid.

57

4

x

Transition Year Maths

Image Produced by Reflection in a Line (Axial symmetry) Experiment

Why does a mirror reverse images from left to right, but not from top to bottom?

NOTE To find the image of a symmetry in the y-axis, the y value remains the same. The x value changes sign.

Mirror Writing Draw a line vertically down the centre of a page. Hold a pen in each hand. Relax and try not to think about what you are doing. Put both pens on the page in the middle and then quickly write your signature. If you are normally right-handed, your normal signature should be written on the right-side, and a mirror image on the left-hand side. This is a reflection in the y-axis.

gnitirW rorriM

Mirror Writing

Leonardo da Vinci wrote much of his notes in mirror writing to make it difficult to read, because he did not want people to think he was mad. Hold a mirror to the side of his writing and it will be reflected back to normal. If you hold the mirror to the bottom of the writing it will simply turn it upside down! Therefore it is false to say that a mirror will not reverse images top to bottom. This false statement exists because we cannot see the image of ourselves when the mirror is in the right position to invert the image from top to bottom. An axial symmetry is a reflection in a line (a mirror reflection). Lines are named with capital letters, and a capital S is used to indicate a symmetry. SY means ‘an axial symmetry in the y-axis’.

58

5. Co-Ordinate Geometry

Draw a plane, with the x-axis from –5 to +5 and the y-axis from –3 to +3. Plot points A (–4,3); B (–4,0); C (–4,–3); D (–1,3); E (–2,0) and lines [AD], [BE] and [AC]. Find the image of A, B, C, D, E, [AD], [BE] and [AC] by a reflection in the y-axis. Call the image points A´, B´, C´, D´, E´. Call the image lines [A´D´] [B´E´] and [A´C´]. Let y be the y-axis. SY (–4,+3) = (4,3) = A´ SY (–4,0) = (4,0) = B´ SY (–4,–3) = (4,–3) = C´ SY (–1,3) = (1,3) = D´ SY (–2,0) = (2,0) = E´ SY [AD] = [A´D´] SY [BE} = [B´E´] SY [AC] = [A´C´]

Example 2:

A

D

y

D´

A´

x

-3

Leonardo Da Vinci: Italian scientist and mathematician, (1452–1519) One of the most famous figures of the Renaissance period in history, Leonardo excelled in a variety of disciplines. His thinking was so advanced that he drew detailed plans to build helicopters, tanks and planes – inventions that could not be built until technology caught up with his ideas hundreds of years later! Despite his many talents, he is probably most famous for painting The Mona Lisa and The Last Supper.

59

Transition Year Maths

Two Axial Symmetries TRY THIS: Get two mirrors and place them at a 90º angle to each other. Look at the image of yourself in one mirror in the second mirror. You will see yourself as others have always seen you. The first mirror reverses the image, the second mirror reverses the reversed image, thus returning it to its original orientation. To do two axial symmetries, we use the word ‘after’. SQ after SY reads axial symmetry in line Q after axial symmetry in the line Y. Therefore the axial symmetry in line Y is completed first. Then the image is reflected in Q.

Example 3:

Draw a plane with an x-axis from –5 to +10 and y from –3 to +3. Y is the y-axis and line Q is the line x = 5

{

A (–4,3) [AD] B (–4,0) lines [AC] C (–4,–3) [BE] D (–1,3) E (–2,0) Find image of points by two reflections. Find SQ after SY (axial symmetry in Q after axial symmetry in Y).

plot points

A

{

D

10

Line Y

This is the original letter F.

x

Line Q

This is the intermediate image, the symmetry in Line Y.

This is the final image, the symmetry of the intermediate image in Line Q.

As you can see, two parallel axial symmetries, like two mirrors, will return the image to its original orientation.

60

5. Co-Ordinate Geometry

Image produced by reflection in a point (central symmetry) In axial symmetries figures are reversed but not inverted. In central symmetry images are reversed and inverted. This is the sort of image found on the film of a camera.

Pin Hole Camera Some of the famous painters used camera obscura to paint extremely detailed and accurate paintings. One of the most famous of these is the Dutch artist Johannes Vermeer (1632 – 1675). The simplest version uses the light from a well-lit subject to shine through a small hole in a wall via a lens to the artist’s screen in a dark room. The image will be a central symmetry. It will be both inverted and reversed left to right.

Example 4:

Draw a plane, with the x-axis from –5 to +5 and the y-axis from –3 to +3. Plot points A (–4,3); B (–4,0); C (–4,–3); D (–1,3); E (–2,0); O (0,0) and lines [AD], [BE] and [AC] Find the image of all the points and lines in a central symmetry in origin O. y

A

D

C´

x

-3

SO (–4,3) SO (–4,0) SO (–4,–3) SO (–1,3) SO (–2,0)

= = = = =

(4,–3) (4,0) (4,3) (1,–3) (2,0)

The image of the letter F is both inverted and reversed left to right.

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Transition Year Maths

PROJECT 5.2

Other Dimensions

We live in a 3-Dimensional world. We can move up and down, forward and back and side to side. Objects like cubes, cylinders, cones or pyramids have volume. Break up into small groups and discuss what the following 3 worlds could possibly look like. Think about them and talk about them before you look them up on the internet. (i) A 2-Dimensional world. (ii) A 1-Dimensional world. (iii) A 4-Dimensional world. (You have to use your imagination!)

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Using the grid provided below plot the following points and lines and reveal a shape: Points

A (0,9);

B (2,2);

G (–7,–9); Lines

[AB];

C (9,2); D (4,–3);

H (–5,–3);

[BC];

[CD];

I (–9,2); [DE];

E (6,–9);

F (0,–4);

J (–3,2)

[EF];

[FG];

[GH];

[HI];

2.

Reflect all the points in Question 1 in the y-axis. What is the new shape?

3.

Reflect all the points in Question 1 in the x-axis. What is the new shape?

62

[IJ];

[JA]

5. Co-Ordinate Geometry

4.

Using the grid below, plot the following points and lines to reveal a word: A (–8,3); B (–8,–3); C (–5,–3); D (–3,3); E (–2,3); F (–1,3); G (–3,–3); H (–2,–3); I (–1,–3); J (1,3); K (4,3); L (1,–3); M (4,–3); N (5,3); O (8,3); P (5,0); Q (8,0); R (5,–3); S (8,–3) [AB]; [BC]; [DF]; [GI]; [EH]; [JL]; [JM]; [MK]; [NR]; [NO]; [PQ]; [RS]

5.

Using the grid below, draw the letter T to the left of line Y (the y-axis). Then draw SY (the image of T by an axial symmetry in Y). Finally draw SQ after SY (the image of T by an axial symmetry in the line x = 5 after an axial symmetry in the y-axis).

63

Transition Year Maths

6.

Recall Mr Flat from the beginning of the chapter (page 56). A person who exists in a 3-Dimensional world drops an apple through the flat world of Mr Flat. Describe what Mr Flat will see. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________

7.

Assume the fourth dimension is time. A person lives in this 4-dimensional world. This person has control over the fourth dimension. They have the ability to change the time in a specific spot. Imagine that this person, who can control time, changes the area outside the room you are now in, back 100 years for a few seconds, then returns it to the present! Describe what you might see. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________

8.

Y is the y-axis, X is the x-axis and O is the origin. Using the points shown draw the orientation of the letter F after the following: (i)

SY

(ii)

SY after SX

64

5. Co-Ordinate Geometry

9.

(i)

Write your name in block capitals, in the top left quadrant below.______

(ii)

Write your name as if it was reflected in the y-axis. ___________________

(iii)

Write your name as if it was reflected in the x-axis. ___________________

(iv)

Write your name as if it was reflected through the origin. _____________

0

x

y

65

Transition Year Maths

6. Pythagoras’ Theorem In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

The hypotenuse is the longest side and is opposite the right angle.

5cm 3cm

52 = 32 + 42 25 = 9 + 16

4cm Pythagoras lived in the sixth century BC. He travelled the world to discover all that was known about Mathematics at that time. He eventually set up the Pythagorean Brotherhood – a secret society which worshipped, among other things, numbers. Pythagoras described himself as a philosopher – a person whose interest in life is to search for wisdom. This brotherhood searched for: Excessive Numbers Numbers whose divisors add up to more than the number. Perfect Numbers Numbers whose divisors add up exactly to the number. Defective Numbers Numbers whose divisors add to less than the number. Slightly Defective Numbers whose divisors add to one less than the number.

Example 1:

Identify which of the following are excessive, perfect, defective or slightly defective: 4, 6, 8, 10 and 12

Remember: For this exercise, when adding divisors you do not include the number itself. (i) 4 – divisors are 1, 2; Sum = 3 ⇒ 4 is slightly defective. (ii) 6 – divisors are 1, 2, 3; Sum = 6 ⇒ 6 is perfect. (iii) 8 – divisors are 1, 2, 4; Sum = 7 ⇒ 8 is slightly defective. (iv) 10 – divisors are 1, 2, 5; Sum = 8 ⇒ 10 is defective. (v) 12 – divisors are 1, 2, 3, 4, 6; Sum = 16 ⇒ 12 is excessive.

PROJECT 6.1

Whole Number right-angled Triangles

Find all the whole number right-angled triangles using the numbers 1 to 30 inclusive. There are at least 9 of them. Remember the long side squared will equal the sum of the squares of the other two sides.

66

6. Pythagoras’ Theorem

The Pythagoreans investigated the following triangle: They knew that 12 = 1 and 12 = 1. Therefore the diagonal squared = 2. But 12 = 1 and 22 = 4 so the diagonal had to be between 1 and 2. To their horror, the Pythagoreans proved the length was not a fraction! This reality upset their picture of reality which was based on number. They wanted an ordered world of real numbers. This length appeared evil to them. It shook the foundations of their numerological philosophy and they tried to keep it a secret. There are stories about Hippasus of Metapontium who leaked the story and was thrown out of a boat to drown for threatening the purity of number.

? 1

1

This number is now called an irrational number. It is a decimal which goes on forever with no repetition.

NOTE The going on forever bit involves infinity, which does still cause some problems.

A Proof of Pythagoras’ Theorem x

Construction: Draw a square of side x + y. Inside it draw a square of side z. Proof: Area of large square is (x + y)2 Area of large square is also the area of the small square (z)2 + area of 4 triangles. 1 2

Area of triangle is (base × height) =

y

z

1 xy 2

1 2

∴(x + y)2 = z2 + 4( xy ) ⇒ (x + y)2 = z2 + 2xy . . . . . . . This is the same square. ⇒ x2 + 2xy +y2 = z2 + 2xy . . . . Multiply out (x + y)2 ⇒ x2 + y2 = z2 . . . . . . . . . . . . 2xy can be cancelled.

x

A Pythagorean Triple is a set of three whole numbers, such that one number squared added to another number squared equals the third number squared.

PROJECT 6.2

Estimating the Height of a Tree

Using only a protractor and a metre stick, how would you estimate the height of a tree?

67

Transition Year Maths

PROJECT 6.3

Sequence of Squares and right-angled Triangles

Continue on this sequence for 50 terms and find 4 different right-angled triangles which are not multiples of each other. Subtract each successive square from the previous 12 22 32 42 52 62 72 82 1 4 9 16 25 36 49 64 one. This gives an odd number, but every so often, \/ \/ \/ \/ \/ \/ \/ one of these odd numbers is 3 5 7 9 11 13 15 itself a square. Therefore 2 52 – 42 = 32 gives 32 + 42 = 52. 3 There are three more to be 52 – 42 = 32 found before 502 is reached!

Pythagoras: Greek mathematician, 600 BC Pythagoras loved truth and wisdom and spent his life searching for these qualities, often by studying numbers. He founded a secretive religious group devoted to the study of numbers. This group believed numbers were the ultimate reality, and everything could be predicted by studying them. It is suspected that the concept of irrational numbers was first discovered by this group but it was so abhorrent to their search for truth that they denied the existence of such numbers. Pythagoras was one of the first philosophers to suggest that the thought process was carried out in the brain, not the heart.

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Study the numbers between 20 and 30 and identify which are defective, slightly defective, perfect and excessive (Hint: there is one perfect number between 20 and 30). ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

68

6. Pythagoras’ Theorem

2.

Draw any right-angled triangle in the middle of a page. Use a ge Lar

construction [a set square or the edges of a book] to produce the right angle. Make the sides any length you wish. Now draw a square on each side. Measure the area of the large square in

Small

millimetres. Measure the area of the two small squares. If you have drawn it accurately the area of the large square should Medium

equal the combined area of the two smaller squares. 3.

Which of these triangles are right-angled triangles? The three numbers are the lengths of the three sides:

4.

(a) 1, 1, 2 ___________

(h) 5, 12, 13 ______

(o) 10, 12, 14 __________

(b) 1, 2, 3 ___________

(i) 5, 13, 14 ______

(p) 12, 14, 16 __________

(c) 2, 3, 4 ___________

(j) 5, 14, 15 ______

(q)

2,

2, 1 __________

(d) 3, 4, 5 ___________

(k) 2, 4, 6 ________

(r)

2,

2, 2 __________

(e) 4, 5, 6 ___________

(l) 4, 6, 8 ________

(s)

2,

2, 3 __________

(f)

5, 10, 11 _________

(m) 6, 8, 10 _______

(t)

2,

2, 4 ___________

(g) 5, 11, 12 _________

(n) 8, 10, 12 ______

You are asked to mark the perimeter of a new football pitch. How would you ensure that the corners are in fact right angles? ____________________________________________________________________ ____________________________________________________________________

5.

These are right-angled triangles. Find x in each case. The last figure is the hypotenuse. (i)

2, 4, x________________________

(vi) 3, 1, x ___________________

(ii)

5 , x , 3 ______________________

(vii) 8, 15, x __________________

(iii) x, 3 , 3 _____________________

(viii)10, x, 26 __________________

(iv) 1, x, 3________________________

(ix) x, 24, 25 __________________

(v)

(x) 9, x, 15 ___________________

x, 12, 18 _____________________

69

Transition Year Maths B

6.

12

9

Angle ABC = 90º Angle BDC = 90º

A

C D Find: (i)

|AC| ____________________

(ii) area  ABC______________ (iii) |DB| ____________________ (iv) |AD| ____________________ (v)

|DC| ____________________

7.

B

Angle ABD = 45º Angle BAD = 45º |BC| = 6 C

A D Find (i)

|AC | ____________________

(ii) |DB | ____________________ (iii) |DC | ____________________ (iv) |AD | ____________________ (v)

area  ABC______________

70

7. Ratio and Fibonacci

7. Ratio and Fibonacci Ratio is a way of showing the connection between two or more numbers. A ratio can be written as a fraction, a decimal or as numbers with : between them.

Example 1:

There are 30 students in a class; 18 are girls and 12 are boys. Express this as a ratio. The ratio of girls to boys is: 18 : 12 or 3 : 2 (simplifying) or 3 2

(fraction) or

1·5 (decimal)

Irrational numbers are numbers which never repeat. They cannot be written as a ratio of two real numbers. Rational numbers are numbers which can be written as a ratio (a fraction) using real numbers. The words rational and ratio are intrinsically connected.

The Golden Ratio PROJECT 7.1

Head Measurement: Length / Width

Do this in pairs. Each person in the class measures the length, l, of their partner’s head. Now measure the width, w, of their head. To find the ratio of length : width, divide the length by the width.

l

Ratio of length : width = l w w

71

Transition Year Maths

PROJECT 7.2

Ratio of Human Body: Bottom / Top

Do this in pairs. Each person measures the distance from navel to top of head, t. Then measure the distance from navel to bottom of foot, b. To find the ratio of the bottom to top, divide b by t. Ratio of bottom: top = b t

t

b

Plot the pairs of data on a scatter graph. What is the correlation? Collect the results from both projects for the whole class and find the average. The resultant ratio should be very close to the Golden Ratio in both cases. This is very close to 1·618 for most people. At least since the Renaissance, many artists and architects have proportioned their works to approximate the Golden Ratio, believing this proportion to be aesthetically pleasing. Leonardo da Vinci’s drawings of the human body (in project box above) emphasised its proportion. This Golden Ratio has many unusual properties. We will take the number as 1·618. Remember, this is an approximation. The number is irrational (see note below). 1 = 0·618 = 1·618 – 1 1·618

You will not find too many numbers which satisfy:

1 = x –1 x

NOTE ON DANGER OF ROUNDING OFF NUMBERS: If numbers are rounded to the nearest whole number, then 1·35 rounds down to 1 and 2·7 rounds up to 3. Therefore 1 + 1 = 3. Because 1·35 + 1·35 = 2·7

1+1=3 72

7. Ratio and Fibonacci

PROJECT 7.3

Pairs of Rabbits

Consider pairs of rabbits. They are immature for their first year and do not breed. From the second year on they breed one new pair of rabbits each year. Starting with one pair of rabbits, how many pairs will there be every year for eight years? In Year One, there will be one pair. They do not breed, so in Year Two, there is still only one pair. In Year Three, the original pair of rabbits breed one new pair. Therefore the project will start: Year One

Year Two

Year Three

1

1

2

Year Four

Continue for 8 years, working out the number of pairs of rabbits.

Fibonacci’s Sequence A sequence in mathematics is a row of numbers with some pattern. The pattern in these should be obvious: (a) 1, 2, 3, 4, 5, 6, 7, . . . . . . . . . . . . . . . (add 1) (b) 2, 4, 6, 8, 10, 12, 14, . . . . . . . . . . . . . (add 2) (c) 1, 4, 9, 16, 25, 36, 49 . . . . . . . . . . . . (square) (d) 1, 2, 5, 14, 41, 122, 365 . . . . . . . . . . (multiply by 3, minus 1) Fibonacci, an Italian mathematican developed a famous sequence in the thirteenth century. Start with 1, 1. Every number is calculated by adding the two previous numbers together: 1 + 1 = 2 1 + 2 = 3 2 + 3 = 5 3 + 5 = 8 5 + 8 = 13 8 + 13 = 21 The sequence is: 1, 1, 2, 3, 5, 8, 13, 21 . . .

PROJECT 7.4

Fibonacci Sequence & Ratios

Write out the first 20 terms of the Fibonacci Sequence. Then write down the ratio of each number to the previous one.

These ratios get closer to the Golden Ratio as the numbers get larger. The ratio of 21 21 : 13 is = 1·615 13

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Transition Year Maths

Other Sequences Examining patterns of blocks is an interesting and useful way of introducing sequences. We will examine the number of white tiles in the following example: Example 2: n=1

n=2

n=3

(i) Draw the next two stages (ii) Write the number of white tiles for the first 6 stages in the pattern as a sequence (iii) How many white tiles are there in stage 10? Can you give a general expression for the number of white tiles for any stage? (i) n = 4

n=5

(ii) 8, 10, 12, 14, 16, 18 (iii) Let n = stage number; the number of tiles is 6 + 2n. Therefore in stage 10 there will be 6 + 2(10) = 26 white tiles.

PROJECT 7.5 n=1

Block Sequences 1 n=2

n=3

(i) Draw the next two stages (ii) Write the number of white tiles for the first 6 stages as a sequence (iii) How many white tiles are there in stage 10 and in stage 50? Can you give a general expression for the number of white tiles in any stage? (iv) Will any stage have 101 tiles? (v) Will any stage have an odd number of tiles?

74

7. Ratio and Fibonacci

PROJECT 7.6 n=1

Block Sequences 2 n=2

n=3

(i) Draw the next two stages (ii) Write the number of white tiles for the first 6 stages as a sequence (iii) How many white tiles are there in stage 10 and in stage 50? Can you give a general expression for the number of white tiles in any stage (iv) Try to explain why (n + 1)2 – 1; n2 + 2n; n(n + 2) would all work as the general expression for the number of tiles in stage n.

PROJECT 7.7 n=1

Block Sequences 3 n=2

n=3

(i) Draw the next two stages (ii) Write the number of white tiles for the first 6 stages as a sequence (iii) How many white tiles are there in stage 10 and in stage 50? Can you give a general expression for the number of white tiles in any stage (iv) What stage will have 101 white tiles?

Ratio: To divide numbers in a certain ratio: Example 3:

Divide €100 between Anna and Eamon in the ratio of 7 : 3 Add 7 + 3 = 10 7 3 Anna gets , Eamon gets 10 10 3 7 × 100 = €30 × 100 = €70, 10 10

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Transition Year Maths

Example 4:

Divide €1,000 in ratio of Mick : Joe : Pat; 4 : 5 : 6 Add 4 + 5 + 6 = 15 4 Mick gets × 1, 000 = €267 rounded to nearest euro. 15 5 Joe gets × 1, 000 = €333 rounded to nearest euro. 15 6 Pat gets × 1, 000 = €400 15

Example 5:

Multiple Ratios The ratio of A : B is 3 : 4 The ratio of B : C is 2 : 9 The ratio of C : D is 3 : 5 Find the ratio of A : D First note A : B is 3 : 4 The first part of the next ratio must be changed to 4 ⇒ B : C is 2 : 9 Multiply both sides by 2 = 4 : 18 The first part of the next ratio must be changed to 18 ⇒ C : D is 3 : 5 Multiply both sides by 6 = 18 : 30 ∴ A:B:C:D = 3 : 4 : 18 : 30 ⇒ A : D is 3 : 30 or 1 : 10

PROJECT 7.8

Newspaper Sheets: Ratio of length to width

A double page from a full-sized newspaper is 1,189 mm wide by 841 mm long. This is called A0 paper. If the longer side is halved, the page becomes 594.5 mm × 841 mm. This is called A1. Continue to half the longer side to get the dimensions of A2, A3, A4, A5, A6, A7 and A8 paper. Find the area of each page and give the ratio of length to width of each sheet. What do you notice about all the ratios?

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7. Ratio and Fibonacci

PROJECT 7.9

Nature’s Sequences : Daisy Petals

The seed patterns in sunflowers, the arrangement of petals in daisies, even the sequence of branches on pine trees all follow Fibonacci’s sequence. Count the number of petals in daisies. It is very often 55, not 56 or 54. These sequences occur in nature because they are irregular. They reduce the lines of weakness in a plant’s physical structure making them stronger and less likely to split or break along lines of weakness.

Leonardo of Pisa (nicknamed Fibonacci) Italian mathematician, (1180–1250) Fibonacci is most famous for the Fibonacci Sequence mentioned on page 73. He travelled widely around North Africa with his father where he noticed the differences between number systems in Europe and those in the Arab world. In 1202 he wrote a book called Liber Abaci which introduced the Hindu-Arabic system of numbers to the West. The Greek and Roman system of solving arithmetic problems was difficult. The new Hindu-Arabic system was much easier and allowed Europe to become the centre of the world of mathematics in the following centuries.

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Transition Year Maths

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Divide €1,000 (to nearest euro) between Fred, Herb and Sally in ratio 1 : 2 : 3 ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

2.

Eamon gets €4,000 and Jane gets €5,000 from a prize of €20,000, Tim gets the rest. In what ratio was the prize divided between Eamon : Jane : Tim? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3.

The ratio of

A : B is 3 : 4 B : C is 2 : 6 C : D is 5 : 15

Find the ratio A : D ____________________________________________________________________ ____________________________________________________________________

4.

The ratio of E : F is 4 : 5, F : G is 6 : 7, G : H is 1 : 9, H : I is 5 : 2 Find the ratio (a) E : G _____________________________________________ (b) E : H _____________________________________________ (c) E : I ______________________________________________

78

7. Ratio and Fibonacci (d) I : G ______________________________________________ (e) I : E ______________________________________________ (f) H : F _____________________________________________

5.

Is there a connection mathematically between the Golden Ratio 1·618 and Fibonacci’s Sequence? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

6.

How many sheets of A8 paper are there in one sheet of A0 paper? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

7.

When a plant is growing outwards, new growth appears at an angle of 137·5° to the previous growth. Can you find any connection between the angle 137·5° and the Golden Ratio 1·618? Why does nature pick 137·5°? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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Transition Year Maths

8.

Algebra

In Chapter 1, we saw that our number system – the decimal system – uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, to make all our numbers. The Romans used the letters I, V, X, L, C, D, M to represent their numbers. The Greeks used

l Π Δ H X M

a vertical bar for 1 Pi . . . . . . . . . . . . . . . . for 5 Delta . . . . . . . . . . . . . for 10 Eta . . . . . . . . . . . . . . . for 100 Chi . . . . . . . . . . . . . . for 1000 Mu . . . . . . . . . . . . . . for 10,000

for inscriptions. These symbols and letters are the numbers in these systems. Algebra is different. Algebra is about using a letter for an unknown number or a variable. A letter, usually x or y, is substituted instead of a number. Always clearly state what each variable represents. This will help avoid confusion. In Example 1 make sure to clearly write down Let x = cost of one apple

Substitution: Example 1:

Example 2:

You are in a shop. A sign says 6 apples for 90 cent. You can probably work out in your head how much one apple costs. But if we want to use maths to find the cost, we use algebra. We let x = cost of one apple. 6 apples costs 90 cent ⇒ 6 multiplied by (cost of one apple) = 90 cent ⇒ 6 multiplied by x = 90 ⇒ 6x = 90 90 or divide by 6: x = = 15 cent each 6 Sign in window of Sally’s fruit shop: 8 apples for €1 Sign in window of Herb’s fruit shop: 6 apples for 72 cent Which shop is cheaper for apples? Let x = cost of one apple in Sally’s shop Let y = cost of one apple in Herb’s shop Sally’s shop: 8x = 100 100 1 x= = 12 cent each 2 8 Herb’s shop: 6y = 72 y=

72 = 12 cent each 6

Herb’s is cheaper.

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8. Algebra

a = 2, b = 3, find a value for: (i) a + b = 2+3=5 (ii) 2a + b = 2(2) + 3 = 4 + 3 = 7 (iii) a2 + b = 22 + 3 = 4 + 3 = 7 (iv) 3(a + b) = 3(2 + 3) = 3(5) = 15 (v) 4a2 + 2b = 4(2)2 + 2(3) = 4(4) + 2(3) = 16 + 6 = 22

Example 3:

Multiplication: When numbers are multiplied, a new number is obtained. But when letters representing unknowns or variables are multiplied, we do not get a new letter. When two or more of the same letter are multiplied we simply square it or cube it and so on. (i) (ii) (iii) (iv) (v)

Example 4:

a × a = a2 c × c × c = c3 3a × 2a = 6a2 5 × a = 5a a × b = ab

Adding & Subtraction: Letters representing unknowns or variables can be added and subtracted. Let

a = cost of one apple 3a = cost of three apples a + 3a = 4a = cost of four apples.

Example 5:

(i) (ii) (iii) (iv) (v) (vi)

a + 4a = 5a 2a + 3a + 3b = 5a + 3b 3a + 5a + 4a2 + 2a2 + b = 8a + 6a2 + b 4c2 + 3c + c = 4c2 + 4c a + b + a2 + b2 + 2a + 3b = 3a + 4b + a2 + b2 5a + 3b – 2a – 6b = 3a – 3b

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Transition Year Maths

Dividing: Letters representing unknowns or variables can be divided. There is one exception. Dividing anything by zero is not allowed because zero will go into any finite quantity an infinite number of times. As we shall see later, calculations involving infinity can cause problems. Example 6:

(i)

5 =1 5

(ii)

5a = 1a = a 5

(iii)

5a =5 a

(iv)

5(a – c ) = 5 provided (a – c) ≠ 0 (a – c )

(v)

4a 3 4 × a × a × a 4a = = 2a = 2a 2 2×a ×a 2

Simplification: Expressions with letters representing unknowns or variables can be simplified: If a = cost of one apple and b = cost of one banana, then 4a + 2b = cost of 4 apples and cost of 2 bananas. This is the same as twice (cost of 2 apples and cost of 1 banana) ∴ 4a +2b = 2(2a + 1b) Example 7:

PROJECT 8.1

(i) (ii) (iii) (iv) (v)

4a + 6c = 2(2a + 3c) 10a + 2b = 2(5a + b) 10a + 20b + 5c = 5(2a + 4b + c) 15a – 3b = 3(5a – b) 4a + 8a2 = 4(a + 2a2) = 4a(1 + 2a)

Find the Error in Proof 1=2

Proof of 1 = 2 This proof must have a flaw in it. Using what you have learnt so far, examine each step and find the error. a=b a2 = ab a2 + a2 – 2ab = ab + a2 – 2ab 2(a2 – ab) = a2 – ab 2=1

(substituting letters for numbers) (multiplication of both sides by a) (add a2 – 2ab to both sides) (simplify both sides) (divide both sides by (a2 – ab))

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8. Algebra

There must be something fundamentally wrong with one of the 5 steps. Look carefully at each of the 5 steps. You should be able to deduce what is wrong when you have studied the 5 processes: substitution, multiplication, adding / subtracting, simplifying and division. (See hint on page 87.)

Solving Equations in Algebra One Equation with One Unknown: Example 8:

4x = 60 Find x. Divide both sides by 4 x = 15

One Equation with Two Unknowns: There are an infinite number of correct solutions Example 9:

3a + 2b = 20 a = 0; b = 10 is a solution. a = 2; b = 7 is a solution. There is no unique solution.

Two Equations with Two Unknowns: These are called simultaneous equations. There will be one unique solution. Example 10:

3x + y = 13 x – 2y = –5 Find a solution which simultaneously solves both these equations. Multiply top line by 2 gives 6x + 2y = 26 Lower line remains the same x – 2y = –5 Add 7x = 21 Divide both sides by 7 x= 3 Substitute x = 3 back into the top line 3(3) + y = 13 9 + y = 13 y= 4 This is one unique solution x = 3, y = 4

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Transition Year Maths

RULE As a general rule a problem can be solved if there are as many, or more, equations as there are unknowns. If there are three unknowns, there must be at least three equations.

Example 11:

Solve for x, y and z (equation 1) x + 3y – 2z (equation 2) 4x – y – z (equation 3) 2x + 2y – 3z

= 1 = –1 = –3

Eliminate z from equation 1 and equation 2. x + 3y – 2z = 1 (equation 1) –8x + 2y + 2z = 2 (multiply equation 2 by –2) (Add) –7x + 5y = 3 (equation 4) Eliminate z from equation 2 and equation 3. –12x + 3y + 3z = 3 (multiply equation 2 by –3) 2x + 2y – 3z = –3 (equation 3) (Add) –10x + 5y = 0 (equation 5) Now use (equation 4) and (equation 5) to solve for x + y 7x – 5y = –3 (multiply equation 4 by – 1) –10x + 5y = 0 (equation 5) (Add) –3x = –3 ⇒x = 1 ⇒ y = 2 (putting x = 1 back into equation 5) ⇒ z = 3 (putting x = 1 and y = 2 back into equation 1)

PROJECT 8.2

Solving for x, y, z and w.

Using a method similar to Example 11, solve for x, y, z, and w x + y + z + w = 14 2x + 4y – 3z + w = 9 3x – 2y – z + 2w = 6 –2x – 3y + 2z + w = 0

84

8. Algebra

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Let a = 1, b = 2 and c = 3, evaluate: (i)

a + b + c ________________________________________________________

(ii)

a + b – 2c _______________________________________________________

(iii) 10a + 5a – 2c ____________________________________________________ (iv)

a2 – b2 + c _______________________________________________________

(v)

a3 + b2 + c _______________________________________________________

(vi)

2(a + b)2 + c _____________________________________________________

(vii) 11 + a + 10c _____________________________________________________ (viii) 10a2 + 5b + 8a2 + 3c + 10a + 2a3 + 4c _________________________________

2.

(ix)

a 4 .b 3 .c 6 a 4 .b 2 .c 2

(x)

a +a +b +c –a __________________________________________________ a .a .a .b .c

____________________________________________________

a 15 .b 14 .c 133 Simplify a 13 .b 14 .c 15

___________________________________________________

____________________________________________________________________

3.

Simplify (i)

a a

____________________________________________________

(ii)

2a ____________________________________________________ a

(iii)

2a 2 ___________________________________________________ a

(iv)

a ___________________________________________________ 2a 2

(v)

a5 ____________________________________________________ a4

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Transition Year Maths

4.

Add

(i)

5a + 3a2 + 2a ___________________________________________

(ii) 6 + 3c + 2c _____________________________________________ (iii) a + b + c + a _____________________________________________

5.

Simplify

a(b + c )– ab + 2ac _____________________________________________ a

____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

6.

x is the cost of one apple; y is the cost of one orange. Write formulas for the total cost of the following and use these formulas to find the cost of an apple and an orange. (i)

An apple and an orange cost 25 cent. ______________________________

(ii) Five apples and eight oranges cost €1·67. ___________________________ (iii) The difference in cost between three apples and two oranges is 5 cent. _______________________________________________________________ (iv) 28 apples cost the same as 22 oranges. ______________________________ (v)

How much do apples cost? ________________________________________

(vi) How much do oranges cost? _______________________________________

7.

11 pears cost €1; 12 oranges cost €1·10. Which are cheaper? ____________________________________________________________________

8.

Solve for x and y:

5x + 3y = 23 2x – 3y = 5

____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

86

8. Algebra

9.

A box contains 150 coins, all €1 or €2 coins. The total value of the coins is €204. How many of each coin are in the box? __________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

10. Solve for x, y, and z:

x + 2y + z = 1 2x – y – z = 2 – 3x – 2y + z = 5

____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

11. Can these three simultaneous equations be solved, and why? x + 2y + z = 1 2x + 4y + 2z = 2 –x – 2y – z = –1 ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

12. Solve for a, b and c 2a + b – c = 3 a – b + 2c = 7 –3a + b – 2c = –11 ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ Hint for Project 8·1: the flaw is in the last step! If a = b then a2 = ab, then a2 - ab = 0. Dividing by 0 is not allowed in maths.

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Transition Year Maths

9.

Pi

π is 3·14159265358979323846264338327950288419716939937510582097494459…. Pi is the ratio of a circle’s circumference and its diameter. It is often represented with the Greek symbol ‘π’. Pi is an irrational number so the number of decimal places never ends and no pattern has yet been found in the decimal places (see above). Pi is usually estimated as π = 3·14 or 3 1 or 7

22 7

Your calculator estimates π = 3·141592654 Knowing π to 35 decimal places allows us to calculate the circumference of the universe to an accuracy of about 1 cm. Computer scientists in 2010 calculated π to 5 trillion places. This has absolutely no practical use, but the search for patterns in numbers like π will always interest mathematicians.

PROJECT 9.1

Estimating π

To find an estimate for π • Get a length of string, a piece of chalk, and a metre stick. • Draw a dot in the middle of the school yard or hall with chalk. Then using the string draw a large circle. • Using the metre stick, measure the length from the centre dot to the circle. This is the radius. Then measure the length of the circumference of the circle.

Result The circumference divided by twice the radius should equal π. What did you get?

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9. Pi

PROJECT 9.2

Pi to 1000 Decimal Places

Look up Pi to 1000 places. Is it totally a random number? Is there any pattern? How far to the first zero? How far until there is 6 identical digits in a row? How often do the digits 1, 2, 3 appear in a row? Work in large groups to find how many of each digit appears in the first 1000 places. Plot the result on a bar chart; is the result what you expected? There are 101 eights, does this surprise you?

PROJECT 9.3

Pi Scatter Graph

Every pupil in the class draw a circle of any size they choose. Then measure the circumference as accurately as possible and measure the diameter as accurately as possible. Tabulate all the pairs of measurements, for the whole class. Plot a scatter graph of the data putting circumference length (in centimetres) on the y-axis and diameter (in centimetres) on the x-axis (see chapter 2). What is the correlation like? In nature π can be found in meandering rivers. If we measure the actual length of a river from the source to the mouth and measure the direct length in a straight line from source to mouth, on average the actual length equals π multiplied by the direct length. Why is this true? Rivers tend to form bends due to the effects of erosion. If they bend too much, the degree of bending is restricted because the bends join up, straightening the river again. Einstein explained that the balance between these two opposing factors leads to an average ratio of π (3·14) between actual and direct length.

PROJECT 9.4

Einstein’s Meander Ratio

Measure the actual length and the direct length of the river in the image above. What is the ratio between them? Look up satellite images of meandering rivers and try to estimate the actual length : direct length ratio for a section of a river. Is it close to 3.14 : 1? Do not be surprised if the ratio is less than 3.14 : 1 for the section you measure. The ratio usually only applies to the full length of a river from source to ocean.

NOTE It has been conjectured that Pi is “normal”, mathematically this means that all digits are equally likely and that it is truly a random number. This conjecture has yet to be proven, to the irritation of some professional mathematicians.

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Transition Year Maths

NOTE Consider a circle of radius 1m: The total distance around the circumference will be 6·28 m≈ 2π m. This leads to the formula: circumference = 2π × radius

1m

The Radian When the distance moved around the circumference is 1 radius, the angle in the centre is called 1 radian. There are about 6·28 radians in one full circle. 6·28 radians = 360º ⇒ 1 radian = 57·3º

NOTE ON CALCULATOR USE On a Casio calculator, a small D indicates the calculator is in degree mode. R indicates radians. To change from one to the other press Mode until offered the choice: Deg 1

Rad 2

Grd 3

Pick 1 for degrees and so on.

On a Sharp calculator, a small DEG , RAD or GRAD indicates the mode. Press the DRG button till the required mode appears on the screen.

Archimedes made a good estimation of Pi by working out the area of a 96-sided polygon fitted inside a circle over 2,000 year ago. The English mathematician John Wallis in 1855 showed that Pi could be calculated from the infinite series below, in which the top part of the fraction goes up by two in pairs, beginning with two 2’s and the bottom part goes up by two in pairs beginning at a single 1. 7 ⎛ 2 2 4 4 6 6 8 8 10 ⎞ π = 2 ⎜ × × × × × × × × . . .⎟ ⎝1 3 3 5 5 7 7 9 9 ⎠

90

9. Pi

PROJECT 9.5

John Wallis’s Equation

Using John Wallis’s equation for Pi, continue the series for a total of 30 fractions and estimate π. Compare the answer with π given at the start of the chapter.

Gabriel’s Horn

1 for values of x between 1 and 10 inclusive. x f (x) is plotted below to 2 decimal places.

Draw a graph of the function f (x) =

x y=

1 x

0·2

1

2

3

4

5

6

7

8

9

10

5

1

0·5

0·33

0·25

0·2

0·17

0·14

0·12

0·11

0·1

y 5 4 3 2 1 1

2

3

4

5

6

7

8

9

10

x

This graph can be turned into a 3D shape by rotating it 360°, a full circle, on the x-axis. This produces a shape like a horn.

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Transition Year Maths

If the horn has a radius of 1 m at the wide end, then the total volume of the horn is π m3. Remember this horn goes on forever to the right, just getting thinner and thinner. This causes a problem. The total volume of the horn is π m3, but the surface area is infinite. The proof of this requires integration (part of calculus).

PROJECT 9.6

Gabriel’s Horn Paradox

This is a thought experiment. You will not be able to work it out on paper or using a calculator. Einstein frequently used this method to solve problems. Consider a shape like Gabriel’s Horn, with a radius of 1 m at the top. Pour 3·14 m3 (π m3) of paint into the container. It will fill completely, right to the brim. Now consider another Gabriel’s Horn of the exact same size, and imagine painting the inside of the horn. The surface area is infinite, so when you have used all the paint that filled the first horn, 3·14 m3, the inside will still not be painted. How can this be? Discuss this problem in small groups.

Albert Einstein: German physicist, (1879–1955) Einstein was born in Germany, the son of an engineer. He was a very bright pupil at school, but as a teenager he found authority in school difficult and never formally graduated. At the age of 16 Einstein performed his famous ‘thought experiment’ in which he visualised travelling along inside a beam of light. He is most famous for special and general relativity and the world’s most famous formula, E = mc2, which led to the development of nuclear power.

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9. Pi

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

π can be computed from an equation: 1 ⎛1 1 1 1 1 1 1 1 1 1 1 ⎞ π = 4 ⎜ – + – + − + − + − + − . . .⎟ ⎝ 1 3 5 7 9 11 13 15 17 19 21 23 ⎠ 1 Use your calculator to calculate π until you reach − 39

NOTE ON CALCULATOR USE This problem can be quickly solved using the x–1 button on the calculator 4

5

x–1

–

7

( x–1

reached, close bracket

1

x–1

–

3

x–1

+

+ . . . When the last number is ) and use = .

2.

Using π = 3·14, find the circumference of a circle of radius 2·5 m. ___________

3.

Using π =

4.

Using the calculator value of π, find the circumference of a circle of radius 2·5 m.

22 , find the circumference of a circle of radius 2·5 m._____________ 7

____________________________________________________________________ 5.

Find the radius of a circle of circumference: (a) 1 m _____________________________________________________________ (b) 5 m _____________________________________________________________ (c) 10 m ____________________________________________________________ (d) 20 m ____________________________________________________________

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Transition Year Maths

6.

Put your calculator into degrees mode and find: (a) sin 10º ___________________________________________________________ (b) sin 20º __________________________________________________________ (c) sin 57·3º _________________________________________________________ (d) sin 114·6º ________________________________________________________

7.

Put your calculator into radian mode and find: (a) sin 1 radian ______________________________________________________ (b) sin 2 radian ______________________________________________________ (c) sin 3 radian ______________________________________________________ (d) sin 4 radian ______________________________________________________

8.

Some of the answers for Question 6 and Question 7 are similar. Why is this? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

9.

What is an irrational number? Is π irrational? _____________________________

10. Assuming the value of π provided at the start of this chapter is correct, which is closer to π: the value of π calculated from the first 15 fractions in John Wallis’s equation or 22 ?________________________________________________________________ 7

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9. Pi

11. Fill out a table of the function f (x) =

1 for all values of x from 1 to 15 (to 2 decimal x

places). Then by plotting the points on a graph with x from 1 to 15 and y from 1 to –1, draw a 2-Dimensional picture of Gabriel’s Horn. What do you notice about the shape? y 1 0·8 0·6 0·4 0·2 0

1

2

3

4

5

6

7

8

–0·2 –0·4 –0·6 –0·8 –1

95

9

10

11

12

13

14

15

x

Transition Year Maths

10.

Probability

Probability is the maths of chance – it tells us how likely an event is to occur. The probability of an event occurring is usually written as a fraction or as a percentage.

Example 1:

Toss one coin, assume it is a fair coin and ignore the chance of it landing on its edge. 1 P (H ) = 2

The probability of getting a head is 1 or 1 in 2 or 50%. 2

The probability of getting a tail is also 1 or 1 in 2 or 50%. 2

There are only two possible outcomes in this example – either a head H or a tail T.

NOTE All probabilities are between 0 and 1 inclusive. 0 = impossible; 1 = certainty

DEFINITION P(Event) =

PROJECT 10.1

Number of times this event occurs Total possible number of outcomes

Toss a Coin

Toss a coin 100 times and count the number of heads and the number of tails. How close is it to 50 : 50? If every student in the class does this exercise, and all the results are added together, the number of heads and tails should be surprisingly similar. If you toss the coin often enough it is 100% certain you will eventually get exactly the same number of heads and tails. But it is also true that if you could toss it enough times you would eventually reach the stage where there are a hundred more heads than tails!

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10. Probability

Consider a deck of cards. There are four suits called hearts, diamonds, clubs and spades. In each of these there are 13 cards – an ace, the numbers 2 to 10 inclusive, and the picture cards: jack, queen and king. This makes a total of 52 cards in an ordinary deck. Some games require one extra card, the Joker. We will not use this extra card. If the cards are boxed (shuffled or mixed) and then 13 cards dealt to each of four people, the chances of a particular person getting 13 clubs are 635,013,559,600 to 1. This figure is achieved by asking how many different ways 13 cards can be picked from a 52 card deck. It is possible to pick approximately 365,013,559,600 different combinations of 13 cards from 52; this is the total possible number of outcomes. Only one of these outcomes contains 13 clubs. Therefore, using the definition from the last page: 1 P(13 clubs) = 365, 013, 559, 600 This will be explained in more detail in the next chapter. Example 2:

Consider a deck of cards. It consists of 52 cards:

Clubs

A

2

3

4

5

6

7

8

9

10

J

Q

K

Diamonds

A

2

3

4

5

6

7

8

9

10

J

Q

K

Hearts

A

2

3

4

5

6

7

8

9

10

J

Q

K

Spades

A

2

3

4

5

6

7

8

9

10

J

Q

K

A = Ace J = Jack Q = Queen K = King (a) The probability of drawing a club from a full deck: = P(Clubs) = (b)

13 52

1 ← number of clubs in deck = 4 ← total number of cards

The probability of drawing a 2 from a full deck: 4 ← number of 2s 1 P(2) = = 52 ← total number 13

Probabilities Multiply 26

1

Pick a card from a pack. The odds of it being a red card are 522 or 2 or 1 in 2. The 4

1

odds of it being a “9” are 52 or 13 or 1 in 13. The odds of the card being a red “9” 1

1

1

are 2 multiplied by 13 = 26 or 1 in 26. The following project illustrates the fact that probabilities multiply, work on it in small groups.

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PROJECT 10.2

The Golf Bet

Two golfers play 18 holes of golf. They place a small bet on the game. They agree to play the 1st hole for €1 and play “double or quits” on the remaining holes. If one golfer wins all 18 holes consecutively, how much will the other player owe him? See the solution at the end of the chapter.

Combining Probabilities - Sample Space If two events happen simultaneously, e.g. you throw two dice, a sample space can be constructed to see clearly the possible outcomes. A sample space involves putting all the possible outcomes of one event on one axis of a grid and all the possible outcomes of a second event on the other axis.

Sample Space When completing a sample Throw 1 space, the value of each individual die, e.g. 1 and 3, may 1 be put in the box or the total 3 + 1 1,1 1 = 4 may be used. Both systems are equally valid ways of 2 2,1 completing a sample space. This total number of events is 36 each time. Then count the number of the outcomes when both dice are the same . . . etc

Example 3:

Throw 2 2

3

4

5

6

1,2

1,3

1,4

1,5

1,6

2,2

2,3

2,4

2,5

2,6

3

3,1

3,2

3,3

3,4

3,5

3,6

4

4,1

4,2

4,3

4,4

4,5

4,6

5

5,1

5,2

5,3

5,4

5,5

5,6

6

6,1

6,2

6,3

6,4

6,5

6,6

Throw two ordinary die. Find the probability of (a)

both dice being the same =

(b)

getting two 4’s =

(c)

getting at least one 6 = 11 36

(d)

not getting a 6 =

(e)

getting a total of 10 =

98

6 1 = 36 6

1 36

25 36 3 1 = 36 12

10. Probability

PROJECT 10.3

Toss 2 Coins

Have every pupil in the class toss two coins 50 times. Add all the results together. The result should be very close to: 25% 2 heads 50% 1 head and 1 tail 25% 2 tails

The Famous Three Door Game Show Problem: Imagine there is a contestant on a tv game show. The contestant is faced with three closed doors and must choose to open one of them. Behind one door is a car and behind the other two doors are goats. If the contestant picks the door with the car they will win the car. The contestant picks one door, but then the game show host opens one of the other two doors to reveal a goat (the host knows where the car is, so is always able to open a door with a goat behind it). The host then asks the contestant if they want to change their mind and switch to the third door (the one they did not choose at first and the one the host did not open). Question: Should the contestant change their mind? See the solution on page 104.

Probability and God In the seventeenth century a famous mathematician called Blaise Pascal argued that the excitement a gambler feels when making a bet is equal to the amount he might win multiplied by the probability of winning it. In life the possible prize of eternal happiness has infinite value and the probability of entering heaven by leading a good life no matter how small is certainly finite. Therefore multiplying an infinite prize by a finite probability results in infinity – justifying a belief in God (according to Pascal)!

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Transition Year Maths

Three or More Events Occurring Simultaneously When more than two events occur the best method to solve problems is to write out in full the possible outcomes. Example 4:

Toss three coins together. What is the probability of getting two heads and one tail? Number of events with 2 heads & 1 tail As usual, P(2H & 1T) = Total number of possible outcomes

Possibe Outcomes: Coin One

Coin Two

Coin Three H

H

T H

H T

T H or H

T H

T T

T Therefore the possible outcomes ← are: 1 … HHH ← 2 … HHT 2 Heads & 1 Tail 3 … HTH ← 2 Heads & 1 Tail 4 … HTT 5 … THH 2 Heads & 1 Tail 6 … THT 7 … TTH 8 … TTT P(2 Heads & 1 Tail) =

3 8

100

}

3 of the 8 possible outcomes contain 2 Heads & 1 Tail

10. Probability

Chances of Two People in a Class Having the Same Birthday: Assume there are 366 different birthdays (including leap year birthdays). Start with one person, the probability of a second person having the same birthday is 1 in 366. The probability of the second person not having the same birthday is 365 in 366. The probability of a third person not matching either of the other two is 364 in 366. Probabilities multiply! Probability of people not having the same birthday: 2 people

365 = 0·997 366

3 people

364 365 × = 0·992 366 366

4 people

363 3 364 365 × × = 0·984 366 366 366

PROJECT 10.4

Probability: having same Birthday

Find when the probability of a group of people not having the same birthday becomes 0·5. Continue the calculations above, including 5 people, 6 people until the probability reaches 0·5. It should be between 22 and 23 people.

This means that in a class of 23 people, the chances are greater than 50 : 50 that two people will have the same birthday. For a class of 30 pupils there is a 70% chance of 2 people having the same birthday.

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Throw an ordinary 6-sided die (sides are 1, 2, 3, 4, 5, 6). (a)

Find the probability of getting a 1._________________________________

(b)

Find the probability of getting a 6._________________________________

(c)

Find the probability of getting an even number. _____________________

(d)

Find the probability of not getting a 6. _____________________________

(e)

Find the probability of getting a 2 or a 4. ___________________________

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2.

3.

Throw a backgammon 6-sided die (sides are 2, 4, 8, 16, 32, 64). (a)

Find the probability of getting a 2._________________________________

(b)

Find the probability of getting a 16.________________________________

(c)

Find the probability of getting a 64.________________________________

(d)

Find the probability of getting a 2 or 4._____________________________

(e)

Find the probability of getting an 8 or a 64. _________________________

Imagine you have an ordinary 52-piece deck of cards. Pick one card. Find the following probabilities.

4.

(a)

P(4 of clubs) ___________________________________________________

(b)

P(ace of hearts) ________________________________________________

(c)

P(even numbered spade) ________________________________________

(d)

P(club or spade) ________________________________________________

(e)

P(not a diamond)_______________________________________________

(f)

P(not a 7) _____________________________________________________

(g)

P(6 or 7) ______________________________________________________

(h)

P(J, Q or K) ____________________________________________________

(i)

P(4 or 8 of clubs) _______________________________________________

Toss a coin three times. If the first and second toss give heads, what are the chances of the third toss being a head? ____________________________________________________________________ ____________________________________________________________________

5.

Draw a sample space for throwing 2 backgammon dies (assuming sides are 2, 4, 8, 16, 32, 64). (a)

Find the probability of both dies being the same. ____________________

(b)

Find the probability of two 4’s. ____________________________________

(c)

Find the probability of at least one 8. ______________________________

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10. Probability

6.

(d)

Find the probability of at least one 6. ______________________________

(e)

Find the probability of no 8’s._____________________________________

(f)

Find the probability of no 6’s._____________________________________

(g)

Find the probability of a total of more than 50. ______________________

Toss two coins and record your result. Repeat 100 times. If the coins have been tossed fairly you should get very close to: 2 heads 25 times; 2 tails 25 times and one head and one tail 50 times. ____________________________________________________________________

7.

8.

Write out all the possible outcomes for tossing 3 coins together. Hence, find (a)

P(3 Heads) ____________________________________________________

(b)

P(2 Heads & 1 Tail) _____________________________________________

(c)

P(1 Head & 2 Tails) _____________________________________________

(d)

P(3 Tails) ______________________________________________________

(e)

P(4 Tails) ______________________________________________________

Write out all the possible outcomes for tossing 4 coins together. Hence find P(2 Head and 2 Tails). ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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9.

Using the digits 1 and 2 only, how many different 3-digit numbers can you make? Find the probability of one being the number 121. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

10. Using the digits 4 and 5 only, how many different 4-digit numbers can you make? Pick one of these numbers at random. What are the chances it has exactly two 5’s in it? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3-DOOR GAME SHOW PROBLEM Yes, she should change her mind because by doing so she increases her chance of winning the car! If she sticks to her original choice of door she has a 1 in 3 chance of winning the car and there is a 2 in 3 chance that the car is behind one of the two other doors. When the host opens one of those two other doors the 2 in 3 chance of the car being behind one of those two doors does not change so the 2 in 3 chance now applies to the unopened door. Therefore by changing she has a 2 in 3 chance of winning the car.

‘The Golf Bet’ Solution Golf Bet Project He will owe €131,072. After hole one he owes €1, then €2, the €4 then €8 etc. The answer can also be got from 217 . The 18th term of the sequence 1, 2, 4, 8, 16, 32 ... is 217 = 131,072. In general: the nth term of the sequence 1, 2, 4, 8, 16, 32 ... = 2n-1

104

11. Combinations and Permutations

11.

Combinations and Permutations

Permutations These are arrangements in which the order matters. Consider three letters a, b, c. How many arrangements of these three letters can be made using each once? There are six possible arrangements of three letters: abc acb bac bca cab cba = 6 permutations How many arrangements of two letters can be made from three letters? ab ac ba bc ca cb = 6 permutations How many arrangements of one letter can be made from three letters? a b c = 3 permutations This can be done mathematically. Example 1:

Example 2:

(i)

How many arrangements of three letters can be made from a, b, c (ii) How many arrangements of two letters can be made from a, b, c (iii) How many arrangements of one letter can be made from a, b, c (i)

3

P3 = 3 × 2 × 1 = 6 permutations

(ii)

3

P2 = 3 × 2 = 6 permutations

(iii)

3

P1 = 3 = 3 permutations

How many arrangements of five letters can be made from the letters in the word FIRST? 5

P5 = 5 × 4 × 3 × 2 × 1 = 120

Think of it like this: There are 5 possibilities for position 1, there are only 4 possibilities left for position 2 because one letter is taken already; and so on.

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Example 3:

How many arrangements of five letters can be made from the letters in the word FIRST if each can be used as often as you like? There are 5 possibilities for position 1. There are still 5 possibilities for position 2. There are still 5 possibilities for position 3 etc. Therefore there are 5 × 5 × 5 × 5 × 5 = 3,125 possible permutations of five things taking five at a time and taking each as often as you wish.

NOTE The calculator has a permutation button nPr . This assumes that each letter is only taken or used once.

Combinations These are groups of things where order does not matter. Consider three letters a, b, c. How many combinations of three letters can be made taking each once? There is only 1. abc = 1 combination How many combinations of two letters can be made from three letters? ab , ac, bc = 3 combinations How many combinations of one letter can be made from three letters? a, b, c = 3 combinations

This can be done mathematically: Example 4:

How many combinations of 3 letters can be made from a, b, c? (ii) How many combinations of 2 letters can be made from a, b, c? (iii) How many combinations of 1 letter can be made from a, b, c?

(i)

C3 =

3 × 2 ×1 = 1 combination 3 × 2 ×1

3

C2 =

3×2 6 = = 3 combinations 2 ×1 2

C1 =

3 = 3 combinations 1

(i)

3

(ii)

(iii)

3

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11. Combinations and Permutations

Example 5:

Fred has a voucher to pick any two of the top 10 PS4 games! How many different combinations of 2 games can he pick? C2 =

10

PROJECT 11.1

10 × 9 90 = = 45 2 ×1 2

Games – Permutations

The top ten PS4 games are called: A, B, C, D, E, F, G, H, I, J. Write down all the possible ways in which Fred can pick two of them! If Example 6 is correct it should be 45.

Example 6:

From 10 games, how many ways can: C0 = 1

0 games be picked =

10

1 game be picked =

10

2 games be picked =

10

3 games be picked =

10

4 games be picked =

10 0

5 games be picked =

10

6 games be picked =

10

7 games be picked =

10 0

8 games be picked =

10

9 games be picked =

10

C1 =

10 = 10 1

C2 =

10 × 9 90 = = 45 2 ×1 2

C3 =

10 × 9 × 8 = 120 3 × 2 ×1

C4 =

10 × 9 × 8 × 7 = 210 4 × 3 × 2 ×1

C5 =

10 × 9 × 8 × 7 × 6 = 252 5 × 4 × 3 × 2 ×1

C6 =

10 × 9 × 8 × 7 × 6 × 5 = 210 6 × 5 × 4 × 3 × 2 ×1

C7 =

10 × 9 × 8 × 7 × 6 × 5 × 4 = 120 7 × 6 × 5 × 4 × 3 × 2 ×1

C8 =

10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = 45 8 × 7 × 6 × 5 × 4 × 3 × 2 ×1

C9 =

10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 10 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 ×1

10 games be picked = 10C10 =

10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 =1 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

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Transition Year Maths

NOTE ON CALCULATOR USE Permutations: Use nPr button. To find 5 permutations from 7 things (assuming each can only be used once) press

7

nPr

5

= 2,520

Combinations: Use

nCr button. To find 5

combinations from 7 things press

7

nCr

5

= 21

NOTE ON CALCULATOR USE Permutations: If getting 7 permutations of 7 things, this is sometimes simply called 7! (7 factorial). Therefore 7

nPr 7

PROJECT 11.2

7 = 5,040

n! = 5,040 also

Pascal’s Triangle – first 11 lines

Pascal’s triangle is constructed by adding each two numbers and writing the sum below. Therefore the first five lines are: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Continue the triangle down for 11 lines. Compare the numbers in line 11 with the answer to Example 6 on page 107. What can you conclude?

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11. Combinations and Permutations

PROJECT 11.3

Pascal’s Triangle – first 20 lines

Write out the first 20 lines of Pascal’s Triangle. Then on a blank sheet draw a triangle with 20 lines marked on it and one space left for each of the numbers of Pascal’s Triangle. Instead of writing all the numbers just shade in the spaces which contain even numbers. Start like this:

1 1

1 2

1 1

6 10

5

1

3

3 4

1

1 1 4

1 5

10

1

When you have completed Chapter 13, Chaos, you should check this triangle. After completing 20 lines, you may recognise the shape!

Pascal’s Triangle arises in connection with the Binomial Theorem in algebra. It makes it easy to multiply out expressions like (x + y)4. 1 1 1 1 1

1 2

3 4

Example 6:

1 3

6

1 4

1

... ... ... ...

(x + y)1 (x + y)2 (x + y)3 (x + y)4

= = = =

x+y x2 + 2xy + y2 x3 + 3x2y1 + 3x1y2 + y3 x4 + 4x3y1 + 6x2y2 + 4x1y3 + y4

Expand (x + y)6 using Pascal’s Triangle. The relevant line of Pascal’s Triangle is 1, 6, 15, 20, 15, 6, 1. (x + y)6 = x6 + 6x5y1 + 15x4y2 + 20x3y3 + 15x2y4 + 6x1y5 + y6 This would take a long time to multiply out.

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Transition Year Maths

Example 7:

Expand (x + y)6 using the nCr button on a calculator. First just write x6y0

x5y1

x4y2

x3y3

x2y4

x1y5

x0y6

Notice that the power of x goes from 6 to 0 while y goes from 0 to 6. Next use nCr to put coefficients in front.

(x + y)6 =

C 0x 6 y 0 + 6C1x 5 y 1 + 6C 2x 4 y 2 + 6C 3x 3 y 3 + 6C 4x 2 y 4 + 6C 5x 1 y 5 + 6C 6x 0 y 6

6

= x 6 + 6x 5 y 1 + 15x 4 y 2 + 20x 3 y 3 + 15x 2 y 4 + 6x 1 y 5 + y 6

Example 8:

Expand (2x + 3y)4 = 4C (2x )4 + 4C (2x )3(3 y )1 + 4C (2x )2(3 y )2 + 4C (2x )1(3 y )3 + 4C (3 y )4 0 1 2 3 4 = 16x 4 + 96x 3 y + 216x 2 y 2 + 216xy 3 + 81y 4

Blaise Pascal: French mathematician, (1623–1662) Pascal studied advanced maths from a young age. He worked with Fermat on probability theory, a new branch of mathematics at the time. This was of great interest to the professional gamblers of the day. At the age of 19 he constructed a mechanical calculator to help his father calculate his tax bill. His greatest contribution to mathematics is his tabular presentation for binomial coefficients, now called Pascal’s Triangle. The unit of pressure in physics is the Pascal and there is also a programming language named after him.

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11. Combinations and Permutations

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Using the letters a, b, c, and d how many permutations can be formed of: (a) one letter _______________________________________________________ (b) two letters ______________________________________________________ (c) three letters _____________________________________________________ (d) four letters ______________________________________________________

2.

Using the letters a, b, c, and d, how many combinations can be formed of: (a) one letter _______________________________________________________ (b) two letters ______________________________________________________ (c) three letters _____________________________________________________ (d) four letters _____________________________________________________

3.

How many arrangements (permutations) of eight letters can be made from the letters in the word ‘HOSPITAL’? ____________________________________________________________________ ____________________________________________________________________

4.

If the letters could be used as often as you wish, how many eight letter permutations could be made from the word ‘HOSPITAL’? ____________________________________________________________________ ____________________________________________________________________

5.

In a class of 26 girls, how many different teams of 5 could be picked? ____________________________________________________________________ ____________________________________________________________________

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Transition Year Maths

6.

There are 24 boys in a class. They do a test in Maths, Irish and English. There is a prize for first place in each test: (a) How many possible different ways could the Maths prize be awarded? _______________________________________________________________ (b) How many possible different ways could all three prizes be awarded? _______________________________________________________________ (c) If the same boy cannot win more than one prize, how many different ways can the three prizes be awarded? _______________________________________________________________

7.

An exam has 10 questions. The students must do any six: (a) How many different groups of six questions can be chosen? _______________________________________________________________ (b) If question 1 must be done with any five other questions, how many different groups of six can now be chosen? _______________________________________________________________ (c) If students have an option of question 9 or question 10 along with any other five questions, how many groups of six questions can be chosen? _______________________________________________________________

8.

A deck of cards contains 13 hearts and 13 diamonds, which are red, and 13 spades and 13 clubs, which are black. How many ways can a ‘hand’ of five cards be dealt? (a) from the full deck of 52 cards? _____________________________________ (b) if it must contain all hearts? _______________________________________ (c) if it must contain no spades? _______________________________________ (d) if it must be all black?_____________________________________________ (e) if it contains the ace of clubs and any four other cards? ________________ (f)

if it cannot contain the ace of hearts? _______________________________

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11. Combinations and Permutations

(g) if it cannot contain any of the four aces?_____________________________ (h) if it must contain the 4, 5, 6 and 7 of clubs? __________________________

9.

Expand (a) (x + y)3 _________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (b) (x + 2)5 _________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (c) (x + 2y)4 ________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (d) (2x – 3y)3 _______________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (e) (1 + 2x)5 ________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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12.

Fermat’s Last Theorem (and other conjectures)

Pierre de Fermat was born in France in 1601 and was probably the world’s best amateur mathematician. He invented and proved many theorems in mathematics. A statement in maths is called a conjecture until it is proven.

Fermat’s Conjecture Fermat’s famous conjecture stated that for x3 + y3 = z3 or x4 + y4 = z4 or x5 + y5 = z5 or xn + yn = zn where n is any number greater than 2; there are no whole number solutions Fermat wrote in the margin of his copy of Arithmetica around 350 years ago, ‘To resolve a cube into the sum of two cubes, or a fourth power to the sum of two fourth powers, or in general any power higher than the second into two of the same kind, is impossible, of which fact I have found a remarkable proof. The margin is too small to contain it.’ All of Fermat’s other theorems have been verified years ago, but this one has proved very difficult. In 1908 Dr Paul Wolfskehl offered a prize of 100,000 Deutschmarks (worth about €1,000,000 today) to anyone who could prove Fermat’s last theorem before 2007. We have already looked at Pythagoras’s Theorem, x2 + y2 = z2. This has easy solutions like 32 + 42 = 52 or 52 + 122 = 132. It was proven 2,600 years ago. In the last 350 years thousands of mathematicians have spent years looking at x3 + y3 = z3. This equation looks simple, but it is very difficult to prove that it has no solution.

PROJECT 12.1

Whole Numbers & Equations

No one has ever found any whole numbers which work in the equation x3 + y3 = z3, but using numbers between 1 and 10 there is one solution which is only wrong by 1. Try to find it. x, y and z are three different numbers between 1 and 10, x3 + y3 = z3 – 1 Most people believe that if Fermat had a proof, it probably had an error in it. A man called Andrew Wiles finally proved the conjecture in 1994, after working on it for almost all of his adult life, often in secret. The proof takes up 130 pages of small print. Only a handful of people in the world would have any chance of understanding it or verifying that it is in fact true. Very few people even with a degree in mathematics could understand the first one of the 130 pages.

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12. Fermat’s Last Theorem (and Other Conjectures)

On 27 June 1997 Andrew Wiles collected the Wolfskehl prize which because of World War inflation was then worth only €40,000. He said, ‘having solved this problem there is a sense of loss, but at the same time there is a sense of freedom. I was so obsessed by this problem that for eight years I was thinking about it all the time, when I woke in the morning to when I went to sleep at night. That is a long time to think of one thing. That particular odyssey is now over. My mind is at rest.’

Prime Numbers A prime number is a natural number which has no proper factors. A prime number has two divisors, itself and 1. Therefore, 1 is not a prime number. Look at all the numbers between 1 and 20 inclusive: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 Remove 1 as it is not prime. Take out the numbers which are divisible by 2 (excluding 2 itself). This leaves: 2, 3, 5, 7, 9, 11, 13, 15, 17, 19 Take out the numbers which 3 divides into (excluding 3 itself), this leaves: 2, 3, 5, 7, 11, 13, 17, 19

Pierre de Fermat: French amateur mathematician, (1601–1665) Known as The Riddler, Fermat formed many theorems. Although he devoted much time to mathematics, he was not a professional mathematician. Instead, he preferred to send letters to his friends with his most recent theorem but he did not send any accompanying proof. In this way, he managed to receive fame as a mathematician without having to prove his theorems. Along with Pascal, he developed the basis for the theory of probability. One mystery which still troubles some mathematicians is ‘Did Fermat have a simple proof for his famous last theorem as he claimed?’ Most mathematicians doubt it but there are some who hope to find fame and glory by finding the original proof.

The numbers 4 divides into are already gone (it divides into 2). Then check 5 and so on. The full list of primes between 1 and 20 inclusive is: 2, 3, 5, 7, 11, 13, 17, 19

PROJECT 12.2

Prime Numbers

List all the prime numbers between 2 and 100 inclusive. There are 25 of them.

There is a lot of interest in prime numbers, especially to cryptographers. Codes based on multiplying large prime numbers together are very difficult to break. To some extent the success of computer-based financial transactions depends on the security provided by such codes.

PROJECT 12.3

Credit Card Security

What property of prime numbers makes them so important for ensuring the security of credit card numbers during online transactions? Working in small groups of 2-4, carry out some research on the internet and write a note on this topic (search using terms like ‘credit card’, ‘security’, and ‘prime numbers’).

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Transition Year Maths

Fermat loved mathematical challenges. He noted that 26 is between 25 (which is 52) and 27 (which is 33). See if you can find any other number sandwiched between a square and a cube? You will not find any. Fermat proved that no other number could have this property. This proof was just one of hundreds of proofs he completed.

PROJECT 12.4

Equations: Power of 4

Write out what 14, 24, 34, 44 and 54 are. Then try all the combinations to verify that none solve the equation, x4 + y4 + z4 = w4

PROJECT 12.5

Equations: Whole Numbers

Write out a value for 14, 24, 34, 44, 54, 64, 74, 84, 94 and 104. Then using a different whole number for x, y, z and w between 1 and 10 find a solution to x4 + y4 = z4 + w4 + 32

Euler’s Conjecture

Christian Goldbach: Prussian mathematician, (1690–1764) Goldbach was born in Kaliningrad, Russia. He worked closely with Leibniz and Euler on number theory and infinite series. He is most famous for his theory as discussed below. British publishers Faber & Faber offered $1,000,000 if anyone could prove the conjecture before April 2002, but the deadline passed and no one came forward with a proof. In a Spanish movie called La Habitacion de Fermat (2007) a young mathematician claims to have solved it. In real life the conjecture remains unsolved.

There was another conjecture called Euler’s Conjecture which stated that there were no solutions to the equation: x4 + y4 + z4 = w4 Euler’s Conjecture was never proved, but was assumed to be true because so many people tried so many numbers without success. Even with years of computer-sifting the lack of a counter example made it likely this conjecture was true. Then amazingly a man called Naom Elkies in 1988 found 2,682,4404 + 15,365,6394 + 18,796,7604 = 20,615,6734 You can check this to 9 places of decimals on your calculator.

Goldbach’s Conjecture Goldbach’s Conjecture was proposed by Christian Goldbach (1690 – 1764). When Andrew Wiles was in Dublin in 2003 he was asked if he was now working on Goldbach’s Conjecture – he did not reply. The conjecture states, ‘Every even number greater than 2 is the sum of two prime numbers’. Pick any even number. 20 can be written as the sum of 13 and 7; two prime numbers.

116

12. Fermat’s Last Theorem (and Other Conjectures)

PROJECT 12.6

Sum of Prime Numbers

Write down all the prime numbers between 1 and 50. Then verify that each even number between 2 and 50 can be written as the sum of two of these prime numbers.

The conjecture is over 250 years old and no one has succeeded in proving it. Proving conjectures like this is what some mathematicians live for. Stalin awarded a prize of 100,000 roubles to the Russian mathematician Ivan Matveyevich Vinogradov in 1914 for partially proving the conjecture.

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Using four different numbers x, y, z and w, all between 1 and 10 inclusive, find a solution to w3 = x3 + y3 + z3 ______________________________________________ ____________________________________________________________________

2.

No one has ever found a solution to x3 + y3 = z3. But there is a solution which is wrong by 2. Find the solution, using whole numbers between 1 and 10 inclusive, to x3 + y3 = z3 – 2. ____________________________________________________________________ ____________________________________________________________________

3.

What did Fermat discover about the number 26? ____________________________________________________________________ ____________________________________________________________________

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Transition Year Maths

4.

Using whole positive numbers between 1 and 10, find solutions to the following: (You may use the same number for different letters in these questions.) (a) x4 + y4 + z4 = w4 + 2 ________________________________________________ (b) 5(x4 + y4 + z4) = w4 + 10 ____________________________________________ (c) x4 + y4 + z4 + 13 = w4 _______________________________________________ (d) x4 + y4 + z4 + 30 = w4 _______________________________________________

5.

Twin primes are prime numbers which differ by 2, e.g. 5 and 7. How many sets of twin primes can you find between 1 and 100? ____________________________________________________________________ ____________________________________________________________________

6.

Evaluate 15, 25, 35, 45, 55, 65, 75, 85, 95 and 105. Then using different whole positive numbers between 1 and 10 solve: x5 + y5 + z5 + 407 = w5 ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

7.

Why are prime numbers becoming popular to study again? ____________________________________________________________________ ____________________________________________________________________

118

13. Chaos

Chaos

13.

Chaos, a new branch of mathematics, began in the 1960s. It explains things like the shape of a snowflake, Jupiter’s great red spot, heartbeats, brain waves, disease epidemics, traffic jams, the pattern of smoke rising or water falling from a tap. Much of the background work for chaos was done by people trying to predict the weather.

← chaos starts ← smooth flow

← smooth flow

← chaos starts

Tap

Kettle

The Weather In the 1950s, there was optimism about weather forecasting. It was hoped we would soon be able to predict the weather for months and years in advance, that aeroplanes would seed clouds to make rain and that scientists would even be able to stop rain. In 1961 a scientist named Edward Lorenz (1917 – 2008) put weather data into computer programs to predict the weather. Then he put the same data into the same programs but rounded a certain figure off to the fourth decimal place. He found the new weather prediction was almost the same for two to three days but then changed dramatically, caused by the effects of chaos. Small changes in the initial conditions of a system can lead to huge differences in the outcome (the weather a few days later in this case). His conclusion has been called the ‘butterfly effect’. This effect claims that if a butterfly flaps its wings in a garden in Brazil it will set in place a sequence of chaotic events which will affect the weather worldwide in a few days’ time, possibly causing a tornado in Texas. Lorenz immediately knew that the future of long-range weather forecasting was doomed.

119

Transition Year Maths

The Chaos Game Can a totally random event give rise to a regular pattern?

PROJECT 13.1

Sierpinski Gasket

We need a die. On sides 1 and 2 of the die stick an A . On sides 3 and 4 of the die stick a B . On sides 5 and 6 of the die stick a C . 1

When the die is thrown, side A has a 3 probability of coming up. 1

Side B has a probability of 3 also 1

Side C has a probability of 3 also Draw an equilateral triangle. Label the corners A, B, and C. Draw a dot anywhere in the triangle. Throw the die, note the letter, move the dot half-way towards the corresponding vertex, getting a new point. Plot this point. Throw the die again and repeat the procedure. Continue the procedure moving half way from each new point to the vertex as indicated by the die. Do this 100 times carefully and note the pattern. Each time, the next point is plotted half-way from the previous point B and the vertex indicated by the die.

A

Result: if it is done 1,000,000 times by computer a very regular patter is obtained: The strange shape is the Sierpinski Gasket. This shape was considered by most mathematicians as belonging to ‘The Gallery of Monsters’ – shapes which were very confusing and useless. Now they are considered very useful to model irregularities of nature.

120

C

13. Chaos

The Fish in the Pond Problem The greatest success a mathematician can have is if he/she can match a real physical situation to a mathematical formula. Einstein achieved this with E = mc2 connecting energy with mass and the speed of light. Biologists, trying to work out how real fish in a real pond will occupy the pond, had to find a function which matched the reality of life, hunger, competition, overpopulation etc. in the pond. Consider a pond. The number of pairs of fish in the pond can vary between 0 (extinction) and 1 (pond is full of fish). A population of 0·1 means the pond is 10% full. A population of 0·5 means the pond is ½ full. The number of fish in the pond can be modelled by the formula: xnext year where x xnext year r

= r x(1–x) = number of fish in the pond this year = number of fish in the pond next year = rate of increase (reproductive rate)

Edward Lorenz: American mathematician and meteorologist, (1917–2008) Lorenz was a meteorologist. His work on weather forecasting led him to accidentally discover the theory of chaos. In 1972 he published a paper entitled Predictability: Does the Flap of a Butterfly’s Wings in Brazil Set off a Tornado in Texas? The Chaos Theory was one of the major scientific discoveries of the twentieth century.

Start with the pond 1% full, therefore x = 0·01 Pick a rate of increase, r = 1·2 and investigate what happens. Today Year 1 Year 2

x = 0·01 . . . . . pond is 1% full x next year = 1·2 (0·01)(1 – 0·01) = 0·01188 . . . pond is 1·18% full x next year = 1·2 (0·01188) (1 – 0·01188) = 0·01409 . . . pond is 1·4% full 0·01667 0·01967 0·02313 0·02712 0·03166 0·03679 0·04252 0·04886 0·05577 0·06319 0·07103 settles at 0·1666 . . . 16·666% full

121

Transition Year Maths

NOTE This calculation may be done easily on most calculators:

=

Ans Clear

Set starting pop. ·01

=

Ans Clear

Ans

(

1

1.

Clear memory

2. 3.

0

= 1·2 Ans

–

Ans

)

Press S ⇔ D to give decimals. Just keep hitting

=

button to get the population.

Expected Results: If rate = 0·8

rate = 2·6

rate = 3·4

rate = 3·9

Time (years)

Time

Time

Time

1 Population

pond full

0 pond empty

• •

• • •

If the rate of increase is 0·8 the population rapidly approaches 0 and the fish become extinct! If the rate of increase is 2·6 the population increases in the pond to about 0·65 and then oscillates for a few years, settling on a fixed population of 0·615 (61·5% full). It is this final population (or populations for oscillating systems) which is plotted at the end of the project. If the rate of increase is 3·4 the population increases to 0·84, drops to 0·45 next year, back to 0·84, back to 0·45. Population never settles but oscillates between 0·84 (84% full) and 0·45 (45% full) forever. If the rate is 3·5 population oscillates from 0·39 to 0·83 to 0·49 to 0·87 every four years forever! If the rate is 3·9 population is chaotic. It never settles and randomly goes up and down irregularly forever.

Therefore the rate of increase determines the eventual outcome of this system governed by the equation xnext = r(x)(1–x).

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13. Chaos

PROJECT 13.2

Fish in the Pond

Give each pupil in the class a different rate of increase for r and ask them to investigate what happens to the population of fish in the pond over 20 years. Use the calculator method provided on page 108 to do the project quickly. You can use the following rates of increase: r = 1·0, 1·5, 2·0, 2·5, 2·7, 2·9, 3, 3·2, 3·3, 3·4, 3·5, 3·6, 3·7, 3·8, 3·9 and 4. The data is collected and plotted on a separate graph with rate from 0 to 4 on the x-axis, and population from 0 to 1 on the y-axis. This is also a great project to do with a spreadsheet program like Microsoft Excel. A file to aid with calculations and graphing is included in the Teachers Resource CD.

Patterns in Nature Everything in nature follows some sort of pattern. The most common are: (a) Stable Systems Example: when a football is kicked it will always end up on the ground. (b) Random Systems Example: the number of millimetres of rainfall in Cork every year. A plot over 200 years shows no pattern. It is chaotic. (c) Oscillators Example: tides: a plot of the height of water in Dublin Bay for every hour for a year would look highly regular. It oscillates between two points. The fish in the pond problem can be any of these depending on the reproductive rate (rate of increase). • If the rate is less than about 3, it’s stable. • If the rate is between 3 and 3·5 it oscillates. • If the rate is greater than 3·7 it is random (or chaotic) usually. You can use your calculator to investigate systems. What happens to the system x2 – 1 if you start with 0·5, over 20 iterations. Most calculators will automatically enter the answer of the previous calculation into the next calculation. If your calculator does not do this, the following exercise will be slow. Enter:

Example 1:

0

·

5

=

Ans Clear Then Ans x2

–

1

=

=

=

=

Just keep repeating this 20 times. Note the number each time. Result: –0·75, –0·44, –0·81, –0·35, –0·88, –0·26, –0·95, –0·10, –0·99, –0·2 0, –1, –0, –1, –0, –1, –0, –1, –0, –1 The system is an oscillator. When it settles, it jumps from – 0 to –1 forever. There is much debate in mathematical circles as to how important the chaos theory is. Does it adequately describe apparently disordered dynamical systems in nature? Is it really as important as calculus or relativity as a mathematical tool? Only time can answer these questions.

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Transition Year Maths

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Investigate the system x2 – 1. Begin with 0·3 over 20 iterations. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

2.

Investigate x2 – 1. Begin with 2. How many iterations will your calculator do? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3.

Investigate x2 – 2. Begin with 0·4 over 20 iterations. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

4.

(a) Investigate x2. Begin with 0·5 over 10 iterations. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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13. Chaos

(b) Investigate x3. Begin with 0·5 over 10 iterations. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ (c) Investigate x –1. Begin with 0·5 over 10 iterations. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

5.

Which of the following are chaotic, which are stable and which are oscillators? (a) Orbit of a comet _________________________________________________ (b) Share price of a stock on stock market taken daily ____________________ (c) Human heartbeat taken every second _______________________________ (d) The price of milk taken every minute for one day_____________________ (e) The amount of daylight taken every day for 10 years __________________ (f)

Ice Ages ________________________________________________________

(g) Evolution _______________________________________________________ (h) Hour hand on clock distance (hand moves every hour) _______________ (i)

6.

Tides __________________________________________________________

Take a number, e.g. 32. Halve it and halve that and halve that and so on. This gives a list of numbers called the orbit of 32. 1 1 1 , , ... 2 4 8 Find the orbit of 1,000, using ‘halve it’ as the iteration. Orbit of 32 = 16, 8, 4, 2, 1,

____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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Transition Year Maths

*** Refer to note on calculator use on page 123 for Questions 7, 8 and 9 ***

7.

Use the iteration 4x(1–x). Begin with x = 0·02. Write out the first 20 terms. Is the system which this iteration describes – stable, chaotic or an oscillator? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

8.

Use the iteration 3x(1–x). Begin with x = 0·02. Write out the first 20 terms. Is the system which this iteration describes – stable, chaotic or an oscillator? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

9.

Use the iteration 2x(1–x). Begin with x = 0·02. Write out the first 20 terms. Is the system which this iteration describes – stable, chaotic or an oscillator? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

126

14. Complex Numbers

14. Complex Numbers Mathematicians have a concept called completeness. It is the need to be able to answer every single question. Historically it is this need for completeness which led the Hindus to discover negative numbers. Later the Greeks developed the idea of irrational numbers. Then during the sixteenth century, an Italian called Rafaello Bombelli came up with the question, ‘If the square root of +1 is both +1 and –1, then what is the square root of –1? He answered the question himself and declared that –1 = i, an imaginary number. In your own life you develop through all these stages of number. N:

Natural numbers – these are whole positive numbers. These are the first numbers people understand, e.g. you are three years old and fighting with your little sister because she has three sweets and you have only two!

Z:

Integers – positive and negative whole numbers. Later in life, perhaps when you are five years old you understand the idea of minus numbers. You may have five sweets but you owe your friend two sweets, so you realise that in fact, you really only have three sweets, 5 – 2 = 3.

R:

Real Numbers – all numbers on the number line. Later again in life you realise that there are fractions and decimals. You may divide a bar of chocolate with eight squares and give your brother three squares and keep five for yourself. He gets 3 of the bar and you get 5 of it. These numbers are real numbers. 8

8

Other real numbers include: 44 1 , –3 1 , 2·114, –3·49, 3 , 6, 9 × 1015, etc. 4

4

this is probably where you are at now!

PROJECT 14.1

Not Real Numbers?

Can you think of any number which does not fall into the category of real numbers? Discuss this problem in small groups of 2-4

Complex Numbers These are numbers with a real and an imaginary part. They are written as a + ib where a and b are real numbers.

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Transition Year Maths

Imaginary Numbers These numbers were discovered by European mathematicians in the sixteenth century. The problem began by asking ‘what is the square root of minus one?’ 1 = 1 since (1)2 = 1 and

1 = –1 since (–1)2 = 1

but

–1 = ?, the problem seems to be impossible!

The solution cannot be +1 or –1 since the square of both these is 1. Bombelli created the number i, an imaginary number. –1 = i The most important property of this number is the fact that i 2 = –1 Examples of complex numbers are: 2 + 3i 4 – 2i 0 + 8i 6 + 0i etc. All ordinary mathematical processes can be carried out on these numbers. The development of numbers through history and the understanding of numbers through growing up are similar. A person who gives up on fractions is stuck in a point in history thousands of years ago as far as the mathematical development of the brain is concerned. A person who is familiar with complex numbers is almost up to date with the evolution of maths. There will most likely never be an end. The universe appears to be infinitely complex. Mathematicians today are working on numbers like a + bj where a, b are both themselves complex numbers. The search for completeness shows no sign of an end. 1. Addition:

Simply add the real parts, then add the imaginary parts separately. 3 + 5i + 6 + 2i 9 + 7i Therefore (3 + 5i) + (6 + 2i) = (9 + 7i)

2. Subtraction:

The best rule is to ‘change the sign on the lower line and add’. 8 − 3i 3 + 2i minus −add

⇒ add

8 − 3i 3 − 2i

11 − 5i Therefore (8 – 3i) – (–3 + 2i) = (11 – 5i)

128

14. Complex Numbers 77++ 22ii 2 − 2i

minus

7 + 2i 7 + ⇒ add –2 + 2i 5 + 4i

3. Multiplication: Each part of the first complex number must be multiplied by each part of the second complex number. Remember i 2 = –1 (i) 4(3 + 2i) = 12 + 8i (ii) 4i(3 + 2i) = 12i + 8i 2 = 12i + 8(–1) = 12i – 8 = – 8 + 12i (iii) (3 + 2i)(4 + 5i) = 12 + 8i + 15i + 10i 2 = 12 + 23i + 10(–1) = 12 + 23i – 10 = 2 + 23i

Division To divide complex numbers we need the concept of the complex conjugate. The conjugate of a complex number is the same number with the sign of the imaginary part changed. The conjugate of 5 + 3i is 5 – 3i. To divide we multiply the top and bottom by the conjugate of the bottom. 5 + 2i 5 + 2i 3 − 4i × = 3 + 4i 3 + 4i 3 − 4i =

15 – 20i + 6i – 8i 2 9 − 12i + 12i − 16i 2

=

15 – 14i – 8(−1) 9 − 16(−1)

=

15 – 14i + 8 9 + 16

=

23 – 14i 25

129

Transition Year Maths Let us consider a question where a complex number appears: Example 1:

Case (i) Real solutions A farmer has 100 m of fence to surround a small vegetable plot. The farmer wants to enclose a rectangular area of 400 m2. How long and wide should it be? x = length y = width x 2 length + 2 width = 100 m 2x + 2y = 100 x + y = 50 y = 50 – x 50 – x ∴ length = x width = 50 – x The area = length by width 400 = x (50 – x) 400 = 50x – x 2 x 2 – 50x + 400 = 0 Solve this equation using the formula for quadratic equations. a = 1, b = –50, c = +400 –b ± b 2 – 4ac 50 ± 2500 – 1600 50 ± 30 = = 2a 2 2

So the length can be 40 m or 10 m. Check this works: 40 + 40 + 10 + 10 = 100 and 40 × 10 = 400

Rafael Bombelli: Italian mathematician, (1526–1572) Bombelli was the first mathematician to successfully deal with the problem of the square root of minus numbers. Before he introduced the letter i for the square root of minus one, these numbers were simply ignored. His book, Algebra, is one of the most remarkable achievements of sixteenth-century mathematics. He discovered complex numbers. The lunar crater ‘Bombelli’ is named after him.

130

14. Complex Numbers

Example 1:

Case (ii) Imaginary solutions A farmer has 100 m of fence to surround a small vegetable plot. The farmer wants to enclose a rectangular area of 650 m2. How long and wide should it be? As with case (i) the area = length × width ⇒ 650 = x(50 – x) ⇒ 650 = 50x – x2 ⇒ x2 – 50x + 650 = 0 a = 1, b = –50, c = +650 –b ± b 2 – 4ac 50 ± 2500 – 2600 = 2a 2 50 ± –100 = 2 50 ± 100 × –1 = 2 50 ± 10i = 2 = 25 ± 5i

The length of the enclosure can be 25 + 5i or 25 – 5i. There are no real solutions! But there are solutions, imaginary ones! Why is this? There are no real solutions because in our normal 3-Dimensional classical view of earth it is impossible to enclose 650 m2 with 100 m of fencing. The best that can be done is to have the length of each side 25 m, a square. This will enclose 25 × 25 = 625 m2. What of our solution, the length can be 25 + 5i metres or 25 – 5i metres. What does this mean? Our thinking has not evolved far enough yet to understand this! These solutions could be a window to a fourth dimension, or related to relativity. If the fence is going fast enough, could it enclose more space? The possibilities are endless. What is true is that just because we cannot understand it does not mean it is nonsense.

131

–1 = i

William Hamilton: Irish mathematician (1805–1865) Hamilton’s great contribution to mathematics is the discovery of quaternions. He was looking for ways of extending complex numbers (which can be viewed as points on a two-dimension plane) to higher dimensions. Whilst walking along the Royal Canal in Dublin with his wife he discovered a solution to fourdimension complex numbers. The formula he discovered is: i 2 = j 2 = k 2 = ijk = –1 The Central Bank of Ireland issued a commemorative €10 coin in 2005 to honour the 200th anniversary of his birth.

Transition Year Maths

The Argand Diagram It is clear that some method of visualising complex numbers would be useful. The German mathematician Carl Fredrich Gauss (1777 – 1855) proposed the Argand diagram. This has the real numbers on the x-axis and the imaginary ones on the y-axis. All complex numbers can be plotted and are usually called z1, z2 etc. Example 2:

Draw an Argand Diagram and plot z1(2 + 4i); z2(–2 –4i); z3(4 + 0i); z4(–3 + 0i); z5(–3 + 2i)

Im

z1

4 3

z5

2 1

z4 -3

-2

-1

z3 1

2

3

Re

-1 -2 z2

-3 -4

The modulus of a complex number |z| is the distance from the point to the origin. z = x 2 + y2

Example 3:

z1 = 2 + 4i = 22 + 4 2 = 4 + 16 = 20 z3 = 4 + 0i = 4 2 + 02 = 16 = 4

132

14. Complex Numbers

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

(a) Add (3 + 5i) and (2 + 2i) __________________________________________ _______________________________________________________________ (b) Add (3 + 5i) and (2 – 2i) __________________________________________ _______________________________________________________________ (c) Add (–3 + 5i) and (2 – 5i) _________________________________________ _______________________________________________________________

2.

(a) Take (2 + 2i) from (3 + 5i) ________________________________________ _______________________________________________________________ (b) Take (2 + 2i) from (3 – 5i)_________________________________________ _______________________________________________________________ (c) Take (–2 – 5i) from (–4 – 3i)_______________________________________ _______________________________________________________________

3.

(a) Multiply (3 + 2i) by 3 _____________________________________________ _______________________________________________________________ (b) Multiply (3 + 2i) by 3i_____________________________________________ _______________________________________________________________ (c) Multiply (3 + 2i) by 3 + 3i _________________________________________ _______________________________________________________________ (d) Multiply (3 – 2i) by (5 – 4i) ________________________________________ _______________________________________________________________ (e) Multiply (3 – 2i) by (4 – 2i) by (1 + i) _______________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

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Transition Year Maths

4.

(a) Divide (3 + 6i) by 3 _______________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ (b) Divide (6 – 3i) by 3 ______________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ (c) Divide (3 + 6i) by (3 + 3i) _________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ (d) Divide (4 + 10i) by (8 – 2i) _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ (e) Evaluate

– 6 – 8i 2 + 3i

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

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14. Complex Numbers

5.

Evaluate

(2 + i) (3 – i) (6 + 3i)(2 – 4i)

____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

6.

Show that 2 – 3i is a root of x2 – 4x + 13 = 0. Work out what the other root is. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

7.

z1 = 1 + 2i, z 2 = –3 + 3i, z 3 = –2 + 0i (a) Find |z 1| ________________________________________________________ _______________________________________________________________ (b) Find |z 2| ________________________________________________________ _______________________________________________________________ (c) Find |z 3| ________________________________________________________ _______________________________________________________________

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Transition Year Maths

(d) Let z4 = z 1 + z 2. Find |z 4| ___________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ z1 . Find |z 5| ____________________________________________ z2 _______________________________________________________________

(e) Let z 5 =

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ (f)

Let z 6 = z 2 × z 3. Find |z 6| __________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(g) Let z 7 =

z 1+ z 2 . Find |z 7| _________________________________________ z3

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

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14. Complex Numbers

(h) Plot z 1, z 2, z 3, z 4, z 5, z 6, and z 7 on an Argand Diagram.

8.

z 1 = 2 + 3i, z 2 = 4 – i, z 3 = z 1 + z 2 (a) Calculate z 3 _____________________________________________________ _______________________________________________________________ _______________________________________________________________ (b) Plot z 1, z 2 and z 3 on an Argand Diagram. What do you notice about the three points?

137

Transition Year Maths

Calculus

15.

Calculus is the ability to calculate the rate of change, known as the derivative, of one quantity with respect to another. Sir Isaac Newton (1642–1727) and Gottfried Leibnitz (1646–1716) discovered calculus in the seventeenth century. It is one of the new branches of mathematics. Newton is accepted as one of the greatest minds in the history of man. He discovered much of the maths and physics we still use today. But he also spent time thinking about what happens after death and the possibility of eternal life. As a result he spent a large part of his life studying alchemy (a form of chemistry). A small part of alchemy is trying to change mercury to gold, the main part is trying to find an elixir to prolong life. There is an Irish connection to the story of calculus. One mile outside Thomastown in County Kilkenny on a bend of the River Nore is Dysart Castle and Church. Bishop George Berkeley was born there in 1685. He opposed the new maths of calculus on the grounds that the small increments used were infinitely small. If they were zero, the whole grounds on which calculus is based is flawed. The connection between infinitely small and zero is still not resolved. Therefore Berkeley’s opposition to calculus is still valid but it is ignored because calculus is so useful. The reality is that calculus works for engineering and physics applications. Differentiation is written

dx Change in x = dt Change in t

RULE FOR SIMPLE DIFFERENTIATION 1. 2. 3.

Look at the number in front of the t Multiply this number by the power of t Reduce the power of t by 1

138

15. Calculus

Example 1:

dx = 10t dt

(a)

x = 5t 2, differentiate

(b)

x = 3t, differentiate

(c)

x = 5t 2 + 3t, differentiate

(d)

x = 6t 4 – 2t 3 + 4t 2 + 3t – 2, dx differentiate = 24t 3 – 6t 2 + 8t + 3 dt

(e)

x = 6t -2,

dx =3 dt dx = 10t + 3 dt

dx = –12t –3 dt

We will look at calculus from the point of view of the motion of a car moving to the right: x is the distance moved. Velocity (v) is defined as the

change in distance dx = time dt

Differentiate distance with respect to time gives velocity. Acceleration (a) is defined as the

change in velocity dv = time dt

Differentiate velocity with respect to time gives acceleration. In summary if motion is defined in terms of time (t) then x is the distance moved. Differentiate once to get velocity (v) Differentiate twice to get acceleration (a)

Bishop George Berkeley: Irish philosopher and mathematician, (1685–1753) Berkeley was born in Dysart Castle in County Kilkenny. In 1734 he published The Analyst which was an attack on the principles of calculus, especially the concept of infinitesimal change. As a result the foundations of calculus were rewritten in a much more rigorous form using limits. Berkeley felt that calculus distanced God from his worshippers. The university and city of Berkeley in America were named after him.

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Transition Year Maths

Example 2:

The manufacturers of a small rocket claim its motion is given by x = 3t 2 + 2t. Therefore the distance x it will move in metres can be found by putting the time t into the formula. After three seconds it will move 3(3)2 + 2(3) = 33 metres. Using calculus we can find its velocity or acceleration at any time t. Motion is defined by

x = 3t 2 + 2t

Velocity is defined by

v=

dx = 6t + 2 dt

Acceleration is defined by

a=

dv =6 dt

Time (t)

Distance (x)

Velocity (v)

Acceleration (a)

t 0 sec 1 sec 2 sec 3 sec

3t 2 + 2t 0m 5m 16 m 33 m

6t + 2 2 m/s 8 m/s 14 m/s 20 m/s

6 6 m/s2 6 m/s2 6 m/s2 6 m/s2

Examine this to see if it is true! • If a car moves 16 m in 2 seconds, its average velocity should be 8 m per sec, and it is 8 m/s after 1 sec, half way between 0 sec and 2 sec. • If a car moves from 5 m to 33 m in the 2 sec from 1 sec to 3 sec, the average velocity should be 33 – 5 28 14 m/s, and it is after 2 sec. = = 2 2 •

If velocity increases from 8 m/s after 1 sec to 14 m/s after 2 sec it is accelerating at 6 m/s per sec ⇒ acceleration = 6 m/s2.

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Car moving to right, motion defined by x = 20t

Example 3: Start t=

0 Sec

x = 0m

1 sec

2 sec

3 sec

4 sec

5 sec

10m 20m 30m 40m 50m 60m 70m 80m 90m 100m

Time (t) Distance (x) t x = 20t 0 sec 1 sec 2 sec 3 sec

0m 20 m 40 m 60 m

Velocity (v) Acceleration (x) v = 20 a=0 0 m/s2 0 m/s2 0 m/s2 0 m/s2

20 m/s 20 m/s 20 m/s 20 m/s

When t = 1 sec, x = 20(1) = 20m (after 1 sec car has moved 20 m) When t = 2 sec, x = 20(2) = 40m (after 2 sec car has moved 40 m) dx Differentiate once to get velocity, therefore v = = 20 dt ⇒ v = 20 m/s at all times Differentiate twice to get acceleration = a=

Velocity Time

dv = 0 ⇒ a = 0 at all times. dt

Differentiation of Products and Quotients The function to be differentiated is not always as simple as the ones we have been doing so far.

THE PRODUCT RULE If the function is the product of two factors Derivative first factor by derivative of second factor = of Product + second factor by derivative of first factor

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Example 4:

x = (t 2 – 3)(2 – 3t 3) dx (t 2 – 3)(–9t 2) + (2 – 3t 3)(2t) = dt = –9t 4 + 27t 2 + 4t – 6t 4 = –15t 4 + 27t 2 + 4t

THE QUOTIENT RULE If the function is one factor over the second factor Derivative bottom × derivative of top – top by derivative of bottom = of Quotient (bottom)2

Example 5:

x=

2t + 5 t2 +1

dx (t 2 + 1)(2)–((2t + 5)(2t ) = dt (t 2 + 1)2

Questions 1.

=

2t 2 + 2 – 4t 2 − 10t (t 2 + 1)2

=

–2t 2 − 10t + 2 (t 2 + 1)2

The following Questions pages are FREE to download from mentorbooks.ie/resources

dx of the following: dt (a) x = 10t 3 _________________________________________________

Find

(b) x = 10t 2

_________________________________________________

(c) x = 10t

_________________________________________________

(d) x = 10

_________________________________________________

(e) x = 4t 2 + 3t + 2

_________________________________________________

(f)

x = 3t –2 + 4t –3

_________________________________________________

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2.

The motion of a rocket is given by x = 200t 2 + 50t. Differentiate once to find an expression for velocity, differentiate twice to find an expression for acceleration. Find the distance travelled, the velocity and the acceleration after 0, 1, 2 and 3 seconds to complete the table below. Time (t)

Distance (x)

Velocity (v)

Acceleration (a)

0 sec 1 sec 2 sec 3 sec

____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3.

If differentiating distance once gives velocity and twice gives acceleration, what might differentiating a third time give? ________________________________________ ____________________________________________________________________

4.

t +3 t2 +1 Find the distance moved after three seconds and the velocity after three seconds

The distance moved in terms of time, x =

(note: it moves backwards) _____________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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5.

Differentiate with respect to t, to find an expression for velocity. Then find the distance moved and velocity after one second: (a)

x = (2t + 5)(4t + 7) _______________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(b)

x = (2t 2 – t + 1)(5t 2 – 3t – 1) ________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(c)

x=

3t _______________________________________________________ t +2 2

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ (d)

x=

4t _______________________________________________________ t +3 2

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

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15. Calculus

(e)

t2 − 2 ______________________________________________________ 2t 2 + 1 _______________________________________________________________

x=

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ (f)

x = 12t 6 + 5t 5 + 4t 4 + 10t 3 − 2t 2 + t + 8 ________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

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16. Interesting Numbers There are many types of interesting numbers, some of which have very practical uses, like the serial number on euro bank notes and ISBN number on books. There are numbers like pi and e which will prove useful in solving maths problems. Finally there are many numbers, some recently named, which are simply interesting, but may also be of use in solving problems:

Digital Root The Digital Root of a non negative integer is calculated by adding all the digits of the number. The digits of this number are then added. This process is repeated until it results in a single-digit.

Example 1:

PROJECT 16.1

Find the digital root of 69,794,698 6 + 9 +7 + 9 + 4 + 6 + 9 + 8 = 58 5 + 8 = 13 1+3=4 The digital root is 4

Euro Bank Notes

All euro bank notes have a serial number which consists of a letter followed by eleven numbers. The letter identifies the country which issued the note (this may not be the country where it was printed). The remaining eleven numbers, when their digital root is calculated gives a check sum also particular to that country. Check that the letter matches the digital root on as many notes as you can.

NOTE Irish notes: code is T, digital root 6 German notes: code is X, digital root 2 French notes: code is U, digital root 5 You can look up the other codes and digital roots on the internet.

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16. Interesting Numbers

PROJECT 16.2

ISBN Numbers

ISBN numbers are international standard book numbers. Use the internet to find out what the different parts of the number on this book and your other textbooks can tell you. The ISBN number can usually be found above the bar code on the back cover of a book.

Narcissistic Numbers These are numbers which equal the sum of their digits raised to the power of the number of digits. 153 is a narcissistic number because it is a 3 digit number which is the sum of the cubes of its digits: 153 = 13 + 53 + 33

PROJECT 16.3

Narcissistic Numbers

Try to find three more 3 digit narcissistic numbers.

PROJECT 16.4

Other Number Types

Use the internet to find out what these numbers are, and give an example of each: Triangular numbers Hexagonal numbers Friedman numbers Weird numbers Strange numbers

Factorials These are straightforward numbers which you will need at senior cycle level. They are denoted by an exclamation mark: n! (n factorial) is the product of all the integers less than or equal to n

Example:

Calculate 7! 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040

They do have some interesting features such as: 0! = 1 It may seem strange that multiplying no numbers together gives an answer of 1. It has to equal 1 or else a lot of other equations and series would not work. For example, the sequence you used in this chapter to calculate e actually begins with 1 . If 0! did not equal 1, this will give a 0! wrong answer for the value of e. 70! is just a bit larger than a googol, the large number mentioned on page 7 of this book. Brown numbers: these are pairs of integers (x, y) which satisfy the condition y! + 1 = x2

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PROJECT 16.5

Brown Numbers

There is a conjecture which states that there are only 3 such pairs of brown numbers (5, 4) (11, 5) and (71,7). Use your calculator to verify that these are brown numbers.

The Number e – An Introduction e is an irrational number. It appears in calculations where values increase exponentially and continually. It helps create the formulas for exponential systems, like the growth of bacteria, the growth of money in a compound interest account or radioactive decay. e1 = 2.71828182845904523

PROJECT 16.6

Calculating e

1 1 1 1 + +… e1 = 1+ + + 1 1× 2 1× 2 × 3 1× 2 × 3 × 4 Write out the next 10 terms of this sequence. Calculate the sum of all the terms and see how close you get to the value of e above.

PROJECT 16.7

Natural Exponential Function

Draw a graph of the function y = ex for x = -2, -1.5, -1, -.5, 1, 1.5, 2, 2.5. Put the x values on the x -axis. The shape you get is called the natural exponential function. When a person gets an infection, bacteria grow at a rapidly increasing rate like this, until the immune system kicks in and prevents further infection.

Divisors Divisors are all the numbers (including the number 1) which divide evenly into a given number, but in the case of friendly and sociable numbers, not including the number itself. Therefore, the divisors of 24 are 1, 2, 3, 4, 6, 8 and 12. They sum to 36.

Friendly Numbers These are pairs of numbers such that each number is the sum of the divisors of the other number. Talismen (good luck charms) sold in the Middle Ages were often

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16. Interesting Numbers

inscribed with these numbers on the grounds that they would promote love. An Arabian mathematician claimed that people could write one of the pair of numbers on one fruit and eat it, then write the second number on another fruit and give it to a lover as a mathematical aphrodisiac! There was only one pair discovered until Pierre de Fermat discovered the pair 17,296 and 18,416 in 1636. Descartes discovered the pair 9,363,584 and 9,437,056 in 1638. A 16-year-old Italian, Nicolo Paganini, found 1,184 and 1,210 in 1867. Computers can be programmed to find larger ones now!

Remember: When dealing with friendly and sociable numbers, the number itself is not considered a divisor. Example 1:

Verify that 1,184 and 1,210 are, in fact, friendly using a calculator. Start by dividing 1,184 by 2, then by 3 etc. The divisors of 1,184 are 1, 2, 4, 8, 16, 32, 37, 74, 148, 296, 592. Sum = 1,210. The divisors of 1,210 are 1, 2, 5, 10, 11, 22, 55, 110, 121, 242, 605. Sum = 1,184.

Example 2: Divisors of

Prove there are no friendly pairs between 10 and 20. 10 = 1, 2, 5, sum = 8 11 = 1 sum = 1 There are no friendly pairs here. 12 = 1, 2, 3, 4, 6 sum = 16 The sum of 12’s divisors = 16 13 = 1 sum = 1 But the sum of 16’s divisors ≠ 12 14 = 1, 2, 7 sum = 10 The sum of 14’s divisors = 10 15 = 1, 3, 5 sum = 9 But the sum of 10’s divisors ≠ 14 16 = 1, 2, 4, 8 sum = 15 The sum of 16’s divisors = 15 17 = 1 sum = 1 But the sum of 15 divisors ≠16 18 = 1, 2, 3, 6, 9 sum = 21 20 = 1, 2, 4, 5, 10 sum = 22

}

PROJECT 16.8

Find the Friendly Pair

Pythagoras and his followers found one friendly pair over 2,000 years ago. Both numbers are between 200 and 300. Work in small groups of 2 - 4 to try to find that pair. You will experience, in a small way, what it is like to discover something. (See hint on page 155.)

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Sociable Numbers: These are groups of three or more numbers which form closed loops. The sum of the divisors of the first give the second, the sum of the divisors of the second give the third and so on until the divisors of the last give the first number.

PROJECT 16.9

Loop of Five

12,496 is the first in a loop of five sociable numbers. Working in small groups of 2 - 4 try to find the other four numbers in this loop. This could take 1 hour or more. (See hint on page 155.)

Factorising Quadratics using: x=

−b ± b 2 − 4ac 2a

Multiply (x + 4) by (x + 5) Each part of the first bracket must be multiplied by each part of the second bracket. ⇒ x by x = x2 x by 5 = 5x 4 by x = 4x 4 by 5 = 20 ∴ Multiply (x + 4)(x + 5) = x2 + 9x + 20 Factorising is the reverse of multiplying. Therefore the factors of x2 + 9x + 20 are (x + 4) and (x + 5). These may not be the only factors. It is easy to multiply two expressions like (x + 4) and (x + 5) to get x2 + 9x + 20 but it is not obvious how to factorise x2 + 9 + 20 to get (x + 4) and (x + 5). Example 1:

RULE The following formula works to factorise all quadratics of the form ax2 + bx + c = 0: The roots are x =

−b ± b 2 − 4ac 2a

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16. Interesting Numbers

Example 2:

Use the quadratic formula to find the factors of x2 + 9x + 20 The number before the x2 is called a, and here a = 1 The number before the x is called b, and here b = 9 The number which has no coefficient of x is called c, and here c = 20 The formula gives the roots as x = −9 + 81 − 80 2 –9 9+ 1 = 2 −9 + 1 = 2 −8 = 2

One root =

−9 ± 81 − 4(1)(20) 2(1) −9 − 81 − 80 2 –9 − 1 = 2 −9 − 1 = 2 −10 = 2

The second root =

= –4 = –5 The factors are (x – (first root)) and (x – (second root)) = (x + 4) = (x + 5) There are other methods for solving quadratic equations.

Example 3:

The area of a rectangle is 77 cm2. One side is 4 cm longer than the other. Find the length and breadth of the rectangle. Let x = short side (breadth), and so the length is x + 4 Area = length by breadth = (x + 4)(x) = x2 + 4x = 77 The quadratic we must solve is x2 + 4x – 77 = 0 a = 1, b = 4, c = –77 Roots are

−4 + 16 − 4(1)(−77) 2 –4 + 324 = 2 −4 + 18 = 2 = 14 2 =7

−4 − 16 − 4(1)(−77) 2 –4 − 324 = 2 −4 − 18 and 2 –22 = 2 and – 11 and

Ignore the negative answer, as length can only be positive. Therefore the breadth (x) is 7 cm and the length (x + 4) is 11 cm. In this chapter we have seen many new types of named numbers, some long established and some new, some useful and some of dubious use. Nevertheless it is always useful to explore them all.

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PROJECT 16.10 Other Types of Number Use the internet to investigate the following types of number and give examples of each: Mersenne numbers, pronic numbers, happy numbers and sad numbers.

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Show that there are no friendly pairs between 1 and 10. ____________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

2.

Find and sum the divisors of 200 (exclude the number 200 itself). _______________ ____________________________________________________________________

3.

Find and sum the divisors of each number between 20 and 30 (including 20 and 30). ____________________________________________________________________ ____________________________________________________________________

4.

Are there any friendly pairs between 20 and 30? ____________________________________________________________________ ____________________________________________________________________

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16. Interesting Numbers

5.

Are there any sociable loops of numbers between 20 and 30? ____________________________________________________________________ ____________________________________________________________________

6.

Multiply

(i) (x + 1) (x + 2) ________________________________________ (ii) (x + 1) (x + 5) ________________________________________ (iii) (x – 4) (x + 5) ________________________________________ (iv) (x – 4) (x – 3) ________________________________________ (v) (2x + 1) (x – 3) _______________________________________ (vi) (2x + 3) (3x + 4) ______________________________________ (vii) (10x – 1) (x + 2) ______________________________________ (viii)(x – 2) (x) ___________________________________________

7.

Factorise (a) x2 + 5x + 6 ______________________________________________________ ____________________________________________________________________ (b) 3x2 + 11x – 4 ____________________________________________________ ____________________________________________________________________ (c) x2 – 9x – 52 _____________________________________________________ ____________________________________________________________________ (d) 2x2 – 5x – 3 _____________________________________________________ ____________________________________________________________________

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8.

The area of a rectangle is 112 cm2. One side is 9 cm longer than the other. Find the length and breadth of the rectangle._____________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

9.

Find the two values of x which solve 3x2 + 5x – 3 = 0, correct to 2 decimal places. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

10. x+1 6

Use Pythagoras’ theorem to write an equation, then solve for x.

x–1 ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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16. Interesting Numbers

Hint for friendly numbers project:

One number is between 200 and 224 and is even.

Hint for sociable numbers project:

Start with 12,496. You should find 19 divisors which add to 14,288. The divisors of 14,288 add to give 15,572. The divisors of 15,572 add to give 14,536. This only leaves one number to find.

Rene Descartes: French academic, (1596–1650) Descartes left the military service to study philosophy and mathematics. He was first to use letters near the end of the alphabet for unknown values. He decided that x values should be on the horizontal axis and y values should be on the vertical axis. All points can be plotted as an ordered pair of numbers in co-ordinate geometry.

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17.

Infinity and Paradoxes

Mathematics cannot be applied to everything, and cannot explain everything. A paradox is a set of statements which contain contradictions. The existence of paradoxes causes problems for mathematical logicians which they cannot solve.

Russell’s Paradox An English man called Bertrand Russell (1872 – 1970) tried to show that all mathematics could be reduced to pure logic, but he came across a contradiction which is now known as Russell’s Paradox. An example of Russell’s Paradox: 20 people live on an island. Some people cut their own hair, the rest use the barber who only cuts the hair of people who do not cut their own hair. 10 people never cut their own hair. Draw a Venn diagram of the people on the island: Let A = people who cut their own hair Let B = people who do not cut their own hair A B The circles are separate because there is no overlap (or intersection) between the two groups of people. In which circle is the barber? He cannot be in circle B because he only cuts the hair of people who do not cut their own hair. Is it possible to work out how many people cut their own hair on this island? If the statements at the start of the example are true, it is impossible to work out how many people cut their own hair. This is a paradox.

PROJECT 17.1

Hair Cutting / Venn Diagram

20 people live on an island; 10 cut their own hair, 3 people sometimes cut their own hair, but otherwise use the barber. The rest always use the barber, who only cuts the hair of those who do not cut their own hair. Discuss this problem in small groups of 2 – 4 and try to agree on a Venn diagram with 2 overlapping circles showing in which circle each of the 20 people should be placed. Is it possible? In which area is the barber? Paradoxes like these have led to philosophies of numbers, such as intuitionism. An intuitionist will claim that there are statements which are neither true nor false.

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17. Infinity and Paradoxes

Schrödinger’s Cat Another paradox which causes problems for logical mathematics is Schrödinger’s Cat. Imagine a cat has been placed inside a box along with a device which operates at random times to release a deadly gas which will kill the cat. There is no way of knowing without opening the box if the cat is alive or dead. Logically it seems the cat must be either alive or dead, so we simply open the box to find out which! According to quantum mechanics the problem is not so simple. As weird as it sounds, quantum mechanics says that before we look inside the box the cat exists in a state of probability (a chance) of being alive (or dead). Therefore, the fate of the cat is not determined until we look inside the box. The instant we look inside the box, one of the possibilities actualises and the other vanishes, leaving the cat either dead or alive. Schröndinger’s Cat is paradoxical because before we look inside the box the cat can be thought of as simultaneously dead and alive.

PROJECT 17.2

Falling Tree in Forest

Consider the classic problem of the tree falling in the centre of a forest: if there is no one there to hear it, does it make any sound? There is no known maths applicable to this problem. Write 50 words on your thoughts on this problem.

Both of the above problems acknowledge the part an observer plays in the problem. There are more and more examples in nature where what we see depends on how we look at it. In future, mathematics will have to consider the role of us, the observer, in solving problems.

Achilles and the Tortoise Achilles and the tortoise decide to race. The tortoise gets a head start of 100 meters. Achilles can walk twice as fast as the tortoise. The race starts. When Achilles gets to the point where the tortoise started, the tortoise will be 50 meters ahead of him. When Achilles covers this 50 metres, the tortoise will have now moved another 25 meters ahead. When Achilles gets to that point, the tortoise will now still be 12.5 meters ahead. This continues for an infinite number of times with Achilles getting closer and closer, but never catching the tortoise as the tortoise moves a shorter and shorter distance ahead each time. In reality we know that the faster man i.e. Achilles will pass out the slower opponent i.e. the tortoise, so what is wrong with the argument?

PROJECT 17.3

Achilles and the Tortoise

Investigate the famous paradox of the warrior Achilles and the tortoise on the internet, and then discuss the problem in small groups. (See the solution to this paradox on page 161)

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Koch Snowflake The Koch Snowflake is a fractal shape which has an infinitely long outline but encloses a normal finite space, first described by Helge von Koch in 1904. Draw a simple equilateral triangle. On each of the three sides place another equilateral triangle exactly one third the size and in the middle of the side. The length of each side is now four thirds of the original length. Now continue the same process with each of the new sides. Every time, the overall length of the shape is four thirds times the previous. This can be done forever making the shape infinitely long but the area of the shape will never grow beyond the bounds of the hexagon (no. 2 below). 2.

1.

3.

4.

5.

Coastline of Ireland How long is the coastline of Ireland? If we used a 10km long measuring stick to measure it, we might get a length of 2,500km. However if we were to ‘zoom in’ closer to the coastline and measure it with a 1km long measuring stick we might get a length of over 15,000km. As we measure it in more and more detail and go in and around every cove it will grow to 25,000km. If we look close enough we can make it 100,000km or any length we wish. The reason for this is that a coastline is also fractal in nature and therefore the length of the outline will depend on the measuring stick (or unit of length) that we use to measure it. Like the Koch Snowflake the coastline of Ireland has a potentially infinite length but it encloses a finite area.

NOTE A fractal is a never-ending pattern that repeats itself at different scales. Although they are very complex they are created by repeating a simple process. Fractals appear everywhere in nature from sea shells to snowflakes to swirling galaxies. Fractals differ from the usual Geometric patterns you will encounter. With normal geometric patterns (like the ones in Chapter 5) the scale of the changes in the pattern is constant. With Fractals the patterns are much more complex and often the pattern replicates at every scale.

Infinity Infinity is an idea that can cause problems in mathemathics due to its paradoxical nature. If we try to imagine an enormous number that comes close to infinity, there are still more than this number of numbers between 0 and 1. The number infinity is a paradox. Call the number NI. This number has some great properties: NI + 1 = NI NI × 2 = NI NI × NI = NI NIN = NI. With these properties in mind let’s look at Hilbert’s Hotel. I

Hilbert’s Hotel David Hilbert, a German mathematician (1862 – 1943), came up with the concept of Hilbert’s Hotel. Hilbert’s Hotel has infinitely many rooms, and it is entirely full. If one more guest arrives, where can he be put? The management simply moves each guest to the next room, the guest in room 1 moves to room 2 etc. No one has to leave, so the latecomer gets room 1. If an infinitely large coachload of guests arrive, can they

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17. Infinity and Paradoxes

all be accommodated? Yes, the management simply asks each guest to move to a room with a number twice as large as the one they are now in. The guest in room 1 moves to room 2, the guest in room 2 to room 4 etc. Now all the odd numbered rooms, an infinity of them, are available.

PROJECT 17.4

David Hilbert: German mathematician (1862–1943) Hilbert is considered to be one of the most influential mathematicians of the last two centuries. He has contributed to number theory, quantum mechanics, general relativity and proof theory. He put forth a list of 23 unsolved problems to the International Congress of Mathematicians in Paris in 1900. Some of these are still not fully solved. He also proposed the Grand Hotel Cigar Mystery where every guest can smoke a cigar without any cigars being brought into the hotel.

Koch Curve

Consider a 1 m-sided equilateral triangle. What is the area it encloses? What is the total length of the perimeter? Now add on the first 3 triangles of a Koch curve. What is the area now? Its new perimeter? Now add on the next layer of 12 smaller equilateral triangles. What is the new area? New perimeter? Continue this process for 10 layers of triangles. What do you notice about the area after 10 layers, and the perimeter after 10 layers?

Questions 1.

The following Questions pages are FREE to download from mentorbooks.ie/resources

Let Nl = the number infinity, what do the following equal: (a) Nl + NI _________________________________________________________ (b) 2Nl ____________________________________________________________ (c) NI + 1,000_______________________________________________________ (d)

1 ____________________________________________________________ N1

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Transition Year Maths

2.

Koch curves and the Sierpinski Gasket are members of the ‘Gallery of Monsters’ in maths. There is another shape called the Sierpinski Carpet in this group. Imagine a square, cut out the centre

1 9

of the square, then continue to cut out the centre

1 9

of the

8 smaller squares that remain. Continue for ever like this. The shape produced is called a Sierpinski Carpet. What will the area of the remaining shape be? What will the total outline length be? ____________________________________________________________________ ____________________________________________________________________

3.

Draw a Sierpinski Carpet in as much detail as you can. Use the instructions in Question 2 to draw it.

4.

Begin with a 1m long line. Remove the middle third, then remove the middle third of each of the remaining segments, and so on. The cantor set is the dust of points which remain. What will the total length of this line be? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

5.

Consider these three sets: (a) ‘The set of large sets’ (b) ‘The set of small sets’ (c) ‘The set of all sets which are not members of themselves’. One of these is a paradox, one is a member of itself and one is not a member of itself. Which is which? Explain why. (a) ________________________________________________________________ (b) ________________________________________________________________ (c) ________________________________________________________________

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17. Infinity and Paradoxes

Achilles and the Tortoise paradox Solution The error in the logic is the assumption that an infinte sum of positive numbers (units of time in this case) is infinite. The sum of an infinite series of numbers may be finite. Therefore since Achilles was moving twice as fast as the tortoise and he took 1 minute to get to the point where the tortoise 1

started, the tortoise will have moved on. It will take him 2 a minute to catch up to where the tortoise had moved to. From there it will still take him a 1

1

further 4 of a minute to get to the next position, then of a minute and so 8 on indefinitely. Add all the times together, it is a sum from 0 to infinity. ∞

1

1

1

1

1

1

1

∑ 2n = 1 + 2 + 4 + 8 + 16 + 32 + 64 = 2 minutes n =0

He will catch him after two minutes. So, even though it appears he will never catch the tortoise because of the infinite units of time, it is in fact, a finite amount of time. In your future studies you will find some series sum to an infinite amount, while others like this, sum to a finite number.

Gabriel’s Horn Paradox Recall the project at the end of Chaper 9 (page 92). You could start to think in terms of the thickness of molecules of paint and the fact that they would not fit down the tube at a certain point, but this is wrong. This is a genuine paradox. There is at this time no explanation, but it is still worth thinking about! There are many things that science and maths have uncovered which cannot be explained.

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Answers Calculator Use 1. 2. 3. 4. 5. 6. 7. 8.

25, 400, 376996, 9 × 1014 125, 8000, 2.3 × 108, 2.7 × 1022 3125, 3.2 × 106, 8.7 × 1013, 2.4 × 1037 9, 31·6, 100, 5,477 31·6, 4·3, 10, 21·5 6, 3.6 × 106, 2.4 × 1018, 2.6 × 1032 5, 10, 10, 5, 1 5, 20, 60, 120, 120

9.

22 51 63 46 2, 571 , , , , 5 10 10 5 25

8. 9. 10. 11. 12.

13. 14. 15. 16. 17.

1,000,000 1010 1111 101010 1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000, 1001, 1002, 1010 11102, 324, 226, 168, 1410, 1212, 1014, E20 0·00001, 0·0001, 0·0011, 0·0101 0·1, 0·01, 0·001, 0·0001 (a) 100010, (b) 110110 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D, 1E, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29 2A 100102, 1216, 1024

10. 210 11. (a) 6 × 101 (b) 3·6 × 103 (c) 8·6 × 105 (d) 3·1 × 107 (e) 3·1 × 109 (f) (i) 1·1 × 1017 18. (ii) 1·9 × 1015 13 (iii) 3·2 × 10 (iv) 1·3 × 1012 (v) 3·6 × 109 2. Statistics (vi) 3·6 × 107 12. (i) ~ 10287 seconds 1. (i) (a) 5 (b) 6 (c) 5 (d) 2·08 (ii) ~ 10-46 tons (ii) (a) 3 (b) 2 (c) 3 (d) 1·22 8 14 (iii) (a) 2 (b) 1 (c) 1 (d) 1·29 13. 4·2 × 10 ; 3·1 × 10 13 22 2. (ii) 49 minutes 14. 6 × 10 ; 3·8 × 10 (iii) 40 - 45 minutes 15. 2·7 × 1025; 1·3 × 1013; 2·7 × 1015 (iv) Normal distribution with positive 16. 22; 1056; 0·0275, 195 skew 3. (ii) A normal distribution 4. (b) Strong negative correlation 5. (b) Weak negative correlation 1. Number Systems 6. (b) No correlation 1. X, XI, XII, XIII, XIV, XV, XVI, XVII,XVIII, XIX, XX, XXI, XXII, XXIII, 7. 90°, 72°, 72°, 36°, 36°, 18°, 18°, 18°

2.

3. 4. 5.

6. 7.

XXIV, XXV, XXVI, XXVII, XXVIII, XXIX, XXX MIM, MM, MMI, MMII, MMIII, MMIV, MMV, MMVI, MMVII, MMVIII, MMIX, MMX, MMXI, MMXII 2,767 (a) CXCIV (b) MCDXCII (c) MMCCXXII (d) MMDCCCXXXVII 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110 1111 11100

3. Graphs 1.

3. 6. 7. 8.

162 162

(a) curve (1 bend); (b) straight line; (c) curve (1 bend); (d) curve (2 bends); (e) curve (2 bends); (f) curve (3 bends) 1; 10; 4 –1, 3; (x + 1) (x – 3) –1.3, 0.2, 3.1, (x + 1.3) (x – 0.2) (x – 3.1) (a) (x – 1) (x + 1) (x – 3) (b) (x – 1) (x + 1) (x + 3) (c) (x – 1) (x + 1) (x + 2)

ANSWERS

(d) (x – 1) (x + 1) (x - 3) (x + 2) (e) (x – 2) (x + 2) (x + 3) (f) (x – 3) (x + 3) (x + 1) 9. –2, –0.3, 2.3, (x + 2) (x + 0.3) (x – 2.3) 10. x2 + y2 = 16; x2 + y2 = 36; x2 + y2 = 2 1 11. (0, 0), 6; (0, 0), 30 ; (0, 0), 3 12. (a) x2 + y2 – 2x – 2y + 1 = 0 (b) x2 + y2 – 30x + 200 = 0 13. (0, –1), 2; (–1, –2), 2; (–1, –1),

2;

5. 6.

16

9 x 10 J 6,428,571,429 days ~ 17 million years 9 × 1013J 9 × 1025J v −u a= t x=

v , u

x=

b + 2c + 3d – 4ea – 5 fa a

y5 sw , x= x = 10 a w+y d + 100e b –a , x= x= 2 2 2c 7. 31/304, 24/297, 38/311, 16/289, –17/256. Northern hemisphere summer. 8. €689·47, €689·47 9. 2356cm3, 633cm3 10. 3.26cm3, 6.51cm3 11. 53.05cm, 65.77cm s 12. r = 4π 13. h =

s – 2πr 2 4 πr

5. Co-ordinate geometry 1. 2. 3. 4. 8.

1.

20 – divisors 10, 5, 4, 2, 1; sum is 22 ⇒ excessive 21 – divisors 7, 3, 1; sum is 11 ⇒ defective 22 – etc.

3.

(d) (h) (m) (r)

5.

20 , 2, 6 , 8 , 180 , 3·16, 17, 24, 7, 12

6.

15,

54,

7.

2 18 ,

7·2, 18 ,

5·4, 18 ,

9·6 18 , 18

7. Ratio and Fibonacci

4. Formulas 1. 2. 3. 4.

6. Pythagoras’ Theorem

Star Star Star L I N E (i) Reversed (ii) Reversed and inverted

6.

€167 €333 €500 4 : 5 : 11 1 : 12 (a) 24 : 35 (b) 24 : 315 (c) 4 : 21 (d) 18 : 5 (e) 21 : 4 (f) 21 : 2 256

7.

360° – 137.5° = 222.5°

1. 2. 3. 4.

360 = 1·618 222·5 Causes irregular patterns which is less likely to split and is therefore stronger.

8. Algebra 1. 2.

6, –3, 9, 0, 8, 21, 42, 61, 162, 1 a2 c2 1

1, 2, 2a, 2a, a 4. 7a + 3a2, 6 + 5c, 2a + b + c 5. 3c 6. x + y = 25; 5x + 8y = 167; 3x – 2y = 5; 28x = 22y; 11 cent; 14 cent 7. Pears 8. 4, 1 9. 96 @ €1 54 @ €2 10. 2, –3, 5 11. No, because the first two equations are effectively the same, one is twice the other. 12. 2, 3, 4 3.

163

Transition Year Maths

9. Pi

8.

3.18918 15.700m 15.714m 15.708m 0·159; 0·796; 1·591; 3·183 0·1736; 0·3420; 0·8415; 0·9092 0·8415; 0·9093; 0·1411; –0·7568 Because 1 radian = 57·3° Yes 10. 22 7 11. Plot (1, 1); (1, –1); (2, 0·5); (2, –0·5); (3, 0·33) . . . and so on

9.

1. 2. 3. 4. 5. 6. 7. 8. 9.

12. Fermat’s Last Theorem 1. 2. 3. 4.

10. Probability 1.

1 in 6; 1 in 6; 1 in 2; 5 in 6; 2 in 6

1 1 1 1 1 ; ; ; ; 6 6 6 3 3 1 1 5 1 3 48 3. ; ; ; ; ; ; 52 52 52 2 4 52 1 4. 2 25 1 1 11 5. ; ; ; 0; ; 1; 36 6 36 36 1 3 3 1 7. ; ; ; ; 0 8 8 8 8 8. 6 16 1 9. 8; 8 6 10. 16; 16

5. 6.

2.

4, 12, 24, 24 4, 6, 4, 1 40,320 88 = 16,777,216 65,780 24; 13,824; 12,144 210; 126; 112

93 = 83 + 63 + 13 53 + 63 = 73 – 2 It’s the only number between a cube and a square (i) 1, 1, 2, 2 (ii) 6, 5, 3, 10 (iii) 3, 3, 3, 4 or 1, 1, 1, 2 (iv) 5, 5, 2, 6 Eight 1, 32, 243, 1024, 3125, 7776. 16807, 32768, 59049, 100,000 95 + 85 + 65 + 407 = 105

8 12 1 ; ; 52 52 26

13. Chaos 1 3

11. Combinations and Permutations 1. 2. 3. 4. 5. 6. 7.

2,598,960; 1,287; 575,757; 65,780; 249,900; 2,349,060; 1,712,304; 48 (a) x3 + 3x2y + 3xy2 + y3 (b) x5 + 10x4 + 40x3 + 80x2 + 80x + 32 (c) x4 + 8x3y + 24x2y2 + 32xy3 + 16y4 (d) 8x3 – 36x2y + 54xy2 – 27y3 (e) 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5

1. 2. 3. 4.

5.

6. 7. 8. 9.

164

–·91, –·17, –·97, –·06, –·99, . . . . . . –1, 0, –1, 0 3, 8, 63, 3968, . . . . . . most calculators will do about 8 –1·84, 1·38, –0·08, –1·99, 1·97, 1·89, 1·6, 0·57, –1·67, . . . . . . (a) 0·25, 0·0625, . . . . . . (b) 0·125, 0·0019, . . . . . . (c) 2, 0·5, 2, 0·5, 2, 0·5, . . . . . . (a) oscillator (f) chaotic (b) chaotic (g) chaotic (c) oscillator (h) stable (d) stable (i) oscillator (e) oscillator 500, 250, 125, 62·5, 31·25, 15·625, . . . . . . 0·078, 0·289, 0·822, 0·585, 0·970, . . . . . . Chaotic 0·059, 0·016, 0·415, 0·728, . . . . . . 0·61, 0·70, 0·61, 0·70, . . . . . . oscillator 0·039, 0·075, 0·139, 0·239, . . . . . . 0·5, 0·5, 0·5, . . . . . . stable

ANSWERS

14. Complex numbers 1. 2. 3. 4. 5.

6.

7. 8.

22, 11, 14, 1, 36, 6, 16, 13, 28, 1, 42 No 5 + 7i ; 5 + 3i ; –1 + 0i No 1 + 3i ; 1 – 7i; –2 + 2i (i) x2 + 3x + 2 9 + 6i ; –6 + 9i; 3 + 15i; 7 – 22i ; 22 – 6i (ii) x2 + 6x + 5 6 22 –36 2 3 1 (iii) x2 + x – 20 1 + 2i; 2 – i; + i ; + i; + i (iv) x2 – 7x + 12 13 13 2 2 34 17 1 1 (v) 2x2 – 5x – 3 + i 6 6 (vi) 6x2 + 17x + 12 (vii) 10x2 + 19x – 2 2 (viii) x2 – 2x 2 + 3i use –b ± b – 4ac 7. (a) (x + 3)(x + 2) 2a (b) (3x – 1)(x + 4) (c) (x – 13)(x + 4) 29 10 , 72 , 5 , 18 , 2, 29 , (d) (2x + 1)(x – 3) 4 36 8. 7, 16 (a) 6 + 2i 9. 0·47, –2·14 10. 9 3. 4. 5. 6.

15. Calculus 1. 2.

30t 2; 20t ; 10; 0; 8t + 3; –6t –3 –12t –4 v + 400t + 50 a = 400m/s2

4.

6m 10

5.

(a) 77m, 50m/s (b) 2m, 17m/s

–26m/s 100

1 m/s 3 3 (d) 1m, 5 m/s 16 (c) 1m,

17. Infinity and Paradoxes 1. 2. 4. 5.

Nl, Nl, Nl, 0 0, ∞ 0m (a) a member of itself because the set itself is a large set (b) not a member of itself because the set itself is not a small set (c) a paradox

10 (e) – 1 , m/s 3 9 (f) 38m, 143m/s

16. Interesting Numbers 1.

2.

The sum of the divisors of the numbers 1 to 10 are 1, 1, 1, 3, 1, 6, 1, 7, 4, 8. There are no friendly pairs. Remember you do not include the number itself 1 + 2 + 4 + 5 + 8 + 10 + 20 + 25 + 40 + 50 + 100 = 265

165

Transition Year Maths

166

ANSWERS

167

Transition Year Maths

168