CONTENTS Introduction
v
1 – Algebra and functions 1: Manipulating algebraic expressions 1 1.1 Manipulating polynomials algebraically 1.2 Expanding multiple binomials 1.3 The binomial expansion 1.4 Factorisation 1.5 Algebraic division 1.6 Laws of indices 1.7 Manipulating surds 1.8 Rationalising the denominator Summary of key points Exam-style questions 1
3 6 11 14 16 22 24 26 28 29
2 – Algebra and functions 2: Equations and inequalities
31
2.1 2.2 2.3 2.4 2.5 2.6
Quadratic functions The discriminant of a quadratic function Completing the square Solving quadratic equations Solving simultaneous equations Solving linear and quadratic simultaneous equations 2.7 Solving linear inequalities 2.8 Solving quadratic inequalities Summary of key points Exam-style questions 2
3 – Algebra and functions 3: Sketching curves 3.1 Sketching curves of quadratic functions 3.2 Sketching curves of cubic functions 3.3 Sketching curves of quartic functions 3.4 Sketching curves of reciprocal functions 3.5 Intersection points 3.6 Proportional relationships 3.7 Translations 3.8 Stretches Summary of key points Exam-style questions 3
4 – Coordinate geometry 1: Equations of straight lines
32 35 37 39 42 46 47 52 56 56
58 59 63 67 72 79 82 86 91 97 98
100
4.1
Writing the equation of a straight line in the form ax + by + c = 0 4.2 Finding the equation of a straight line using the formula y – y1 = m(x – x1)
101 103
4.3
Finding the gradient of the straight line between two points 4.4 Finding the equation of a straight y−y x−x line using the formula y − y1 = x − x1 2 1 2 1 4.5 Parallel and perpendicular lines 4.6 Straight line models Summary of key points Exam-style questions 4
105 109 112 118 124 124
5 – Coordinate geometry 2: Circles
126
5.1 Equations of circles 5.2 Angles in a semicircle 5.3 Perpendicular from the centre to a chord 5.4 Radius perpendicular to the tangent Summary of key points Exam-style questions 5
127 134 138 144 151 151
6 – Trigonometry
154
6.1 Sine and cosine 6.2 The sine rule and the cosine rule 6.3 Trigonometric graphs 6.4 The tangent function 6.5 Solving trigonometric equations 6.6 A useful formula Summary of key points Exam-style questions 6
155 158 166 169 172 174 177 177
7 – Exponentials and logarithms
179
7.1 The function ax 7.2 Logarithms 7.3 The equation ax = b 7.4 Logarithmic graphs 7.5 The number e 7.6 Natural logarithms 7.7 Exponential growth and decay Summary of key points Exam-style questions 7
180 184 189 191 196 201 204 209 209
8 – Differentiation
212
8.1 The gradient of a curve 8.2 The gradient of a quadratic curve 8.3 Differentiation of x² and x3 8.4 Differentiation of a polynomial 8.5 Differentiation of xn 8.6 Stationary points and the second derivative 8.7 Tangents and normals Summary of key points Exam-style questions 8
213 217 222 225 228 232 237 240 240
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CONTENTS
9 – Integration 9.1 Indefinite integrals 9.2 The area under a curve Summary of key points Exam-style questions 9
10 – Vectors 10.1 Definition of a vector 10.2 Adding and subtracting vectors 10.3 Vector geometry 10.4 Position vectors Summary of key points Exam-style questions 10
11 – Proof 11.1 Proof by deduction 11.2 Proof by exhaustion 11.3 Disproof by counter example Summary of key points Exam-style questions 11
12 – Data presentation and interpretation 12.1 Measures of central tendency and spread 12.2 Variance and standard deviation 12.3 Displaying and interpreting data Summary of key points Exam-style questions 12
13 – Probability and statistical distributions 13.1 Calculating and representing probability 13.2 Discrete and continuous distributions 13.3 The binomial distribution Summary of key points Exam-style questions 13
14 – Statistical sampling and hypothesis testing 14.1 Populations and samples 14.2 Hypothesis testing Summary of key points Exam-style questions 14
15 – Kinematics 15.1 The language of kinematics 15.2 Equations of constant acceleration 15.3 Vertical motion
243 244 249 258 258
260 261 266 269 278 283 283
285 286 289 294 298 298
299 301 308 317 339 339
342 343 350 359 365 365
368 369 374 389 389
391 392 395 402
15.4
Displacement–time and velocity–time graphs 15.5 Variable acceleration Summary of key points Exam-style questions 15
16 – Forces
408 416 422 422
425
16.1 Forces 16.2 Newton’s laws of motion 16.3 Vertical motion 16.4 Connected particles 16.5 Pulleys Summary of key points Exam-style questions 16
426 429 438 441 449 455 455
Exam-style extension questions
460
Answers*
487
1 Algebra and functions 1: Manipulating algebraic expressions 2 Algebra and functions 2: Equations and inequalities 3 Algebra and functions 3: Sketching curves 4 Coordinate geometry 1: Equations of straight lines 5 Coordinate geometry 2: Circles 6 Trigonometry 7 Exponentials and logarithms 8 Differentiation 9 Integration 10 Vectors 11 Proof 12 Data presentation and interpretation 13 Probability and statistical distributions 14 Statistical sampling and hypothesis testing 15 Kinematics 16 Forces
487 490 497 511 515 518 521 527 533 535 540 544 549 552 555 560
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 education@harpercollins.co.uk
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11
11
PROOF
If you are asked to ‘prove’ (or disprove) a mathematical statement, what does this actually mean? The Oxford English Dictionary defines 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 simplifies 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 flagged 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 specific 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
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11 PROOF
›
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.
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 INFORMATION Proof by deduction is the drawing of a conclusion by using the general rules of mathematics.
Example 1 Show that n2 − 8n + 17 is positive for any integer n.
Solution Complete the square. n2 − 8n + 17 = (n − 4)2 + 1 which is always positive for any integer n.
PROOF The phrase ‘show that’ indicates that a proof is required.
Example 2 A 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 12 . Show that x 2 − 13x + 30 = 0.
Solution There are two possibilities for the balls to be different. The first one could be green and the second red or the first one could be red and the second green. The probabilities of each will be the same. 286
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Proof by deduction
Since there are x green balls and (x + 6) balls in total, the probability that the first ball is green is x . x+6 If a green ball is taken (and not replaced), the probability that the second ball is red is 6 , since there is 1 fewer ball in the bag. x+5 The probability that the balls are different is therefore given by x × 6 × 2 since there are two ways this can happen. The x+6 x+5 question states that this probability is equal to 12 . Hence x × 6 × 2 = 12 x+6 x+5 Multiply both sides by 2. x × 6 ×4=1 x +6 x +5 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.
Example 3 Demonstrate that the sum of the squares of two consecutive odd positive integers is always even.
Solution Let 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.
11.1
Since there are (x + 6) balls to start with, when one ball is removed there will be (x + 5) balls in the bag.
PROOF When 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.
PROOF The words ‘demonstrate that’ have been used to indicate that a proof is required.
(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. 287
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11 PROOF
Example 4 Prove that the sum of any four consecutive integers a, b, c and d, where a < b < c < d, is equal to cd − ab.
Solution Since 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.
Consecutive integers increase by 1 from term to term.
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.
Exercise 11.1A PF
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.
PS PF
page 540
x cm
PS
(3x – 1) cm
(x + 3) cm
(2x + 5) cm Prove that 4x2 + 7x − 186 = 0 and hence deduce that the perimeter of the shape is 68 cm.
288
M
Modelling
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PS
Problem solving
PF
Proof
CM
Communicating mathematically
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11.2
Proof by exhaustion
PF
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.
PS PF PS PF PS PF PS
PF PS
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. PF
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.
PS
Then So Factorising: Dividing leaves
ab = b2 ab − a2 = b2 − a2 a(b − a) = (b + a)(b − a) a=b+a
As b = a, substituting gives a = a + a So
a = 2a
If a = 1 then
1 = 2.
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 finite number of cases
›
prove that the statement is true for all of the cases.
prove that the list of cases identified is exhaustive – that is, that there are no other cases
KEY INFORMATION Proof by exhaustion requires splitting the statement into a finite, exhaustive set of cases, all of which are tested. 289
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11 PROOF
Example 5 Suppose n is an integer between 24 and 28 inclusive. Prove that n is not a prime number.
Solution A 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.
PROOF To use proof by exhaustion, you need to list all of the possible integers and test each case, one by one.
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.
Example 6 Elif 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.
Solution If 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. Elifâ&#x20AC;&#x2122;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. 290
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Proof by exhaustion
11.2
Example 7 Suppose x and y are even positive integers less than 8. Prove that their sum is divisible by 2.
Solution x = {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
PROOF
Case 9: x + y = 6 + 6 = 12 = 2 × 6
Proof by exhaustion is really only a feasible method to use when the number of cases to be tested is small.
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.
Example 8 Prove by exhaustion that there are only two distinct triangles with sides of integer length that have a perimeter of 10.
Solution The 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
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11 PROOF
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}.
There cannot be any possibilities which only have integers of 4 or more because the sum would be at least 12.
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 2, {1, 2, 7}: 1 + 2 < 7, so two sides sum to less than the third side. Case 3, {1, 3, 6}: 1 + 3 < 6, 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 5, {2, 2, 6}: 2 + 2 < 6, so two sides sum to less than 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.
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.
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11.2
Proof by exhaustion
Example 9 Suppose n is an integer between 2 and 10 inclusive. Prove that n2 − 2 is not divisible by 5.
Solution Case 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. Case 3, n = 4: n2 − 2 = 42 − 2 = 14, which is not a multiple of 5. Case 4, n = 5: n2 − 2 = 52 − 2 = 23, which is not a multiple of 5. Case 5, n = 6: n2 − 2 = 62 − 2 = 34, which is not a multiple of 5. Case 6, n = 7: n2 − 2 = 72 − 2 = 47, which is not a multiple of 5. Case 7, n = 8: n2 − 2 = 82 − 2 = 62, which is not a multiple of 5. Case 8, n = 9: n2 − 2 = 92 − 2 = 79, which is not a multiple of 5. Case 9, n = 10: n2 − 2 = 102 − 2 = 98, which is not a multiple of 5. If n is an integer between 2 and 10 then n2 − 2 is not divisible by 5.
TECHNOLOGY Using a spreadsheet software package, you could generate the values for more cases and explore what the value of n is when n2 − 2 is first a multiple of 5.
Exercise 11.2A PF
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.
PS
PF
page 541
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.
PS PF PS PF PS PF PS PF PS PF PS
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11 PROOF
PF
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.
PS PF PS PF
10 Prove that if n is a positive integer less than 10 then n6 − n is a multiple of 2.
PS
11.3 Disproof by counter example To prove something is false it is only necessary to find 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’.
KEY INFORMATION A counter example is an exception to the rule.
Example 10 Sawda 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.
Solution If 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.
Stop and think
TECHNOLOGY Using a spreadsheet software package you could use the formula n2 − n + 5 to generate the terms in the sequence.
What is the first 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.
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Disproof by counter example
11.3
Example 11 Disprove that if x and y are consecutive odd numbers then one of x or y must be prime.
Solution Let 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 12 Georgina 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.
Solution The interior angle (in degrees) of a regular polygon is given by 180 − 360 , where n is the number of sides. n 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 − 360 = 128 74 , 7 which is not an integer, so the heptagon is a counter example.
Stop and think
Note that, since the formula is 180 − 360 , any value of n n which is not a factor of 360 will be a counter example.
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?
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11 PROOF
Example 13 If x is irrational and y is irrational and x ≠ y, disprove the statement that x is irrational. y
Solution An 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. 2 x = y 2 2 1 = 2 1 is rational so if x is irrational and y is irrational and x ≠ y 2 then x is not necessarily irrational. y
Stop and think
If x and y are irrational, can you generalise the qualities of x and y so that xy and x are rational? Can you prove your generalisation? y
Exercise 11.3A PS
1
page 543
Simon says that:
a all rectangles are squares
CM
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. PF
2
PS
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.
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11.3
Disproof by counter example
PF
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.
PS
PF PS
PF PS
PF PS PF PS PF PS PF PS
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11 PROOF
SUMMARY OF KEY POINTS
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Questions involving proof are flagged 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 finite, exhaustive set of cases, all of which are tested:
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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:
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A counter example is an exception to the rule.
EXAM-STYLE QUESTIONS 11 PF
page 543
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. Suppose x and y are even positive integers less than 8. Prove by exhaustion that their difference is divisible by 2.
[3 marks]
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3
PS PF
4
PS PF
5
PS PF
6
PS PF
7
PS PF
8
PS
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 marks]
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] 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] Show that the statement ‘n2 − n + 1 is a prime number for all values of n’ is untrue for n > 1. a b Use the diagram to prove Pythagoras’ theorem algebraically.
a
c b
[4 marks]
c c b
c a b PF
9
a
Disprove the statement that if x and y are real numbers, if x2 > y2 then x > y.
[6 marks] [3 marks]
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