Edexcel A-level Mathematics - Collins Educationresources.collins.co.uk/PFD/A-Level Maths_Book 1Chapter11_pre-endorsement.pdfconjecture and derive results about angles and sides, including

Edexcel A-level

Mathematics Helen Ball

Kath HipkissMichael KentChris Pearce

Student Book Year 1 and AS

CONTENTSIntroduction v

1 – Algebra and functions 1: Manipulating

algebraic expressions 1 1.1 Manipulating polynomials algebraically 3 1.2 Expanding multiple binomials 6 1.3 Th e binomial expansion 11 1.4 Factorisation 14 1.5 Algebraic division 16 1.6 Laws of indices 22 1.7 Manipulating surds 24 1.8 Rationalising the denominator 26 Summary of key points 28 Exam-style questions 1 29

2 – Algebra and functions 2: Equations

and inequalities 31 2.1 Quadratic functions 32 2.2 Th e discriminant of a quadratic function 35 2.3 Completing the square 37 2.4 Solving quadratic equations 39 2.5 Solving simultaneous equations 42 2.6 Solving linear and quadratic

simultaneous equations 46 2.7 Solving linear inequalities 47 2.8 Solving quadratic inequalities 52 Summary of key points 56 Exam-style questions 2 56

3 – Algebra and functions 3:

Sketching curves 58 3.1 Sketching curves of quadratic functions 59 3.2 Sketching curves of cubic functions 63 3.3 Sketching curves of quartic functions 67 3.4 Sketching curves of reciprocal functions 72 3.5 Intersection points 79 3.6 Proportional relationships 82 3.7 Translations 86 3.8 Stretches 91 Summary of key points 97 Exam-style questions 3 98

4 – Coordinate geometry 1: Equations

of straight lines 100 4.1 Writing the equation of a straight line

in the form ax + by + c = 0 101 4.2 Finding the equation of a straight line

using the formula y – y1 = m(x – x1) 103

4.3 Finding the gradient of the straight line between two points 105

4.4 Finding the equation of a straight line using the formula y y

y yx xx x

1

2 1

1

2 1

−− = −

− 109 4.5 Parallel and perpendicular lines 112 4.6 Straight line models 118 Summary of key points 124 Exam-style questions 4 124

5 – Coordinate geometry 2: Circles 126 5.1 Equations of circles 127 5.2 Angles in a semicircle 134 5.3 Perpendicular from the centre to a chord 138 5.4 Radius perpendicular to the tangent 144 Summary of key points 151 Exam-style questions 5 151

6 – Trigonometry 154 6.1 Sine and cosine 155 6.2 Th e sine rule and the cosine rule 158 6.3 Trigonometric graphs 166 6.4 Th e tangent function 169 6.5 Solving trigonometric equations 172 6.6 A useful formula 174 Summary of key points 177 Exam-style questions 6 177

7 – Exponentials and logarithms 179 7.1 Th e function ax 180 7.2 Logarithms 184 7.3 Th e equation ax = b 189 7.4 Logarithmic graphs 191 7.5 Th e number e 196 7.6 Natural logarithms 201 7.7 Exponential growth and decay 204 Summary of key points 209 Exam-style questions 7 209

8 – Differentiation 212 8.1 Th e gradient of a curve 213 8.2 Th e gradient of a quadratic curve 217 8.3 Diff erentiation of x² and x3 222 8.4 Diff erentiation of a polynomial 225 8.5 Diff erentiation of xn 228 8.6 Stationary points and the second derivative 232 8.7 Tangents and normals 237 Summary of key points 240 Exam-style questions 8 240

iii

04952_P00i_viii.indd iii04952_P00i_viii.indd iii 10/01/2017 17:2410/01/2017 17:24

9 – Integration 243 9.1 Indefi nite integrals 244 9.2 Th e area under a curve 249 Summary of key points 258 Exam-style questions 9 258

10 – Vectors 260 10.1 Defi nition of a vector 261 10.2 Adding and subtracting vectors 266 10.3 Vector geometry 269 10.4 Position vectors 278 Summary of key points 283 Exam-style questions 10 283

11 – Proof 285 11.1 Proof by deduction 286 11.2 Proof by exhaustion 289 11.3 Disproof by counter example 294 Summary of key points 298 Exam-style questions 11 298

12 – Data presentation and

interpretation 299 12.1 Measures of central tendency

and spread 301 12.2 Variance and standard deviation 308 12.3 Displaying and interpreting data 317 Summary of key points 339 Exam-style questions 12 339

13 – Probability and statistical

distributions 342 13.1 Calculating and representing

probability 343 13.2 Discrete and continuous distributions 350 13.3 Th e binomial distribution 359 Summary of key points 365 Exam-style questions 13 365

14 – Statistical sampling and

hypothesis testing 368 14.1 Populations and samples 369 14.2 Hypothesis testing 374 Summary of key points 389 Exam-style questions 14 389

15 – Kinematics 391 15.1 Th e language of kinematics 392 15.2 Equations of constant acceleration 395 15.3 Vertical motion 402

15.4 Displacement–time and velocity–time graphs 408

15.5 Variable acceleration 416 Summary of key points 422 Exam-style questions 15 422

16 – Forces 425 16.1 Forces 426 16.2 Newton’s laws of motion 429 16.3 Vertical motion 438 16.4 Connected particles 441 16.5 Pulleys 449 Summary of key points 455 Exam-style questions 16 455

Exam-style extension questions 460

Answers* 487 1 Algebra and functions 1: Manipulating

algebraic expressions 487 2 Algebra and functions 2: Equations

and inequalities 490 3 Algebra and functions 3: Sketching

curves 497 4 Coordinate geometry 1: Equations of

straight lines 511 5 Coordinate geometry 2: Circles 515 6 Trigonometry 518 7 Exponentials and logarithms 521 8 Diff erentiation 527 9 Integration 533 10 Vectors 535 11 Proof 540 12 Data presentation and interpretation 544 13 Probability and statistical distributions 549 14 Statistical sampling and hypothesis

testing 552 15 Kinematics 555 16 Forces 560

Exam-style extension question answers

563

Formulae 578

Glossary 580

Index 585

* Short answers are given in this book, with full worked

solutions for all exercises, large data set activities, exam-style

questions and extension questions available to teachers by

emailing [email protected]

iv

CONTENTS

04952_P00i_viii.indd iv04952_P00i_viii.indd iv 10/01/2017 17:2410/01/2017 17:24

If you are asked to ‘prove’ (or disprove) a mathematical statement, what does this actually mean? The Oxford English Dictionary defi nes proof as ‘The action or an act of testing something; a test, a trial, an experiment … An operation to check the correctness of an arithmetical calculation.’

Can you show that if x and y are even integers then the sum of x and y will also be an even integer? For a group of mathematical statements to constitute a proof you need to ensure that the proof will work for all possibilities and not just one or two. For this proof, how could you ensure that x and y are even integers? Well, an even integer is a whole number that is divisible by 2. So if we say that m, n, p ... are integers, then 2m will be an even integer and so will 2n, 2p, etc. Now you can let x = 2m and y = 2n. If you now sum x and y you get x + y = 2m + 2n. This simplifi es to 2(m + n), which is an even number. You have now proved that if x and y are even integers then the sum of x and y will also be an even integer.

If you are asked to ‘show’, ‘prove’ or ‘demonstrate’, you are being asked to prove that something is true. The word ‘prove’ may not actually appear in what you are being asked to do but you need to understand that this is what is required of you. You may also be asked to ‘disprove’ something, or to ‘show by using a counter example’ that something is false. Throughout this book, questions involving proof are fl agged as such and proof is discussed in context in the margin.

LEArNING oBJECTIVES

You will learn how to:

› understand and use the structure of mathematical proof, proceeding from given assumptions through a series of logical steps to a conclusion

› use proof by deduction

› use proof by exhaustion

› use disproof by counter example.

ToPIC LINKS

Most chapters of this book demonstrate proofs related to specifi c topics within those chapters. For example, Chapter 2 Algebra and functions 2: Equations and inequalities includes the proof for the quadratic formula by completing the square. In this chapter you will practise using different types of proof.

PrIor KNoWLEDGE

You should already know how to:

› argue mathematically to show that algebraic expressions are equivalent, and use algebra to support and construct arguments and proofs

11

Proof11

04952_P285_298.indd 285 1/10/17 11:12 AM

11.1 Proof by deduction

Proof by deduction is the most commonly used form of proof throughout this book – for example, the proofs of the sine and cosine rules in Chapter 6 Trigonometry. Proof by deduction is the drawing of a conclusion by using the general rules of mathematics and usually involves the use of algebra. Often, 2n is used to represent an even number and 2n + 1 to represent an odd number.

KEY INforMATIoNProof by deduction is the drawing of a conclusion by using the general rules of mathematics.

› apply angle facts, triangle congruence, similarity and properties of quadrilaterals to conjecture and derive results about angles and sides, including Pythagoras’ theorem and the fact that the base angles of an isosceles triangle are equal, and use known results to obtain simple proofs

› use vectors to construct geometric arguments and proofs.

You should be able to complete the following questions correctly:

1 Show that the sum of any three consecutive odd numbers is always a multiple of 3.

2 Show that the sum of two consecutive positive integers is always an odd number.

3 Show that the difference between the squares of any two consecutive positive integers is equal to the sum of the two integers.

4 Prove that the sum of the squares of any two even integers is always a multiple of 4.

5 Prove that (2n − 1)2 − (2n + 1)2 is a multiple of 8.

Example 1Show that n2 − 8n + 17 is positive for any integer n.

SolutionComplete the square.

n2 − 8n + 17 = (n − 4)2 + 1

which is always positive for any integer n.

ProofThe phrase ‘show that’ indicates that a proof is required.

Example 2A bag has x green balls and 6 red balls. Two balls are taken from the bag at random. The probability that the balls are different colours is 1

2. Show that x 2 − 13x + 30 = 0.

SolutionThere are two possibilities for the balls to be different. The fi rst one could be green and the second red or the fi rst one could be red and the second green. The probabilities of each will be the same.

11 Proof

286

227878 A-Level Maths_Chapter 11.indd 286 12/5/16 6:12 PM

Since there are x green balls and (x + 6) balls in total, the

probability that the first ball is green is xx 6+ .

If a green ball is taken (and not replaced), the probability that

the second ball is red is x

65+ , since there is 1 fewer ball in the bag.

The probability that the balls are different is therefore given by x

x 6+ ×

x6

5+ × 2 since there are two ways this can happen. The

question states that this probability is equal to 12.

Hence xx 6+ ×

x6

5+ × 2 = 12

Multiply both sides by 2.x

x + 6 × 6

5x + × 4 = 1

Multiply both sides by (x + 6)(x + 5).

x × 6 × 4 = (x + 6)(x + 5)

Simplify.

24x = x2 + 11x + 30

Subtract 24x from both sides.

0 = x2 − 13x + 30, as requested.

Since there are (x + 6) balls to start with, when one ball is removed there will be (x + 5) balls in the bag.

ProofWhen you are asked to show a result, the result should be the conclusion of your working and be the last line of working of a series of logical steps which follow on from one another.

Example 3Demonstrate that the sum of the squares of two consecutive odd positive integers is always even.

SolutionLet 2n be an even number. Then 2n − 1 is the odd number which precedes 2n and 2n + 1 is the odd number which succeeds 2n.

(2n − 1)2 + (2n + 1)2 = 4n2 − 4n + 1 + 4n2 + 4n + 1

= 8n2 + 2

= 2(4n2 + 1)

which has a factor of 2 and so is even.

Alternatively:

Let 2n be an even number. Then 2n + 1 is the odd number which immediately succeeds 2n and 2n + 3 is the second odd number which succeeds 2n.

(2n + 1)2 + (2n + 3)2 = 4n2 + 4n + 1 + 4n2 + 12n + 9

= 8n2 + 16n + 10

= 2(4n2 + 8n + 5)

which has a factor of 2 and so is even.

ProofThe words ‘demonstrate that’ have been used to indicate that a proof is required.

11.1 Proof by deduction

287


Exercise 11.1A page 549

1 Prove that the expression n3 − n is always the product of three consecutive integers for n ⩾ 2.

2 The diagram shows a shape with an area of 201 cm2.

x cm

(x + 3) cm

(2x + 5) cm

(3x – 1) cm

Prove that 4x2 + 7x − 186 = 0 and hence deduce that the perimeter of the shape is 68 cm.

PF

PS

1

2

PF

PS

Example 4Prove that the sum of any four consecutive integers a, b, c and d, where a < b < c < d, is equal to cd − ab.

SolutionSince the integers are consecutive, they can all be written in terms of the same variable.

Let the smallest integer, a, be equal to n.

Hence b = n + 1, c = n + 2 and d = n + 3.

The sum of the four integers, a + b + c + d, is given by n + (n + 1) + (n + 2) + (n + 3).

n + (n + 1) + (n + 2) + (n + 3) = n + n + 1 + n + 2 + n + 3

= 4n + 6

The expression cd − ab can be rewritten as (n + 2)(n + 3) − n(n + 1).

(n + 2)(n + 3) − n(n + 1) = n2 + 5n + 6 − (n2 + n)

= n2 + 5n + 6 − n2 − n

= 4n + 6

Since both resulting expressions are equal to 4n + 6, the sum of any four consecutive integers a, b, c and d, where a < b < c < d, is equal to cd − ab.

Consecutive integers increase by 1 from term to term.

M Modelling PS Problem solving PF Proof CM Communicating mathematically

11 Proof

288


11.2 Proof by exhaustion

Proof by exhaustion is sometimes called the ‘brute force method’. It is called this because the steps involved are as follows:

› divide the statement into a fi nite number of cases

› prove that the list of cases identifi ed is exhaustive – that is, that there are no other cases

› prove that the statement is true for all of the cases.

3 Using completion of the square, prove that n2 − 6n + 10 is positive for all values of n.

4 Prove that any square number is either a multiple of 4 or one more than a multiple of 4.

5 A cyclic quadrilateral has all four vertices on the circumference of a circle. Prove that the opposite angles of a cyclic quadrilateral have a sum of 180°.

6 Six people, Alf, Bukunmi, Carlos, Dupika, Elsa and Frederique, are to be seated around a circular table for a dinner party. Alf is to sit next to Bukunmi. Carlos must not sit next to Dupika but must sit next to Elsa. Ignoring rotations and reflections, show that there are eight possible seating arrangements.

7 The following diagram shows a large square with side length a containing four congruent right-angled triangles and a smaller square. The base of each triangle is of length b.

a

b

Show that the area of the four right-angled triangles plus the smaller square is equal to the area of the large square.

8 Look at this ‘proof’ which seems to show that 1 = 2. Identify the flaw in the workings.

Let a = b where a, b > 0.

Then ab = b2

So ab − a2 = b2 − a2

Factorising: a(b − a) = (b + a)(b − a)

Dividing leaves a = b + a

As b = a, substituting gives a = a + a

So a = 2a

If a = 1 then 1 = 2.

3

4

5

PF

PS

PF

PS

PF

PS

PF

PS

6

7

8

PF

PS

PF

PS

KEY INforMATIoNProof by exhaustion requires splitting the statement into a fi nite, exhaustive set of cases, all of which are tested.


289


Example 5Suppose n is an integer between 24 and 28 inclusive. Prove that n is not a prime number.

SolutionA prime number is an integer which has two and only two distinct factors. 1 is not a prime number.

n = {24, 25, 26, 27, 28}

Case 1, n = 24: the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24 so 24 is not prime.

Case 2, n = 25: the factors of 25 are 1, 5, 25 so 25 is not prime.

Case 3, n = 26: the factors of 26 are 1, 2, 13, 26 so 26 is not prime.

Case 4, n = 27: the factors of 27 are 1, 3, 9, 27 so 27 is not prime.

Case 5, n = 28: the factors of 28 are 1, 2, 4, 7, 14, 28 so 28 is not prime.

If n is an integer between 24 and 28 then n is not a prime number.

ProofTo use proof by exhaustion, you need to list all of the possible integers and test each case, one by one.

Example 6Elif has been asked to list all the possible arrangements of the word PEEP. She has written EEPP and PEPE.

Prove by exhaustion that there are 6 distinct ways of arranging the letters in the word PEEP.

SolutionIf the first letter is P and the second letter is P, then the final two letters must both be E.

Elif writes PPEE.

If the first letter is P and the second letter is E, then the final two letters must be P and E.

Elif writes PEPE and PEEP.

If the first letter is E and the second letter is P, then the final two letters must be P and E.

Elif writes EPEP and EPPE.

If the first letter is E and the second letter is E, then the final two letters must both be P.

Elif writes EEPP.

Elif’s final solution is PPEE, PEPE, PEEP, EPEP, EPPE, EEPP.

Elif has considered all the possibilities and shown that there are 6 distinct ways of arranging the letters.

11 Proof

290


Example 7Suppose x and y are even positive integers less than 8. Prove that their sum is divisible by 2.

Solutionx = {2, 4, 6} and y = {2, 4, 6}

Case 1: x + y = 2 + 2 = 4 = 2 × 2

Case 2: x + y = 2 + 4 = 6 = 2 × 3

Case 3: x + y = 2 + 6 = 8 = 2 × 4

Case 4: x + y = 4 + 2 = 6 = 2 × 3

Case 5: x + y = 4 + 4 = 8 = 2 × 4

Case 6: x + y = 4 + 6 = 10 = 2 × 5

Case 7: x + y = 6 + 2 = 8 = 2 × 4

Case 8: x + y = 6 + 4 = 10 = 2 × 5

Case 9: x + y = 6 + 6 = 12 = 2 × 6

As 2 is a factor in each case then if x and y are even integers less than 8, their sum is divisible by 2.

ProofProof by exhaustion is really only a feasible method to use when the number of cases to be tested is small.

Example 8Prove by exhaustion that there are only two distinct triangles with sides of integer length that have a perimeter of 10.

SolutionThe question requires you to show that of all the possible sets of three integers that have a sum of 10, only two sets can be used to draw a triangle. It is simpler to start by listing all the possible ways that three integers can have a sum of 10 before considering whether or not each set can represent the sides of a triangle. The order of the integers does not matter.

To do this systematically, begin by considering possibilities including a 1. If one integer is 1, then the other two add up to 9. You can have 1, 1, 8 (two 1s and an 8), 1, 2, 7 (a 1, a 2 and a 7), 1, 3, 6 and 1, 4, 5. There are no other possibilities with a 1.

Case 1: 1, 1, 8

Case 2: 1, 2, 7

Case 3: 1, 3, 6

Case 4: 1, 4, 5


291


Now consider possibilities including a 2 but no 1. If one integer is 2, then the other two add up to 8. You can have 2, 2, 6 or 2, 3, 5 or 2, 4, 4.

Case 5: 2, 2, 6

Case 6: 2, 3, 5

Case 7: 2, 4, 4

Now consider possibilities including a 3 but no 1 or 2. If one integer is 3, then the other two add up to 7. You can have 3, 3, 4 only.

Case 8: 3, 3, 4

Hence there are eight sets of three integers with a sum of 10: {1, 1, 8}, {1, 2, 7}, {1, 3, 6}, {1, 4, 5}, {2, 2, 6}, {2, 3, 5}, {2, 4, 4} and {3, 3, 4}.

However, in order to be the sides of a triangle, any pair of sides must sum to more than the third side. Therefore, if any pair does not sum to more than the third side, then that set of numbers cannot represent the sides of a triangle. If the sides add up to less than the third side then they cannot reach to meet. If the sides add up to the same as the third side then they all lie on a straight line.

Consider each set of integers individually.

Case 1, {1, 1, 8}: 1 + 1 < 8, so two sides sum to less than the third side.



Case 4, {1, 4, 5}: 1 + 4 = 5, so two sides sum to the third side.


Case 6, {2, 3, 5}: 2 + 3 = 5, so two sides sum to the third side.

Case 7, {2, 4, 4}: 2 + 4 > 4 and 4 + 4 > 2, so any pair of sides sums to more than the third side.

Case 8, {3, 3, 4}: 3 + 3 > 4 and 3 + 4 > 3, so any pair of sides sums to more than the third side.

Therefore {2, 4, 4} and {3, 3, 4} are the only sets of numbers that give two distinct triangles with sides of integer length that have a perimeter of 10.

There cannot be any possibilities which only have integers of 4 or more because the sum would be at least 12.

Alternatively, you could observe that a triangle which satisfies this property cannot have a side length of 5 or more since this is half the perimeter or more. {2, 4, 4} and {3, 3, 4} are the only two sets for which all three integers are below 5.

11 Proof

292



1 Three people, Alf, Bukunmi and Carlos, are to be seated along one edge of a table for a dinner party. Alf is to sit next to Bukunmi. Show using proof by exhaustion that there are four possible seating arrangements.

2 Prove that there are two prime numbers between 20 and 30 inclusive.

3 Prove that no square number below 100 ends in 7.

4 Show that the expression n2 + 3n + 19 is prime for all positive integer values of n below 10.

5 Suppose x and y are odd positive integers less than 7. Prove that their sum is divisible by 2.

6 Prove by exhaustion that there are only two distinct triangles with sides of integer length that have a perimeter of 11.

7 Prove by exhaustion that there are five distinct quadrilaterals with sides of integer length that have a perimeter of 10.

PF

PS

1

2

3

4 4

PF

PS

PF

PS

PF

PS

5

6

7

PF

PS

PF

PS

PF

PS

Example 9Suppose n is an integer between 2 and 10 inclusive. Prove that n2 − 2 is not divisible by 5.

SolutionCase 1, n = 2: n2 − 2 = 22 − 2 = 2, which is not a multiple of 5.

Case 2, n = 3: n2 − 2 = 32 − 2 = 7, which is not a multiple of 5.








If n is an integer between 2 and 10 then n2 − 2 is not divisible by 5.

TECHNoLoGYUsing a spreadsheet software package, you could generate the values for more cases and explore what the value of n is when n2 − 2 is fi rst a multiple of 5.


293


11.3 Disproof by counter example

To prove something is false it is only necessary to fi nd one exception (even though there may be more). This is called a counter example. For example, when learning to spell you might have been told ‘i before e except after c’. Turning this into a mathematical proof you could say ‘disprove, by counter example, that i does not always precede e after letters other than c in the English language’. A counter example would be the word ‘weird’.

8 A tetronimo is a 2D shape made from four identical squares joined by their edges. Show that, ignoring rotations and reflections, there are five possible tetronimos.

9 Prove that there are only six possible half-time scores for a match which finishes 2−1.

10 Prove that if n is a positive integer less than 10 then n6 − n is a multiple of 2.

PF

PS

PF

PS

8

9

10 PF

PS

KEY INforMATIoNA counter example is an exception to the rule.

Example 10Sawda notes that 12 − 1 + 5 = 5 and 22 − 2 + 5 = 7, which are both prime.

Show that it is untrue that n2 − n + 5 is prime for all positive integer values of n.

SolutionIf Sawda continues with n = 3 and n = 4, the results are:

32 − 3 + 5 = 11

42 − 4 + 5 = 17

Both of these are prime so are not counter examples.

However, 52 − 5 + 5 = 25, and 25 is not prime so n = 5 is a counter example.

Note that if n = 5, then all three terms are multiples of 5, so this was likely to give an answer which was not prime (unless it was a negative answer, 0 or 5 itself).

Also note that n2 + n + 41 is prime for all integer values of n below 40.

TECHNoLoGYUsing a spreadsheet software package you could use the formula n2 − n + 5 to generate the terms in the sequence.

Stop and think What is the fi rst integer value of n for which n2 + n + 41 is not prime? Can you show this mathematically? Construct similar equations but with different prime numbers with similar qualities.

11 Proof

294


Example 11Disprove that if x and y are consecutive odd numbers then one of x or y must be prime.

SolutionLet x = 25 and y = 27; then x and y are consecutive odd numbers.

A prime number is an integer which has two and only two distinct factors. 1 is not a prime number.

Let x = 25 − the factors of 25 are 1, 5, 25 so 25 is not prime.

Let y = 27 – the factors of 27 are 1, 3, 9, 27 so 27 is not prime.

If x and y are consecutive odd numbers then one of x or y is not necessarily a prime number.

In this example you could have methodically worked through each pair of consecutive odd numbers until you found a pair where one of the numbers was not prime.

Example 12Georgina draws two regular polygons. The pentagon has an interior angle of 108° and the decagon has an interior angle of 144°. Georgina conjectures that the interior angle of any regular polygon will be an integer number of degrees. Find a counter example to show that Georgina’s conjecture is false.

SolutionThe interior angle (in degrees) of a regular polygon is given by 180 −

n360, where n is the number of sides.

For the pentagon and decagon that Georgina drew, n = 5 and n = 10 respectively.

For a regular triangle (i.e. an equilateral triangle), n = 3, and the interior angle is 60°.

For a regular quadrilateral (i.e. a square), n = 4, and the interior angle is 90°.

For a regular hexagon, n = 6, and the interior angle is 120°.

None of these is a counter example to Georgina’s conjecture that the interior angle is an integer.

However, for a regular heptagon with n = 7, 180 − 3607

= 128 47 ,

which is not an integer, so the heptagon is a counter example.

Note that, since the formula is 180 −

n360, any value of n

which is not a factor of 360 will be a counter example.

Stop and think How many regular polygons have an interior angle which is an integer number of degrees? How can you check that you have found them all?


295



1 Simon says that:

a all rectangles are squares

b every integer less than 10 is positive

c when a number is doubled the answer is larger

d for every integer n, n3 is positive

e all prime numbers are odd

f log (A + B) = log A + log B

g (a + b)2 = a2 + b2

Find a counter example for each of Simon’s statements.

2 Patrick has noticed that 20 > 02, 25 > 52 and 210 > 102. Initially he makes a generalisation that 2n > n2 for positive integer values of n but then notices that 22 = 22. He then amends his generalisation to 2n ⩾ n2. Prove that Patrick’s generalisation is still incorrect.

1

a

b

c

d

e

f

g

2

PS

CM

PF

PS

Stop and think If x and y are irrational, can you generalise the qualities of x and y so that xy and xy

are rational? Can you prove your generalisation?

Example 13If x is irrational and y is irrational and x ≠ y, disprove the statement that x

y is irrational.

SolutionAn irrational number is a real number that cannot be expressed as a fraction.

Let =x 2

Let =y 8

y can also be expressed as 2 2.

=xy

22 2

= 12

12

is rational so if x is irrational and y is irrational and x ≠ y

then xy

is not necessarily irrational.

11 Proof

296


3 Harrison says that the expression x2 + 8x + 15 is positive or zero for all integer values of x. Prove by counter example that the expression x2 + 8x + 15 is not positive or zero for all integer values of x.

4 Five people, Alf, Bukunmi, Carlos, Dupika and Elsa, are to be seated around a circular table for a dinner party. Alf is to sit next to Bukunmi. Carlos must not sit next to Dupika but must sit next to Elsa. Carlos says, ‘If Alf and Bukunmi are sitting together, then that means I have to sit next to Dupika.’ Ignoring rotations and reflections, prove by counter example that Carlos is incorrect.

5 Elaine is learning her times tables. She observes that when she multiplies 3 by 11 the answer is 33 and when she multiplies 5 by 7 the answer is 35, and suggests that the product of any two prime numbers is odd. Find a counter example to disprove her suggestion.

6 Disprove the statement that if x and y are positive integers and x + y is even then x and y must both be even.

7 If n is prime, disprove that for all positive integers n is odd.

8 If x is irrational and y is irrational and x ≠ y, disprove the statement that xy is irrational.

9 Disprove the statement that, if x and y are real numbers, if x2 = y2 then x = y.

PF

PS

PF

PS

4

5

6

7 7

PF

PS

PF

PS

PF

PS

8

9 9

PF

PS

PF

PS


297


11 Proof

Summary of key points

› Questions involving proof are fl agged throughout this book.

› Proof by deduction is the drawing of a conclusion by using the general rules of mathematics.

› Proof by exhaustion requires splitting the statement into a fi nite, exhaustive set of cases, all of which are tested:

› In any proof by exhaustion, it is necessary to be systematic and to list all the possibilities in as structured an order as you can.

› Disproof by counter example:

› A counter example is an exception to the rule.

Exam-style questions 11 page 552

1 Prove that (3n + 5)2 − (3n − 5)2 is a multiple of 12 for all positive integer values of n. [4 marks]

2 Elif has been asked to arrange the letters in the word SUMS. She has written MUSS and SMUS. Prove by exhaustion that there are 12 distinct ways of arranging the letters in the word SUMS. [3 marks]

3 Suppose x and y are even positive integers less than 8. Prove by exhaustion that their difference is divisible by 2. [5 marks]

4 Prove that the sum of n consecutive integers is a multiple of n when n is odd but not when n is even. [5 marks]

5 A number is palindromic if it reads the same backwards and forwards, for example 252 and 1001. Show by exhaustion that 343 is the only palindromic three-digit cube number. [3 marks]

6 Jayne writes that sin 2q = 2 sin q for any angle q. Prove by counter example that sin 2q does not always equal 2 sin q. [2 marks]

7 Show that the statement ‘n2 − n + 1 is a prime number for all values of n’ is untrue for n > 1. [4 marks]

8 Use the diagram to prove Pythagoras’ theorem algebraically.

[6 marks]

9 Disprove the statement that if x and y are real numbers, if x2 > y2 then x > y. [3 marks]

1

2

3

PF

PS

PF

PS

PF

PS

4

5

6

7

PF

PS

PF

PS

PF

PS

PF

PS

8

9 9

PF

PS

PF

PS

a b

b a

a

a

b

b

c

c

c

c

298


Documents

Edexcel A-level Mathematics - Collins Educationresources.collins.co.uk/PFD/A-Level Maths_Book 1Chapter11_pre-endorsement.pdfconjecture and derive results about angles and sides, including