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Solution Manual for Concise Introduction To Pure Mathematics, 4th Edition

Solution Manual for Concise Introduction To Pure Mathematics, 4th Edition is your guide to textbook mastery, offering detailed solutions to every chapter's exercises.

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    solutions MA nu A l F o R A Concise Introduction to Pure Mathematics, Fourth Edition Martin l iebeck by K25658_SM_Cover.indd 1 17/12/15 1:17 pm

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    Chapter 1 Solutions 1. TTFTFFTFT 2. (a) Notice that B = { x | x real , − 2 < x < 2 } and D = { 0 } . So the only pair, neither of which is contained in the other, is D , E . (b) X cannot be D , but could be B , C or E . 3. (a) Not valid. (b) Valid. Let C be the statement “I eat chocolate” and D the statement “I am depressed”. We are given C ⇒ D . Therefore ¯ D ⇒ ¯ C . We are told that ¯ D is true, so I am not eating chocolate. (Actually I am, it’s delicious.) (c) Valid. Let E be the statement “movie was made in England”, W be “movie is worth seeing”, I be “movie reviewed by Ivor”. Then we are given ¯ W ⇒ ¯ E and W ⇒ I . Hence ¯ I ⇒ ¯ W ⇒ ¯ E . Therefore the movie was not made in England. 4. The implications between these statements are (a) ⇔ (d) ⇔ (e). 5. (a) True (b) False (c) False (d) False: for example if a = b = 2 then ab is a square but a , b are not. (e) True: if a , b are squares then a = m 2 , b = n 2 with m , n integers, so ab = m 2 n 2 = ( mn ) 2 , a square. 6. (a) By contradiction: suppose √ 6 − √ 2 ≤ 1. Squaring both sides gives 6 + 2 − 2 √ 12 ≤ 1, hence 7 ≤ 2 √ 12, hence 49 ≤ 48, a contradiction. Therefore √ 6 − √ 2 > 1. (b) Suppose n 2 is even. If n is odd then n 2 is odd, by Example 1.2. Hence n 1 K25658_SM_Cover.indd 5 17/12/15 1:17 pm

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    2 SOLUTIONS TO CHAPTER 1 cannot be odd, so n is even. (c) Observe that m 3 − m = m ( m 2 − 1 ) = ( m − 1 ) m ( m + 1 ) , a product of three consecutive integers. Of three consecutive integers, exactly one must be divis- ible by 3, and at least one must be even. Hence their product is divisible by 3 × 2 = 6. 7. We produce counterexamples to each of these statements: (a) 2 4 − 2 = 14 is not divisible by 4. Hence the statement is false when n = 2 , k = 4. (b) 7 is not the sum of three squares. 8. We are given that 8881 is not a prime, so there are integers a , b greater than 1 such that 8881 = ab . Choose such a factorization with a as small as possible. Then a is prime (otherwise it would have a smaller factor, contradicting the choice of a being as small as possible). Also a ≤ b , so 8881 = ab ≥ a 2 , and so a ≤ √ 8881. Since 97 2 > 8881, this implies that a < 97, and since the next prime down from 97 is 89, this means that a ≤ 89. 9. (a) Negation: ∃ n ∈ Z such that n is a prime number and n is even. This negation is true, as n = 2 is an even prime. (b) Negation: ∃ n ∈ Z such that ∀ a , b , c , d , e , f , g , h , ∈ Z , n  = a 3 + b 3 + c 3 + d 3 + e 3 + f 3 + g 3 + h 3 . The original statement is true — i.e. every integer is the sum of 8 cubes. This is tricky to prove. Here is a proof. Let n be any integer. By Q6(c), n 3 − n is a multiple of 6. Write n 3 − n = 6 x with x ∈ Z . Observe that ( x + 1 ) 3 + ( x − 1 ) 3 = 2 x 3 + 6 x , and hence n = n 3 − 6 x = n 3 − ( x + 1 ) 3 − ( x − 1 ) 3 + 2 x 3 . Thus n is in fact the sum of 5 cubes. (Notice that it is crucial in this proof to allow negative cubes. If we insist on all of the cubes a 3 , . . . , h 3 being non-negative (i.e. a ≥ 0 , . . . , h ≥ 0), then the negation is true — for example, 23 is not the sum of 8 non-negative cubes.) (c) Negation: ∀ x ∈ Z , ∃ n ∈ Z such that x = n 2 + 2. This negation is false, for example x = 4 is not of the form n 2 + 2. So the original statement is true. (d) Negation: ∀ x ∈ Z , ∃ n ∈ Z such that x = n + 2. This negation is true: take n = x − 2. (e) Negation: ∃ y ∈ { x | x ∈ Z , x ≥ 1 } such that 5 y 2 + 5 y + 1 is not prime. This negation is true: e.g. if y = 12 then 5 y 2 + 5 y + 1 = 781, which is not prime as it is a multiple of 11. K25658_SM_Cover.indd 6 17/12/15 1:17 pm

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    SOLUTIONS TO CHAPTER 1 3 (f) Negation: ∃ y ∈ { x | x ∈ Z , x 2 < 0 } such that 5 y 2 + 5 y + 1 is not prime. This negation is false, as the set in question is the empty set. So the original statement is true. 10. Let x be a real number which is less than every positive real number. Suppose for a contradiction that x is positive, i.e. x > 0. Then x > 1 2 x > 0. But this means that x is not less than the positive real number 1 2 x , which is a contradiction. Therefore x cannot be positive. 11. (a) Let x = a 1 . Then a x = 3. Since a x > a x − 1 > ... > a 1 ≥ 1, it follows that x ≤ 3. Easily see that only x = 2 is possible. Thus a 1 = 2. (b) Since a 1 = 2, have a 2 = a a 1 = 3, and then a 3 = a a 2 = 6. Therefore a 6 = a a 3 = 9. As a 3 < a 4 < a 5 < a 6 , follows that a 4 = 7 , a 5 = 8. Then a 7 = a a 4 = 12, a 8 = 15, a 9 = 18. (c) Work as in (b) to see that a 10 , a 11 , .... is 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 30 , 33 , 36 , 39 , 42 , 45 , 48 , 51 , 54 , 55 , 56 , ......, 81 , 84 , 87 , ...., 162 , 163 , 164 , ...., 243 , 246 , 249 , ..... and so on. In particular, a 100 = 181. K25658_SM_Cover.indd 7 17/12/15 1:17 pm

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    Chapter 2 Solutions 1. (a) By contradiction. Suppose √ 3 is rational, so √ 3 = m n where m , n are integers and m n is in lowest terms. Squaring, we get m 2 = 3 n 2 . Thus m 2 is a multiple of 3, and so by Example 1.3, m is a multiple of 3. This means m = 3 k for some integer k . Then 3 n 2 = m 2 = 9 k 2 , so n 2 = 3 k 2 . Therefore n 2 is a multiple of 3, hence so is n . We have now shown that both m and n are multiples of 3. But m n is in lowest terms, so this is a contradiction. Therefore √ 3 is irrational. (b) By contradiction again. Suppose √ 3 = r + s √ 2 with r , s rational. Squar- ing, we get 3 = r 2 + 2 s 2 + 2 rs √ 2. If rs  = 0 this gives √ 2 = 3 − r 2 − 2 s 2 2 rs . Since r , s are rational this implies that √ 2 is rational, which is a contradiction. Hence rs = 0. If s = 0 then r 2 = 3, so r = √ 3, contradicting the fact that √ 3 is irra- tional by (a). Therefore r = 0 and 3 = 2 s 2 . Writing s = m n in lowest terms, we have 3 n 2 = 2 m 2 . Now the proof of Proposition 2.3 shows that m and n must both be even, which is a contradiction. 2. (a) Suppose x = √ 2 + √ 3 / 2 is rational. Then x 2 = 2 + 3 2 + 2 √ 3, hence √ 3 = 1 2 ( x 2 − 7 2 ) . As x is rational this implies √ 3 is rational, a contradiction by Q1(a). Hence √ 2 + √ 3 / 2 is irrational. (b) By (a), 1 + √ 2 + √ 3 / 2 is the sum of a rational and an irrational, hence is irrational by Proposition 2.4(i). (c) We have 2 √ 18 − 3 √ 8 + √ 4 = 6 √ 2 − 6 √ 2 + 2 = 2, which is rational. (d) Let x = √ 2 + √ 3 + √ 5. Then ( x − √ 2 ) 2 = ( √ 3 + √ 5 ) 2 , which gives x 2 − 6 − 2 √ 15 = 2 x √ 2. Squaring again, ( x 2 − 6 ) 2 + 60 − 4 ( x 2 − 6 ) √ 15 = 8 x 2 , hence √ 15 = (( x 2 − 6 ) 2 − 8 x 2 + 60 ) / 4 ( x 2 − 6 ) . Therefore if x is rational, then so is √ 15. But √ 15 is irrational by the hint in the question. Hence so is x . (e) This is sneaky one. Observe that ( √ 2 + √ 3 ) 2 = 5 + 2 √ 6, so √ 5 + 2 √ 6 = √ 2 + √ 3, and hence √ 2 + √ 3 − √ 5 + 2 √ 6 = 0, which is rational. 5 K25658_SM_Cover.indd 9 17/12/15 1:17 pm

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    6 SOLUTIONS TO CHAPTER 2 3. (a) True: if x = m / n and y = p / q are rational, so is xy = mp / nq . (b) False: for a counterexample take the irrationals √ 2 and − √ 2. Their product is − 2, which is rational. (c) False: the product of the two irrationals √ 2 and 1 + √ 2 is √ 2 + 2, which is irrational. (d) True: we prove it by contradiction. Suppose there is a rational a  = 0 and an irrational b such that c = ab is rational. Then b = c a , and since a and c are rational, this implies that b is rational, a contradiction. 4. (a) Let c = x + a x + b and suppose c is rational. Then x + a = c ( x + b ) , which gives x ( c − 1 ) = a − bc . If c  = 1 then x = a − bc c − 1 , which is rational since a , b , c are rational. As x is irrational, this implies that c = 1, hence a = b . (b) Let c = x 2 + x + √ 2 y 2 + y + √ 2 and suppose c is rational. Multiplying up gives √ 2 ( c − 1 ) = x 2 + x − cy 2 − cy . If c  = 1 this gives √ 2 = x 2 + x − cy 2 − cy c − 1 , which is rational. Hence c = 1, which implies x 2 + x = y 2 + y . This yields ( x − y )( x + y + 1 ) = 0, hence either x = y or x + y = − 1. 5. By contradiction. Let α = √ 2 + √ n , and suppose α is rational. Then α − √ 2 = √ n . Squaring both sides, α 2 + 2 − 2 α √ 2 = n , so 2 α √ 2 = α 2 + 2 − n . Since clearly α  = 0, we can divide through by 2 α to get √ 2 = ( α 2 + 2 − n ) / 2 α . As α is rational, this implies that √ 2 is rational, which is a contradiction. Hence α is irrational. 6. Let a and b be two different real numbers with b > a . Choose a positive integer n such that b − a > 1 n . Then there is a rational of the form m n lying between a and b . Also, choose a positive integer m such that b − a > √ 2 m . Then there is a number of the form k √ 2 m lying between a and b , where k is an integer; by Proposition 2.4(ii), k √ 2 m is irrational unless k = 0, in which case a < 0 and b > 0 and we apply the above argument replacing a by 0. 7. Let a n = √ n − 2 + √ n + 2. Then a 2 n = ( n − 2 )+( n + 2 )+ 2 √ ( n − 2 )( n + 2 ) = 2 n + 2 √ n 2 − 4. We are given that a n is an integer. This implies √ n 2 − 4 is ra- tional. By the hint given, this means that n 2 − 4 must be a perfect square, i.e. n 2 − 4 = m 2 for some integer m . Then n 2 − m 2 = 4. Staring at the list of squares 0 , 1 , 4 , 9 , 16 , . . . , we see that the only way the two squares n 2 , m 2 can differ by 4 is to have n 2 = 4 , m 2 = 0. Hence n = 2, so a n = 2. 8. Let r , b , g be the numbers of red, blue and green salamanders at some point in time. If a red and a green meet, these numbers change to r − 1 , b + 2 , g − 1, so K25658_SM_Cover.indd 10 17/12/15 1:17 pm

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    SOLUTIONS TO CHAPTER 2 7 the difference between b and r is increased by 3; if two reds meet, the numbers change to r − 2 , b + 1 , g + 1, and again the difference between b and r increases by 3; and so on — you can easily check that whenever two salamanders meet, the difference between b and r either stays the same, or increases or decreases by 3. Initially, r − b is 15 − 7 = 8. This cannot be changed by adding and subtracting multiples of 3 to 30 − 0 = 30. Hence it is not possible for all the salamanders to be red at some point. K25658_SM_Cover.indd 11 17/12/15 1:17 pm

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    Chapter 3 Solutions 1. 1812 999 . 2. We know that √ 2 is irrational. By Proposition 3.5, periodic decimal expres- sions are rational, hence irrationals have non-periodic decimal expressions. Therefore √ 2 is non-periodic. 3. (a) This number is 0 . 12340123401234 .... which is periodic, hence rational. (b) Suppose the given number is rational. Then the decimal must be pe- riodic, so for some n there is a sequence a 1 a 2 ... a n of n digits which repeats itself. But if we go far enough along the decimal, there will be a sequence of at least 2 n consecutive zeroes. The sequence a 1 a 2 ... a n must appear within this sequence of zeroes, hence must be just 000 ... 0. This means the decimal ends 0000 .... (going on for ever), which is plainly false. Hence (by contradiction) the number is irrational. (c) This is 1 . 1001000010000001 .... . There are ever-increasing sequences of zeroes, so the argument for (b) shows that this number is irrational. 4. Let α be the cube root of 2. Since ( 1 . 25 ) 3 = ( 5 4 ) 3 = 125 64 < 2, we know α > 1 . 25. Also ( 1 . 3 ) 3 > 2, so α < 1 . 3. Therefore α is 1 . 3, correct to 1 decimal place. 5. Let x = 100 9899 . It’s easy to check that 100 + x + 100 x = 10000 x . Writing this equation out in full, using the fact that x = 0 . 010102 . . . , and spacing it out a bit to make the point, we get 100 + 0 . 01 01 02 03 05 ... + 1 . 01 02 03 05 08 ... = 101 . 02 03 05 08 13 ... 9 K25658_SM_Cover.indd 13 17/12/15 1:17 pm

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    10 SOLUTIONS TO CHAPTER 3 You can see that, at least near the beginning of the decimal expansion of x , the numbers formed by looking at successive pairs of digits are being forced to obey the rule defining the Fibonacci sequence. Eventually one has to start carrying digits and the whole thing goes wrong, but it does work for a while (in fact up to the term 55 in the Fibonacci sequence). If you try for example x = 1000 / 998999, then it works for even longer. 6. Consider the proof of Proposition 3.4. If one of the remainders is 0, then the decimal expression ends with repeated zeroes and the period is 1. If not, the remainders are all among 1 , 2 , . . . , n − 1, and so the period is at most n − 1. The first few n for which the period is n − 1 are n = 2 , 7 , 17 , 19 , 23 , 29. Notice that they are prime numbers! You will find an explanation for this in Chapter 26 (see Proposition 26.8 and Exercise 14). 7. Suppose x = m n is a rational which has decimal expression ending in repeat- ing zeroes. Say x = a 0 . a 1 a 2 . . . a k 000 . . . . Then x = A 10 k , where A is an integer (in fact A = a k + 10 a k − 1 + . . . + 10 k a 0 ). So x = A 2 k 5 k . Cancelling out common factors of A and 2 k 5 k , we see that the denominator n is of the form 2 a 5 b . Conversely, if n = 2 a 5 b , then m n is equal to p 10 k for some integers p , k , so the decimal expression for m n has repeating zeroes from the k + 1 th position at least. 8. Suppose 1 n has decimal expansion of period 1, and let r be the repeating digit in the decimal, repeating from the k + 1 th decimal place. Then as in the previous solution, 1 n = A 10 k + r 9 10 − k − 1 . So the denominator n must divide 9 · 10 k + 1 . Hence n = 2 a 5 b 3 c where a , b are arbitrary and c ≤ 2. Conversely, for such numbers n the period of 1 n is 1. K25658_SM_Cover.indd 14 17/12/15 1:17 pm

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    Chapter 4 Solutions 1. ( 50 ) 3 / 4 ( 5 √ 2 ) − 1 / 2 = ( 2 . 5 2 ) 3 / 4 ( 5 . 2 − 1 / 2 ) − 1 / 2 = 2 3 / 4 5 3 / 2 5 − 1 / 2 2 1 / 4 = 2 1 5 1 = 10. 2. This is again 10. 3. 2 617 ; 3 ( 3 332 ) . 4. 2 1 / 2 3 1 / 3 = ( 2 3 ) 1 / 6 ( 3 2 ) 1 / 6 = ( 2 3 3 2 ) 1 / 6 . So take n = 72 , t = 1 / 6. 5. Note that 10000 100 = ( 100 2 ) 100 = 100 200 . So 100 10000 is bigger. Suppose 2 1 / 2 ≥ 3 1 / 3 . Taking 6th powers, this implies 2 3 ≥ 3 2 , which is false. Hence 2 1 / 2 < 3 1 / 3 . 6. Squaring both sides gives x +( 2 − 2 x ) − 2 x 1 / 2 ( 2 − 2 x ) 1 / 2 = 1, hence 2 x 1 / 2 ( 2 − 2 x ) 1 / 2 = − x + 1; squaring again gives 4 x ( 2 − 2 x ) = x 2 − 2 x + 1, hence 9 x 2 − 10 x + 1 = 0, which has solutions x = 1 and x = 1 9 . Of these, only x = 1 satisfies the original equation. 7. x + y ≥ 2 √ xy ⇔ ( x + y ) 2 ≥ 4 xy ⇔ x 2 + y 2 + 2 xy ≥ 4 xy ⇔ ( x − y ) 2 ≥ 0, which is true. Equality holds when x = y . 8. Take a = b = 2 , c = 3 and d = 2 , e = 1, f = any. 8. Taking sixth powers of both sides of the equation y 4 / 3 = x 5 / 6 gives y 8 = x 5 . Hence y = 2 5 , x = 2 8 is a solution. It is the smallest solution, since if 1 < y < 2 5 and y is an integer, then x = y 8 / 5 is not an integer. So the bill was £ 512 . 32. Pretty expensive meal. 11 K25658_SM_Cover.indd 15 17/12/15 1:17 pm

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    Chapter 5 Solutions 1. If y < 0 then − y > 0, so x . ( − y ) > 0, hence − xy > 0, hence xy < 0. If a > b > 0 then a − b > 0, 1 a > 0 and 1 b > 0 (using Example 4.4). Hence the product ( a − b ) . 1 a . 1 b > 0. This means a − b ab > 0, so 1 b − 1 a > 0, so 1 b > 1 a . 2. The inequality is x 2 + x + 1 − x − 1 2 x − 1 ≥ 0, which is x 2 ( 2 x + 1 ) 2 x − 1 ≥ 0. This is true provided x ≤ − 1 2 or x > 1 2 or x = 0. 3. This translates to 3 x 2 − 4 x + 1 < 0, which is ( 3 x − 1 )( x − 1 ) < 0, which is true provided 1 3 < x < 1. 4. (a) The inequality is the same as 2 x + 1 x − 3 < 0, i.e. 2 x 2 − 3 x + 1 x < 0. Factoring, this is ( 2 x − 1 )( x − 1 ) x < 0. So for the inequality to hold, either one or three of the numbers 2 x − 1 , x − 1 , x must be negative. This holds precisely when x < 0 or 1 2 < x < 1. (b) By the formula for the roots of a quadratic, there are two distinct real solutions if and only if “ b 2 − 4 ac > 0”, that is, t 2 − 12 > 0. This occurs when t > √ 12 or t < − √ 12. 5. Inequality translates to ( u − 1 )( v − 1 ) 1 + uv > 0, which is true if 0 < u , v < 1. The other ranges for which it is true are: u , v > 1; or u , v < 1 and uv > − 1; or u < 1 , v > 1 and uv < − 1; or u > 1 , v < 1 and uv < − 1. 6. If x , y ≥ 0 then | x | = x , | y | = y and | xy | = xy = | x | | y | . If x ≥ 0 , y < 0 then | x | = x , | y | = − y and | xy | = − xy = | x | | y | . The other cases are similar. 7. (i) The inequality says either x + 5 ≥ 1 or x + 5 ≤ − 1, i.e. either x ≥ − 4 or x ≤ − 6. (ii) The inequality is true if and only if ( x + 5 ) 2 > ( x − 2 ) 2 , i.e. x 2 + 10 x + 13 K25658_SM_Cover.indd 17 17/12/15 1:17 pm

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    14 SOLUTIONS TO CHAPTER 5 25 > x 2 − 4 x + 4, i.e. x > − 3 2 . (iii) Note that x 2 + 2 x + 3 = ( x + 1 ) 2 + 2, which is always positive, so the right hand side of the inequality is always x 2 + 2 x + 3. When x ≥ − 5, the inequality is x 2 + 2 x + 3 > x + 5, i.e. ( x + 2 )( x − 1 ) > 0, which is true provided x > 1 or x < − 2. And when x < − 5 the inequality is x 2 + 2 x + 3 > − x − 5, i.e. x 2 + 3 x + 8 > 0, which is always true. Hence the range for which the inequality holds is x > 1 or x < − 2. 8. Let x , y be the sides of a rectangle of area A = xy . The perimeter P is 2 ( x + y ) . By Example 5.14, P ≥ 4 √ xy = 4 √ A . The perimeter of a square of side √ A is equal to 4 √ A , so this is smallest possible perimeter. 9. First apply Example 5.14 taking x = ( a 1 a 2 ) 1 / 2 , y = ( a 3 a 4 ) 1 / 2 , to get ( a 1 a 2 a 3 a 4 ) 1 / 4 ≤ 1 2 (( a 1 a 2 ) 1 / 2 + ( a 3 a 4 ) 1 / 2 ) . By Example 5.14, ( a 1 a 2 ) 1 / 2 ≤ 1 2 ( a 1 + a 2 ) and ( a 3 a 4 ) 1 / 2 ≤ 1 2 ( a 3 + a 4 ) . Apply- ing these to the above inequality gives the result. To prove the Arithmetic-Geometric Mean Inequality for n = 8, just repeat the above argument: ( a 1 a 2 . . . a 8 ) 1 / 8 ≤ 1 2 (( a 1 . . . a 4 ) 1 / 4 + ( a 5 . . . a 8 ) 1 / 4 ) by Example 5.14 ≤ 1 2 ( 1 4 ( a 1 + · · · + a 4 ) + 1 4 ( a 5 + · · · + a 8 )) = 1 8 ( a 1 + · · · + a 8 ) . Repeating this argument we can prove the Arithmetic-Geometric Mean In- equality for any case where n is a power of 2. 10. (i) In Question 9, take a 1 = x 4 , a 2 = a 3 = a 4 = y 4 . (ii) As in (i) we have x 3 y ≤ 3 4 x 4 + 1 4 y 4 . Add this to the inequality in (i). 11. Squaring both sides of x + y + z = 0, we get x 2 + y 2 + z 2 + 2 ( xy + yz + xz ) = 0. Hence xy + yz + xz = − 1 2 ( x 2 + y 2 + z 2 ) ≤ 0. 12. Let n be a 1000-digit number. Then n ≥ 10 999 . But the sum of the 1000th powers of the digits of n is at most 1000 × 9 1000 , which is much smaller than 10 999 . So n cannot be a Smallbrain number. Remark: Would this have worked for 100-digit numbers ? Where does the trick stop working ? K25658_SM_Cover.indd 18 17/12/15 1:17 pm

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