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Solutions to the exercises

Solutions to the exercises
1.1 (b) (c) (d) (e) (f )
(a) True: -2 is an integer.
False: 5 is a natural number. False: 1.3 is the rational number False:
1 2 13 10

(b)
.

is not a natural number.

True: - is a real number. True: 2 is a rational number ( 2 ). 1

1.2 (a) True: 1 is a member of the set given. (b) True: the set {-9} is a member of the set given. (c) False: the number 9 belongs to the set given,
but the set {9} does not.

(c)

(d) False: (0, 1) is not a member of the set given. (e) True: the set {0, 1} is a member of the set given.

1.3 (b) (c) (d) (e)

(a) {k Z : -2 < k < 1000}
{x R :2 x 7} {x Q : x > 0 and x2 > 2} {2n : n N} {2k : k Z}

(d)

1.4 (a) l = {(x, y) R2 : y = 2x +5} (b) 1.7 (a) {(x, y) R2 :0 x 2, 1 y 3} (b) {(x, y ) R2 : x 0, y = 2x2 +1} 1.8 (a) The set B consists of the solutions of the equation x2 + x - 6 = 0, which we can write as (x - 2)(x +3) = 0. So B = {2, -3} = A.
(b) A = {k Z : k is odd and 2 < k < 10}
= {3, 5, 7, 9}, B = {n N = {2, 3, 5, Hence A = B , because 9 A : n is a prime number and n < 10} 7}. either because 2 B but 2 A, or / but 9 B . /

1.5 (a) C = {(x, y) R2 :(x - 1)2 +(y +4)2 = 9} (b)

1.6 (a)

of each point of A: 5 - 4 â 2 = -3, 1 - 4 â 1 = -3, -3 - 4 â 0 = -3. This shows that each element of A is an element of B , so A B .

1.9 (a) We calculate x - 4y using the coordinates

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Unit I2 Mathematical language

(b) The set A is the interior of the unit circle, and
B is the half-plane consisting of all p oints with negative y -coordinate. So A B , because, for example, the point ( 1 , 1 ) b elongs to A but not to B . 22

1.15 (a) (1, 7) [4, 11] = (1, 11]. (b) The domain of f is the set
{x R : x2 - 9 > 0} = {x R : x < -3 or x > 3}, that is, (-, -3) (3, ).

1.10 We showed that A B in the solution to Exercise 1.9(a). Also, for example, the point (9, 3) lies in B , since 9 - 4 â 3 = -3, but does not lie in A. Therefore A is a prop er subset of B . 1.11 First we show that A B .

Let (x, y ) A; then x = t2 and y = 2t, for some
t R. Hence y 2 = 4t2 = 4x. So (x, y ) B , and so
A B .
Next we show that B A.
Let (x, y ) B . We must show that (x, y ) A. Let
2 t = 1 y ; then x = 1 y 2 = 1 y = t2 , and y = 2t. So 2 4 2 (x, y ) = (t2 , 2t) A, and so B A.
Since A B and B A, it follows that A = B .


(c)

1.16 (a) (1, 7) [4, 11] = [4, 7).
(b)

1.12
k Subsets of {1, 2, 3, 4} of size k 0 1 2 3 4
{1}, {2}, {3}, {4}
{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}
{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}
{1, 2, 3, 4}


1.17 (a) (1, 7) - [4, 11] = (1, 4) and [4, 11] - (1, 7) = [7, 11].
(b)

The table shows that the set {1, 2, 3, 4} has 1 + 4 + 6 + 4 + 1 = 16 subsets in all.

1.13
10 3

10 2 =

=

10! 3! 7! 11! 11 = 3 3! 8! 10 Hence 2

10 â 9 10! = = 5 â 9 = 45, 2! 8! 2 10 â 9 â 8 = = 10 â 3 â 4 = 120, 3â2 11 â 10 â 9 = = 11 â 5 â 3 = 165. 3â2 10 11 + = . 3 3

n! n and 1.14 (a) n - k = (n - k )! k ! n! n n n , so = . = n-k k k k !(n - k )!

1.18 (a) (b) True: (c) False: (d) True: (e) False: (f ) True: (g) False:

False: 0 is not a natural number. 0 is a rational number. -0.6 is a real number. 37 is an integer. 20 is not a member of the set given. the set {1, 2} is the same as the set {2, 1}. does not contain any elements.

(b) We can interpret the identity as follows.
n k is the number of ways of choosing k elements n from n, which is the same as , the number n-k of ways of excluding k elements from n.

1.19 (a) The elements are 3, 4, 5, 6. Note that 2 and 7 are not included.
(b) The elements are -1, -4. These are the solutions of the equation. (c) The only element is 5. The equation has two solutions, -5 and 5, but only 5 N.

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Solutions to the exercises

1.20 In each case, you may have found a different expression for the same set.
(a) {k Z : -20 < k < -3} (b) {3k : k Z, k = 0} (c) {x R : x > 15}

(b)

1.21 (a)
The circle is not part of the set.

(c)

(b)
The edges of the square belong to the set. equation (x +2)2 +(y - 3)2 = 13, so A B . A is not a subset of B .

1.23 (a) (0, 0), (0, 6) and (-4, 6) all satisfy the
(b) The point (1, 0) belongs to A but not to B , so (c) If x = 2 cos t and y = 3 sin t, then
y2 x2 + = 1, 4 9

(c)

so A B .

an arbitrary element of A; then x2 +4y 2 < 1. Since x2 0 for all x R, this implies that 4y 2 < 1, and hence y 2 < 1 . Hence y < 1 . Thus (x, y ) B . 4 2 To confirm that A is a proper subset of B , we must show that there is an element of B that does not lie in A. The p oint (1, -1), for example, lies in B , since -1 < 1
, but does not lie in A, since 2 12 +4(-1)2 = 5,
which is not less than 1. Therefore A is a proper subset of B .

1.24 We must first show that A B . Let (x, y) be

1.22 (a)

1.25 (a) 1, -1, 2 are the three solutions of x3 - 2x2 - x +2 = 0, so A = B .
(b) We showed in the solution to Exercise 1.23(c) that A B . x2 y 2 + = 1, then (x/2,y /3) lies on the unit circle, If 4 9 so we can find t [0, 2] such that x/2 = cos t and y/3 = sin t. Hence x = 2 cos t and y = 3 sin t, so B A. Since A B and B A, it follows that A = B . (c) The set B contains some negative numbers
(for example, -1) which cannot b e expressed as p, q N. Hence A = B . p for q

The line is not part of the set.


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Unit I2 Mathematical language

1.26 (a) A B = {0, 2, 4, 5, 6},

A B = {4},
A - B = {0, 2}.


(b) A B = (-5, 17],

A B = [2, 3],
A - B = (-5, 2).


(c) A B = B,
A B = A,
A - B = .


Now x = y - 1 is real, since y 1, and satisfies f (x) = y , as required. (Alternatively, x = - y - 1 is real and satisfies f (x) = y .) Thus f (R) [1, ). Since f (R) [1, ) and f (R) [1, ), it follows that f (R) = [1, ), so the image of f is [1, ), as expected. The interval [1, ) is not the whole of the codomain R, so f is not onto.

2.1 (a) This is a translation of the plane that
moves each p oint to the right by 2 units and up by 3 units.

(b) This function is a reflection of the plane in the x-axis. This suggests that f (R2 ) = R2 . We know that f (R2 ) R2 , so we must show that f (R2 ) R2 .
Let (x ,y ) R2 . We must show that there exists (x, y ) R2 such that f (x, y ) = (x ,y ); that is, x = x, y = -y. Rearranging these equations, we obtain x = x , y = -y . Let (x, y ) = (x , -y ); then (x, y ) R2 and f (x, y ) = (x ,y ), as required. Thus f (R2 ) R2 . Since f (R2 ) R2 and f (R2 ) R2 , it follows that f (R2 ) = R2 , so the image of f is R2 , as expected. The codomain of f is also R2 , so f is onto.

(b) This is a reflection of the plane in the x-axis. (c) This is a rotation of the plane through /2
anticlockwise about the origin.

2.2 Only diagram (b) corresponds to a function.
Diagram (a) does not correspond to a function, as
there is no arrow from the element 3.
Diagram (c) does not correspond to a function, as
there are two arrows from the element 1.

2.3 The images of the elements of A are
f (0) = 9, f (1) = 8, f (2) = 7, f (3) = 6, f (4) = 5, f (5) = 4, f (6) = 3, f (7) = 2, f (8) = 1, f (9) = 0. So the image of f is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} = A.

2.4 Only diagram (a) corresponds to an onto
function.
Diagram (b) does not even corresp ond to a function,
as there is no arrow from the element 4.
Diagram (c) corresp onds to a function that is not onto, as there is no arrow going to the element 1.

2.6 Only diagram (c) corresponds to a one-one function. Diagram (a) corresp onds to a function that is not one-one, as there are two arrows going to the element 3. Diagram (b) does not even correspond to a function, as there is no arrow from the element 2. 2.7 (a) This function is not one-one since, for example, f (2) = f (-2) = 1 + 4 = 5.
(b) This function is a reflection of the plane in the x-axis, so we exp ect it to b e one-one. We now prove this algebraically. Suppose that f (x1 ,y1 ) = f (x2 ,y2 ); then (x1 , -y1 ) = (x2 , -y2 ). Thus x1 = x2 and -y 1 = -y2 . So y1 = y2 . Hence (x1 ,y1 ) = (x2 ,y2 ), so f is one-one.

2.5 (a) The sketch of the graph of f below suggests that f (R) = [1, ).

Let x R; then f (x) = 1 + x2 . Since x2 0, we have
1+ x2 1 and so f (R) [1, ).
We must show that f (R) [1, ).
Let y [1, ). We must show that there exists x R
such that f (x) = y ; that is, 1 + x2 = y .

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Solutions to the exercises

2.8 (a) In Exercise 2.7 we saw that f is not one-one, so f does not have an inverse function.
(b) In Exercise 2.7 we saw that f is one-one, so f
has an inverse function.
In Exercise 2.5 we saw that the image of f is R2 and,
for each (x ,y ) R2 , we have (x ,y ) = f (x , -y ). So f -1 is the function f -1 : R2 - R2 (x ,y ) - (x , -y ). This can be expressed in terms of x and y as f -1 : R2 - R2 (x, y ) - (x, -y ). (In this case, f -1 is actually equal to f .)

2.10 (a) The rule of g f is
(g f )(x) = g (f (x)) = g (-x)
= 3(-x)+ 1
= -3x +1.
Thus g f is the function g f : R - R x - -3x +1.

(b) The rule of f g is
(f g )(x) = f (g (x)) = f (3x +1)
= -(3x +1)
= -3x - 1.
Thus f g is the function f g : R - R x - -3x - 1.

(c) This is a linear function, which suggests that it is one-one. First we confirm this algebraically. Suppose that f (x1 ) = f (x2 ); then 8x1 +3 = 8x2 +3, so 8x1 = 8x2 , and hence x1 = x2 . Thus f is one-one, and so it has an inverse function. We now find the image of f . We susp ect that its image is R, so we now prove this algebraically. Let y b e an arbitrary element in R. We must show that there exists an element x in the domain R such that f (x) = y ; that is, 8x +3 = y. Rearranging this equation, we obtain y-3 . x= 8 This is in R and satisfies f (x) = y , as required. Thus the image of f is R. Hence f -1 is the function f -1 : R - R
y - 3
. y - 8 This can be expressed in terms of x as f -1 : R - R
x - 3
. x - 8

2.11 The rule of f g is
(f g )(x, y ) = f (g (x, y )) = f (-x, y ) = (-x, -y ). Thus f g is the function f g : R2 - R2
(x, y ) - (-x, -y ).
(In this case, f g = g f .)

2.12 The rule of g f is
(g f )(x) = g (f (x)) = g (3x +1)
3
= (3x +1)+ 2 1 . = x +1 The domain of g f is {x [-1, 1] : f (x) R -{-2}}. If x [-1, 1], then f (x) R -{-2} unless f (x) = -2. Now f (x) = -2 when 3x +1 = -2, that is, when x = -1. So the domain of g f is [-1, 1] -{-1} = (-1, 1]. Thus g f is the function g f :(-1, 1] - R 1 . x - x +1

2.9 (a) The function
g :[0, ) - R
x - |x|
is a restriction of f that is one-one. (There are many other possibilities.)

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Unit I2 Mathematical language

2.13 For each x R, we have
f (g (x)) = f (x - 3) = (x - 3) + 3 = x; that is, f g = iR . For each x R, we have g (f (x)) = g (x +3) = (x +3) - 3 = x; that is, g f = iR . Since g f = iR and f g = iR , it follows that g is the inverse function of f . through 3/2 anticlockwise ab out the origin.

Rearranging these equations, we obtain x = y and y = -x . So, for each (x ,y ) R2 , we have (x ,y ) = f (y , -x ), thus f (R2 ) R2 . Since f (R2 ) R2 and f (R2 ) R2 , it follows that f (R2 ) = R2 , so f is onto.

(b)


2.14 (a) This function is a rotation of the plane
(b) This function is a translation of the plane that

moves each p oint to the left by 2 units and up by 1 unit.

2.15 (a)

The graph above suggests that f (R) = R. We now
prove this algebraically.
Let x R; then 7 - 3x R, so f (R) R.
Let y R; then we want to find x R such that
y = 7 - 3x. 7-y , which is in R, so for each y R This gives x = 3 7-y . So f (R) R. we have y = f 3 Since f (R) R and f (R) R, it follows that f (R) = R, so f is onto.

(c) (b)

Exercise 1(c)) so we exp ect to find that f (R2 ) = R2 . Let (x, y ) R2 ; then f (x, y ) = (-y, x) R2 , so f (R2 ) R2 . We must now show that f (R2 ) R2 . Let (x ,y ) R2 . We must show that there exists (x, y ) R2 such that f (x, y ) = (x ,y ), that is, x = -y and y = x.

2.16 (a) This function is a rotation (see

The graph above suggests that f (R) = [-1, ). We now prove this algebraically. Let x R; then f (x) = x2 - 4x +3 = (x - 2)2 - 1 -1. So f (R) [-1, ).
Let y [-1, ). We must show that there exists
x R such that f (x) = y , that is,
x2 - 4x +3 = y. This means that (x - 2)2 = y +1, and we can take x = 2 + y + 1, which is in R since y +1 0.

64


Solutions to the exercises

So, for each y [-1, ), we have y = f (2 + y +1). Hence f (R) [-1, ). Since f (R) [-1, ) and f (R) [-1, ), it follows that f (R) = [-1, ). Since f (R) = R, f is not onto.

(c) The graph in the solution to Exercise 2.16(c)
suggests that f is not one-one. To show that this is so, we find two points in the domain of f with the same image. For example, f (0) = f (4) = 3,
so f is not one-one.


(d)

(d) The graph in the solution to Exercise 2.16(d) suggests that f is one-one. We prove this algebraically. Suppose that f (x1 ) = f (x2 ); then 2x1 +3 = 2x2 +3. Thus x1 = x2 , so f is one-one.

above suggests that f ([0, 1]) = [3, 5]. We this algebraically. 1]. Then 0 x 1, so 0 2x 2, so 5. Hence f (x) [3, 5]. Thus [3, 5]. 5]; then we want to find x [0, 1] such y-3 . Now that y = 2x + 3. This gives x = 2 y-3 3 y 5, so 0 y - 3 2, so 0 1. Thus 2 y-3 [0, 1], as required. So for each y [3, 5] we 2 y-3 y-3 , where [0, 1]. So have y = f 2 2 f ([0, 1]) [3, 5].
Since f ([0, 1]) [3, 5] and f ([0, 1]) [3, 5], it follows
that f ([0, 1]) = [3, 5]. So f is not onto.
The graph now prove Let x [0, 3 2x +3 f ([0, 1]) Let y [3,

2.18 (a) We have shown in Exercise 2.17(a) that f is one-one, so f has an inverse, and we have shown in the solution to Exercise 2.16(a) that (x ,y ) = f (y , -x ), so the inverse of f is the function f -1 : R2 - R2 (x ,y ) - (y , -x ). This can be expressed in terms of x and y as f -1 : R2 - R2 (x, y ) - (y, -x).
(b) We have shown in the solutions to Exercises 2.16(b) and 2.17(b) that f is one-one and that 7-y y=f , for y R. 3 Hence f has an inverse f -1 : R - R
7 - y
. y - 3 This can be expressed in terms of x as f -1 : R - R
7 - x
. x - 3 (c) We have shown in Exercise 2.17(c) that f is not one-one, so f does not have an inverse. (d) We have shown in the solutions to Exercises 2.16(d) and 2.17(d) that f is one-one and that the image of f is [3, 5]. We also showed that y-3 y=f , for y [3, 5]. 2 Hence f has an inverse f -1 :[3, 5] - [0, 1] y-3 . y - 2 This can be expressed in terms of x as f -1 :[3, 5] - [0, 1]
x - 3
x - . 2

2.17 (a) This function f is a rotation of the plane, so we exp ect f to be one-one. We now prove this algebraically. Suppose that f (x1 ,y1 ) = f (x2 ,y2 ); then (-y1 ,x1 ) = (-y2 ,x2 ), so -y1 = -y2 and x1 = x2 . Thus (x1 ,y1 ) = (x2 ,y2 ), so f is one-one.
(b) The graph in the solution to Exercise 2.16(b)
suggests that f is one-one. We prove this algebraically. Suppose that f (x1 ) = f (x2 ); then 7 - 3x1 = 7 - 3x2 . Thus x1 = x2 , so f is one-one.

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Unit I2 Mathematical language

2.19 (a) Since any number in the domain of g has an image under g which is in R, and hence in the domain of f , the domain of f g is the domain of g . Also, 1 1 =7-3 . (f g )(x) = f 2-4 2-4 x x Hence the composite is the function f g : R -{2, -2} - R 3 . x - 7 - 2 x -4 (b) Since any point in the domain of g has an image under g which is in R2 , and hence in the domain of f , the domain of f g is the domain of g . Also, (f g )(x, y ) = f (y, x) = (-x, y ). Hence the composite is the function f g : R2 - R2 (x, y ) - (-x, y ).
not a solution of the equation 3x + 5 = 0'.

3.4 (a) The two implications are `if the product mn is odd, then b oth m and n are odd', and `if both m and n are odd, then the product mn is odd'. Both implications are true, so the equivalence is true.
(b) The two implications are `if the product mn is even, then both m and n are even', and `if both m and n are even, then the product mn is even'. The first implication is false, and the second is true. The equivalence is false.

3.5 (a) Suppose that n is an even integer. Then n = 2k , where k Z, so n2 = (2k )2 = 4k 2 = 2(2k 2 ). Hence n2 is even, as required.
(b) Let m and n be multiples of k. Then m = ka and n = kb, where a and b are integers. Hence m + n = ka + kb = k (a + b). Since a + b is an integer, we deduce that m + n is a multiple of k , as required. (c) Suppose that one of the pair m, n is even and
the other is odd. Then one of them is equal to 2k and the other to 2l + 1, for some integers k and l. Then m + n = 2k +(2l +1) = 2(k + l)+ 1, which shows that m + n is odd.

3.1 (a) The negation can be expressed as `x =

3 5

is

(b) The negation can be expressed as ` is greater than or equal to 5'. (c) The negation can be expressed as `there is no
integer that is divisible by 3 but not by 6', or, alternatively, `every integer that is divisible by 3 is also divisible by 6'.

(d) Let n be a positive integer. We note that
n2 + n = n(n +1). Either n or n + 1 must be even, so n2 + n is even, as required. (Alternatively, the implication can be proved by considering two separate cases: the case where n is even, and the case where n is odd. However the above proof is shorter and simpler.)

(d) The negation can be expressed as `there is a real
number x that does not satisfy the inequality x2 0'. of the integers m and n is even'.

(e) The negation can be expressed as `at least one (f ) The negation can be expressed as `the integers
m and n are both even'. x2 - 2x +1 = 0, then (x - 1)2 = 0'. This is true.

3.6 (a) Taking m = 1 and n = 3 provides a

3.2 (a) The statement can be rewritten as `if
then n3 is odd'. This is true.

counter-example, since then m + n = 4, which is even.

(b) The statement can be rewritten as `if n is odd, (c) The statement can be rewritten as `if a given
integer is divisible by 3, then it is also divisible by 6'. This is false.

-3 < 2 but ((-3)2 - 2)2 = (9 - 2)2 = 72 = 49, which is not less than 4.

(b) The number -3 is a counter-example, because

(c) We look for a counter-example. Here is a table for the first few values of n.
n
n

1

2

3

(d) The statement can be rewritten as `if x > 2,
then x > 4'. This is false. and n are both odd'. The given statement is true, and its converse is false. pair m, n is even and the other is odd'. The given statement and its converse are both true.

4 +1 5 17 65 Since 43 + 1 = 65 is not a prime number, it provides a counter-example, so this implication is false.

3.3 (a) The converse is `if m + n is even, then m
(b) The converse is `if m + n is odd, then one of the

3.7 (a) Let P (n) be the statement
Then P (1) is true, since 1 = 1 1(1 + 1). 2 Let k 1, and assume that P (k ) is true: 1+ 2 + ··· + k = 1 k (k +1). 2 1+ 2 + ··· + n = 1 n(n +1). 2

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Solutions to the exercises

We wish to deduce that P ( 1+2+ ··· + k +(k +1) Now 1+ 2+ ··· + k +(k +1) = 1
k (k +1)+ (k +1) 2 = (k +1)( 1
k +1)
2 = 1
(k +1)(k +2). 2

k +1) is true: = 1 (k +1)(k +2). 2
(by P (k ))


(b) Let P (n) be the statement 5n < n!.
Then P (12) is true, b ecause 512 = 2.44 â 108 and 12! = 4.79 â 108 , b oth to three significant figures. Now let k 12, and assume that P (k ) is true; that is, 5k < k !. We wish to deduce that P (k + 1) is true; that is, 5(k+1) < (k +1)!. Now 5k+1 = 5 â 5k
< 5 â k ! (by P (k ))
< (k +1)k !
= (k +1)! ,
where we have used the fact that k 12, so k +1 13 > 5. Hence P (k ) P (k +1), for k = 12, 13,... . Hence, by mathematical induction, P (n) is true, for n = 12, 13,... .

Thus, for k = 1, 2,..., P (k ) P (k +1). Hence, by mathematical induction, P (n) is true for n = 1, 2,....

(b) Let P (n) be the statement
13 +23 + ··· + n3 = 1 n2 (n +1)2 . 4 Then P (1) is true, since 13 = 1 and 1 12 (1 + 1)2 = 1. 4 Let k 1, and assume that P (k ) is true: 13 +23 + ··· + k3 = 1 k 2 (k +1)2 . 4 We wish to deduce that P (k +1) is true: 13 +23 + ··· + k3 +(k +1)3 = 1 (k +1)2 (k +2)2 . 4 Now 13 +23 + ··· + k3 +(k +1)3 = 1
k 2 (k +1)2 +(k +1)3 (by P (k )) 4 = (k +1)2 ( 1
k 2 +(k +1))
4 =
1
4( 1
4(

3.9 (a) Suppose that there exist real numbers a and b with ab > 1 (a2 + b2 ). Then a2 - 2ab + b2 < 0; 2 that is, (a - b)2 < 0. This is a contradiction, so our supposition must be false. Hence there are no such real numbers a and b.
(b) Suppose that there exist integers m and n with
5m +15n = 357. The left-hand side of this equation is a multiple of 5, so the right-hand side is also. But this is a contradiction, so our supposition must be false. Hence there are no such integers m and n.

k +1) (k +4k +4)


2

2

= k +1)2 (k +2)2 . Thus, for k = 1, 2,..., P (k ) P (k +1). Hence, by mathematical induction, P (n) is true for n = 1, 2,....

3.10 Suppose that n = a +2b, where a and b are positive real numbers. Suppose also that a < 1 n and 2 b < 1
n. Then 4
n This a< have
1 = a +2b < 1 n +2( 4 n) = n.
2 contradiction shows that the supposition that 1 1 2 n and b < 4 n must b e false; that is we must 1 1 a 2 n or b 4 n.

3.8 (a) Let P (n) be the statement `4 +1 is a multiple of 5'. Then P (2) is true, because 42â2-3 +1 = 41 +1 = 5. Now let k 2, and assume that P (k ) is true; that is,
42k-3 + 1 is a multiple of 5. We wish to deduce that P (k + 1) is true; that is, 42(k+1)-3 +1 = 42k-1 + 1 is a multiple of 5. Now 42k-1 +1 = 42 42k-3 +1 = 16 â 42k-3 +1 = 15 â 42k-3 +42k-3 +1. The first term here is a multiple of 5, and 42k-3 +1 is a multiple of 5, by P (k ). Therefore 42k-1 +1 is a multiple of 5. Hence P (k ) P (k +1), for k = 2, 3,.... Hence, by mathematical induction, P (n) is true, for n = 2, 3,....

2n-3

3.11 (a) We prove the contrapositive implication, which is n is odd n3 is odd. Suppose that n is odd. Then n = 2k + 1 for some integer k . Then n3 = (2k +1)3
= (2k + 1)(4k 2 +4k +1)
= 8k 3 +12k 2 +6k +1
= 2(4k 3 +6k 2 +3k )+1,
which is odd.

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(b) We prove the contrapositive implication, which
is `if at least one of m and n is even, then mn is even'. Supp ose that at least one of m and n is even; without loss of generality, we can take it to be m (since otherwise we can just interchange m and n). Then m = 2k for some integer k . Hence mn = 2kn, which is even.

We have (f g )(x) = f 1+ 1 x = 1 1 1+ x =x -1

(c) Let n be an integer which is greater than 1. We
prove the contrapositive implication, which is `if n is not a prime number, then n is divisibley at least b one of the primes less than or equal to n'. Suppose that n is not a prime numb er. Then n = ab for some integers a, b, where 1 < a, b < n. By the result of Example 3.15, at least one of a and b is less than or equal to n. This number has a prime factor, which must also be less than or equal to n, and this prime factor must also be a factor of n. This proves the required contrapositive implication.

and f g : R -{0} - R -{0}. Also, 1 1 =1+ (g f )(x) = g x-1 1/(x - 1) =1+x-1=x and g f : R -{1} - R -{1}. Hence, from Strategy 2.1, g is the inverse of f .

3.15 (a) The converse is as follows.
If m - n is an even integer,
then m and n are both even integers.


(b) The original statement is true.
Suppose that m and n are both even; then m = 2p, n = 2q, where p, q are integers. Then m - n = 2p - 2q = 2(p - q ), which is even. The converse is false. For example, 7 - 3 = 4 is even,
but 7 and 3 are both odd.


3.12 (a), (b) and (c) all have the same meaning.
(d) and (f ) have the same meaning.
(You may like to show that (a) is true, and hence
that (b) and (c) are true; that (d) and (f ) are true,
but (e) is false.)


3.13 132 = 169 and 172 = 289, so we need check
only the primes 2, 3, 5, 7, 11, 13.
221 is divisible by 13 (221 = 13 â 17), so it is not
prime.
223 is not divisible by any of 2, 3, 5, 7, 11 and 13, so it is prime.

3.14 (a) This statement is true.
We have n3 - n = n(n2 - 1) = n(n - 1)(n +1). Either n is even or n + 1 is even, so n3 - n is even.

3.16 (a) Let P (n) be the statement
1 1 1 n-1 + + ··· + = . 1â2 2â3 (n - 1)n n Then P (2) is true, since 1 2-1 1 = = . 1â2 2 2 Assume that P (k ) is true: 1 k-1 1 1 + + ··· + = . 1â2 2â3 (k - 1)k k We wish to deduce that P (k +1) is true: 1 1 1 1 + + ··· + + 1â2 2â3 (k - 1)k k (k +1) k = . k +1 Now 1 1 1 1 + + ··· + + 1â2 2â3 (k - 1)k k (k +1) k-1 1 = + (by P (k )) k k (k +1) (k - 1)(k +1)+ 1 = k (k +1) k k2 = . = k (k +1) k +1

(b) This statement is false.
For example, 6 + 4 is a multiple of 5, but 6 and 4 are not multiples of 5.

(c) This statement is false.
For example, if = /2, then sin 2 = sin = 0, but 2sin = 2 sin(/2) = 2.

(d) This statement is false.
For example, f (0) = f (2) = 1.

(e) This statement is true.
We show that f g is the identity on R -{0} and that g f is the identity on R -{1}.

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Solutions to the exercises

Thus, for k = 2, 3,..., P (k ) P (k +1). Hence, by mathematical induction, P (n) is true for n = 2, 3,....

(b) Let P (n) be the statement
32n - 1 is divisible by 8. Then P (1) is true, since 32 - 1 = 9 - 1 = 8, which is divisible by 8. Assume that P (k ) is true: 32k - 1 is divisible by 8. We wish to deduce that P (k +1) is true: 32( Now 32(
k+1)

If n = 3k +2, then n2 = 9k 2 +12k +4 = 3(3k 2 +4k +1) + 1, which is not divisible by 3. Hence the contrapositive is true. Hence the original statement is true.

3.19 (a) This statement is false.

For example, -4 < 3, but (-4)2 32 .


(b) This statement is false.

For example, if x = 1, then x2 - x = 0, not 2.


(c) This statement is true.

One value of x satisfying x2 - x = 2 is x = 2.


- 1 is divisible by 8.

(d) This statement is false.
x2 - x = -1 x2 - x +1 = 0
3 x - 1 + 4 = 0, 2 which is not possible for any real x. 2

- 1 = 32 32k - 1 = 9 â 32 k - 1 = 8 â 32k +(32k - 1), which is also divisible by 8, since P (k ) is true. Thus, for k = 1, 2,..., P (k ) P (k +1). Hence, by mathematical induction, P (n) is true for n = 1, 2,....

k+1)

(e) This statement is false.
For example, if x = y = 1, then x/y and y/x are both
the integer 1.


(f ) This statement is true.

We prove it by mathematical induction.
Let P (n) b e the statement
12 +22 + ··· + n2 = 1 n(n + 1)(2n +1). 6 Then P (1) is true, since 2â3 1 = 1 = 12 . 6 â 1 â (1 + 1)(2 + 1) = 6 Assume that P (k ) is true: 12 +22 + ··· + k2 = 1 k (k + 1)(2k +1). 6 We wish to deduce that P (k +1) is true: 12 +22 + ··· +(k +1)2
= 1
(k +1)(k + 2)(2k +3). 6 Now, 12 +22 + ··· + k2 +(k +1)2 = 1
k (k + 1)(2k +1)+ (k +1) 6 = =
1
6 (k 1
6 (k 1
6 (k

3.17 Suppose that the given statement is false; that is, there are real numbers a and b for which (a + b)2 < 4ab. Then a2 +2ab + b2 < 4ab, so a2 - 2ab + b2 < 0, so (a - b)2 < 0. But (a - b)2 is a square, so cannot be negative. This is a contradiction, so the given statement must b e true. Hence (a + b)2 4ab for all real numbers a and b. 3.18 (a) The contrapositive is as follows.
If n is not divisible by 3, then n2 is not divisible by 3.

2

(by P (k ))

+1)(k (2k +1)+6(k +1))
+ 1)(2k 2 +7k +6)


(b) Suppose that n is not divisible by 3. Then
n = 3k +1 or n = 3k +2, for some integer k . If n = 3k +1, then n2 = 9k 2 +6k +1 = 3k (3k +2)+ 1, which is not divisible by 3.

+1)(k + 2)(2k +3). = Hence P (k ) P (k +1), for k 1. Hence, by mathematical induction, P (n) is true for n = 1, 2,....

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Unit I2 Mathematical language

(g) Let P (n) be the statement
1 1 1 1 . 1- ··· 1 - = 2 3 n 2n Then P (2) is false, since 1 - 1 = 1. 2 4 Hence the statement is false. (In fact, as you can check, P (k ) is true P (k + 1) is true, for k 2; that is, step 2 of a proof by mathematical induction works, even though step 1 does not. The correct expression for the product is 1/n.) 1-

4.3 (a) For n = 5, the Geometric Series Identity is
a5 - b5 = (a - b)(a4 + a3 b + a2 b2 + ab3 + b4 ).

(b) If n is an odd positive integer, then
(-b)n = (-1)n bn = -bn , so an - (-b)n = an + bn . By Theorem 4.2, an - (-b)n = (a - (-b))(an-1 + an-2 (-b)+ ··· + a(-b)n-2 +(-b)n-1 ), so, since n - 1 is even and n - 2 is odd, an + bn = (a + b)(an-1 - an-2 b + ··· - abn-2 + bn-1 ), as required. For n = 5, we have a5 + b5 = (a + b)(a4 - a3 b + a2 b2 - ab3 + b4 ).

(a + b)9 is 9! 9â8â7â6 9 = = = 126. 4 4! 5! 4â3â2â1 (1 + 2x)5 is 5! 5 11 â (2x)4 = 4 4! 1! = (5 â = 80x4 so the required coefficient
n

4.1 (a) By Theorem 4.1, the coefficient of a5 b4 in

(b) By Theorem 4.1, the term involving x4 in
â 24 x4 16)x4 , is 80.

4.4 Using and r = 1 , 2 1 1+ + 2 =1+
=1

1 - (1) 2 1- 1 2

the corollary to Theorem 4.2, with a = 1 we obtain 1 1 + ··· + n-1 4 2 1 12 1 n-1 + ··· + 2 2+ 2
n

4.2 (a) By Theorem 4.1, with b replaced by -b,
(a - b) = (a +(-b))n n n = an + 0 1 + = n 0 n k a
n-k

= 2 1 -


1 2n
1
= 2 - n-1 . 2

a

n-1

(-b)+ ··· n n (-b)
n

(-b)k + ··· + n 1 a
n-1

an -
k

b + ··· n
n b. n

+(-1)

n k n 1

a

n-k k

b + ··· +(-1)n

is a factor of p(x) if and only if p(-3) = 0, that is, 0 = (-3)3 + c(-3)2 +6(-3) + 36 = -27 + 9c - 18 + 36 = 9c - 9. This equation has just one solution, c = 1, so x +3 is a factor of p(x) if and only if c = 1.


4.5 By the Polynomial Factorisation Theorem, x +3

(b) If a = 1 and b = 1, then a - b = 0, so we obtain
0= n 0 -
n

+ ··· +(-1)k

n + ··· k


+(-1) For n = 4, 4 0= 0 =1- = 0, as expected.

n . n this identity is 4 4 - + 1 2 4+6 - 4+1

4.6 (a) (i) Since all the roots are integers, the only possible roots are the factors of 4, that is, ±1, ±2, ±4. Considering these in turn, we obtain the following table.

x 1 -1 2 0 -2 4 -4 p(x) 2 0 -16 20 -108

-

4 3

+

4 4

So the only solutions are x = -1 and x = 2. In fact, x3 - 3x2 +4 = (x +1)(x - 2)(x - 2).

(ii) Since all the roots are integers, the only possible roots are the factors of -15, that is, ±1, ±3, ±5, ±15. Considering these in turn, we obtain the following table.
x 1 -1 3 -3 5 ··· p(x) 0 -48 0 -192 0 ···

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Solutions to the exercises

We do not need to work out any more values, as we already have three roots: x = 1, x = 3 and x = 5. So x3 - 9x2 +23x - 15 = (x - 1)(x - 3)(x - 5).

(b) A suitable equation is
(x - 1)(x - 2)(x - 3)(x +3) = 0, that is, x4 - 3x3 - 7x2 +27x - 18 = 0.

4.10 (a) Putting x = 2, we 16 + 4 - 26 + 6 = 0, so x - 2 2x3 + x2 - 13x +6 = (x - = (x -

obtain is a factor. Hence 2)(2x2 +5x - 3) 2)(x + 3)(2x - 1).

4.7 (a) (a +3b)
= + 4 0

4

a4 + 4 3

4 1

a3 (3b)+ 4 4 (3b)

4 2
4

a2 (3b)

2

(b) Trying integer values which are factors of 10, we find that x = 1 is a root, so x - 1 is a factor. Hence x3 +6x2 +3x - 10 = (x - 1)(x2 +7x + 10)
= (x - 1)(x +2)(x +5),
3 so the solutions of x +6x2 +3x - 10 = 0 are 1, -2 and -5. (c) x2 + x = y 2 + y x2 - y 2 + x - y = 0.
We note that x = y is a solution, so x - y is a factor. Hence x2 - y 2 + x - y = (x - y )(x + y +1), so x = y or x = -y - 1.

a(3b)3 +

= a4 +4a3 3b +6a2 9b2 +4a27b3 +81b4 = a4 +12a3 b +54a2 b2 + 108ab3 +81b4





(b) (1 - t)
= 7 0 + +

7

+ 7 3 7 6

7 1

(-t)+ 7 4 7 7

7 2

(-t)

2

(-t)3 + (-t)6 +

(-t)4 + (-t)
7

7 5

(-t)

5

= 1 - 7t +21t2 - 35t3 +35t4 - 21t5 +7t6 - t

7

4.8 (a) The coefficient of a3 b7 is
10 7 15 13 10 â 9 â 8 = 120. = 3â2 22 = 15 â 14 â 4 = 420. 2

4.11 (a) If the sum of the roots is 0 and the product is -30, then the cubic p olynomial must b e of the form x3 + cx +30, for some c R.
If x = 3 is a root, then
27 + 3c +30 = 0, so c = -19.
Hence the polynomial is x3 - 19x + 30.

(b) We know that x - 3 is a factor. Hence
x3 - 19x +30 = (x - 3)(x2 +3x - 10) = (x - 3)(x - 2)(x +5),
so the other two roots are 2 and -5.

(b) The coefficient of x13 is


4.9 (a) This is a geometric series with a = 3,
r = - 1 and n = 12 terms, so its sum is 3 3 (1 - - 1 3 1- -
12 1 3


)

= 3 â 3 (1 - 4 = 9
(1 - 4 2.25.

1 12 3

)

1 12 3

)

a common ratio r = = 1 and n + 1 terms, so its b sum is a n+1 1- a n+1
b b 1 -
. = a b-a b
1- b


(b) This is a geometric series with first term 1,

71