Instructor’s Solutions Manual, Section 0.1

Problem 1

Solutions to Problems, Section 0.1
The problems in this section may be harder than typical problems found in the rest
of this book.
√
1 Show that 67 + 2 is an irrational number.
√
solution Suppose 67 + 2 is a rational number. Because

√

2 = ( 67 +

√

2) − 76 ,

√
this
implies
that
2 is the difference of two rational numbers, which implies√that
√
2 is a rational number, which is not true. Thus our assumption that 67 + 2 is
√
a rational number must be false. In other words, 67 + 2 is an irrational number.

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Instructor’s Solutions Manual, Section 0.1

2 Show that 5 −

Problem 2

√

2 is an irrational number.
√
solution Suppose 5 − 2 is a rational number. Because

√

2 = 5 − (5 −

√

2),

√
this
√ implies that 2 is the difference of two rational numbers, which implies√that
2 is a rational number, which is not true. Thus our √
assumption that 5 − 2 is
a rational number must be false. In other words, 5 − 2 is an irrational number.

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Instructor’s Solutions Manual, Section 0.1

Problem 3

√
3 Show that 3 2 is an irrational number.
√
solution Suppose 3 2 is a rational number. Because
√

√
3 2
2=
,
3

√
this
implies
that
2 is the quotient of two rational numbers, which implies
√
√ that
2 is a rational number, which is not true. Thus √
our assumption that 3 2 is a
rational number must be false. In other words, 3 2 is an irrational number.

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Instructor’s Solutions Manual, Section 0.1

4 Show that

√
3 2
5

Problem 4

is an irrational number.

solution Suppose

√
3 2
5

is a rational number. Because

√

√
3 2 5
2=
· ,
5
3


√
this implies that 2 is the product of two rational numbers, which implies
that
√
√
3 2
2 is a rational number, which is not true. Thus our assumption that 5 is a
rational number must be false. In other words,

√
3 2
5

is an irrational number.

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Instructor’s Solutions Manual, Section 0.1

Problem 5

√
5 Show that 4 + 9 2 is an irrational number.
√
solution Suppose 4 + 9 2 is a rational number. Because
√
√
9 2 = (4 + 9 2) − 4,

√
this implies
that 9 2 is the difference of two rational numbers, which implies
√
that 9 2 is a rational number. Because
√

√
9 2
2=
,
9

√
this
√ implies that 2 is the quotient of two rational numbers, which implies that
√
2 is a rational number, which is not true. Thus our assumption
that
4
+
9
2
√
is a rational number must be false. In other words, 4 + 9 2 is an irrational
number.

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Instructor’s Solutions Manual, Section 0.1

Problem 6

6 Explain why the sum of a rational number and an irrational number is an
irrational number.
solution We have already seen the pattern for this solution in Problems 1 and 2.
We can repeat that pattern, using arbitrary numbers instead of specific numbers.
Suppose r is a rational number and x is an irrational number. We need to show
that r + x is an irrational number.
Suppose r + x is a rational number. Because
x = (r + x ) − r,
this implies that x is the difference of two rational numbers, which implies that
x is a rational number, which is not true. Thus our assumption that r + x is a
rational number must be false. In other words, r + x is an irrational number.

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Instructor’s Solutions Manual, Section 0.1

Problem 7

7 Explain why the product of a nonzero rational number and an irrational number
is an irrational number.
solution We have already seen the pattern for this solution in Problems 3 and 4.
We can repeat that pattern, using arbitrary numbers instead of specific numbers.
Suppose r is a nonzero rational number and x is an irrational number. We need
to show that rx is an irrational number.
Suppose rx is a rational number. Because

x=

rx
,
r

this implies that x is the quotient of two rational numbers, which implies that
x is a rational number, which is not true. Thus our assumption that rx is a
rational number must be false. In other words, rx is an irrational number.
Note that the hypothesis that r is nonzero is needed because otherwise we
would be dividing by 0 in the equation above.

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Instructor’s Solutions Manual, Section 0.1

Problem 8

8 Suppose t is an irrational number. Explain why

1
t

is also an irrational number.

solution Suppose 1t is a rational number. Then there exist integers m and n,
with n = 0, such that
1
m

= .
t
n
Note that m = 0, because

1
t

cannot equal 0.

The equation above implies that
t=

n
,
m

which implies that t is a rational number, which is not true. Thus our assumption
that 1t is a rational number must be false. In other words, 1t is an irrational
number.

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Instructor’s Solutions Manual, Section 0.1

Problem 9

9 Give an example of two irrational numbers whose sum is an irrational number.
√

√
solution Problem 7 implies that 2 2 and 3 2 are irrational numbers. Because

√

√
√
2 + 2 2 = 3 2,

we have an example of two irrational numbers whose sum is an irrational
number.

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Instructor’s Solutions Manual, Section 0.1

Problem 10

10 Give an example of two irrational numbers whose sum is a rational number.
solution Note that

√

2 + (5 −

√

Thus we have two irrational numbers (5 −
sum equals a rational number.


2) = 5.

√

2 is irrational by Problem 2) whose

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Instructor’s Solutions Manual, Section 0.1

Problem 11

11 Give an example of three irrational numbers whose sum is a rational number.
solution Here is one example among many possibilities:

(5 −

√

2) + (4 −

√

√
2) + 2 2 = 9.

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Instructor’s Solutions Manual, Section 0.1

Problem 12

12 Give an example of two irrational numbers whose product is an irrational
number.
solution Here is one example among many possibilities:

(5 −

√ √
√
2) 2 = 5 2 − 2.

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Instructor’s Solutions Manual, Section 0.1

Problem 13

13 Give an example of two irrational numbers whose product is a rational number.
solution Here is one example among many possibilities:

√ √
√ 2
(3 2) 2 = 3 · 2 = 3 · 2 = 6.

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Instructor’s Solutions Manual, Section 0.1

Problem 14

√
14 Explain why ( 2)3 is an irrational number.
solution Note that

√
√ √ √
( 2)3 = ( 2 2) 2
√
= 2 2.
√
Thus ( 2)3 is the product of a nonzero rational number and an irrational
√ 3
number. Now Problem 7 implies that 2 is irrational.

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Instructor’s Solutions Manual, Section 0.1

Problem 15

15 Suppose t is an irrational number. Explain why at least one of t2 and t3 is
irrational.
solution Suppose t2 and t3 are both rational. Because

t=

t3
,
t2

this implies that t is rational, which is a contradiction. Thus at least one of t2
and t3 is an irrational number.

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Instructor’s Solutions Manual, Section 0.2

Exercise 1

Solutions to Exercises, Section 0.2
For Exercises 1–4, determine how many different values can arise by inserting one
pair of parentheses into the given expression.
1 19 − 12 − 8 − 2
solution Here are the possibilities:
19(−12 − 8 − 2) = −418

19 − (12 − 8) − 2 = 13

19(−12 − 8) − 2 = −382

19 − (12 − 8 − 2) = 17

19(−12) − 8 − 2 = −238


19 − 12 − 8(−2) = 23

(19 − 12) − 8 − 2 = −3

19 − 12(−8) − 2 = 113

19 − 12 − (8 − 2) = 1

19 − 12(−8 − 2) = 139

Other possible ways to insert one pair of parentheses lead to values already
included in the list above. Thus ten values are possible; they are −418, −382,
−238, −3, 1, 13, 17, 23, 113, and 139.

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Instructor’s Solutions Manual, Section 0.2

Exercise 2

2 3−7−9−5
solution Here are the possibilities:
3(−7 − 9 − 5) = −63
3(−7 − 9) − 5 = −53
3(−7) − 9 − 5 = −35

(3 − 7) − 9 − 5 = −18
3 − 7 − (9 − 5) = −8

3 − (7 − 9) − 5 = 0
3 − (7 − 9 − 5) = 10
3 − 7 − 9(−5) = 41
3 − 7(−9) − 5 = 61
3 − 7(−9 − 5) = 101
Other possible ways to insert one pair of parentheses lead to values already
included in the list above. For example,

(3 − 7 − 9) − 5 = −18.
Thus ten values are possible; they are −63, −53, −35, −18, −8, 0, 10, 41, 61, and
101.

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Instructor’s Solutions Manual, Section 0.2

Exercise 3

3 6+3·4+5·2
solution Here are the possibilities:

(6 + 3 · 4 + 5 · 2) = 28
6 + (3 · 4 + 5) · 2 = 40

(6 + 3) · 4 + 5 · 2 = 46
6 + 3 · (4 + 5 · 2) = 48
6 + 3 · (4 + 5) · 2 = 60
Other possible ways to insert one pair of parentheses lead to values already
included in the list above. Thus five values are possible; they are 28, 40, 46, 48,

and 60.

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Instructor’s Solutions Manual, Section 0.2

Exercise 4

4 5·3·2+6·4
solution Here are the possibilities:

(5 · 3 · 2 + 6 · 4) = 54
(5 · 3 · 2 + 6) · 4 = 144
5 · (3 · 2 + 6 · 4) = 150
5 · (3 · 2 + 6) · 4 = 240
5 · 3 · (2 + 6 · 4) = 390
5 · 3 · (2 + 6) · 4 = 480
Other possible ways to insert one pair of parentheses lead to values already
included in the list above. For example,

(5 · 3) · 2 + 6 · 4 = 54.
Thus six values are possible; they are 54, 144, 150, 240, 390, and 480.

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Instructor’s Solutions Manual, Section 0.2

Exercise 5


For Exercises 5–22, expand the given expression.
5 ( x − y)(z + w − t)
solution

( x − y)(z + w − t)
= x (z + w − t) − y(z + w − t)
= xz + xw − xt − yz − yw + yt

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Instructor’s Solutions Manual, Section 0.2

Exercise 6

6 ( x + y − r )(z + w − t)
solution

( x + y − r )(z + w − t)
= x (z + w − t) + y(z + w − t) − r (z + w − t)
= xz + xw − xt + yz + yw − yt − rz − rw + rt

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Instructor’s Solutions Manual, Section 0.2

Exercise 7


7 (2x + 3)2
solution

(2x + 3)2 = (2x )2 + 2 · (2x ) · 3 + 32
= 4x2 + 12x + 9

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Instructor’s Solutions Manual, Section 0.2

Exercise 8

8 (3b + 5)2
solution

(3b + 5)2 = (3b)2 + 2 · (3b) · 5 + 52
= 9b2 + 30b + 25

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Instructor’s Solutions Manual, Section 0.2

Exercise 9

9 (2c − 7)2
solution

(2c − 7)2 = (2c)2 − 2 · (2c) · 7 + 72

= 4c2 − 28c + 49

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Instructor’s Solutions Manual, Section 0.2

Exercise 10

10 (4a − 5)2
solution

(4a − 5)2 = (4a)2 − 2 · (4a) · 5 + 52
= 16a2 − 40a + 25

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