Solution manual for precalculus 3rd edition by axler

Instructor’s Solutions Manual, Section 0.1 Problem 1Solutions to Problems, Section 0.1 The problems in this section may be harder than typical problems found in the rest Solution Manual

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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

Solution Manual for Precalculus 3rd Edition by Axler

Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler

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2 Show that 5 −√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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3 Show that 3√2 is an irrational number.

solution Suppose 3√2 is a rational number Because

√

2 = 3√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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√

2 is a rational number, which is not true Thus our assumption that 3√52 is a rational number must be false In other words, 3√52 is an irrational number.

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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

√

2 = 9√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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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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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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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 6= 0, such that

1

t =

m

n Note that m 6= 0, because 1t 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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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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10 Give an example of two irrational numbers whose sum is a rational number.

2 + ( 5 −√2 ) = 5.

Thus we have two irrational numbers (5 −√2 is irrational by Problem 2) whose sum equals a rational number.

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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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12 Give an example of two irrational numbers whose product is an irrational number.

( 5 −√2 ) √

2 = 5√2 − 2.

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13 Give an example of two irrational numbers whose product is a rational number.

( 3√2 ) √

2 = 3 ·√22= 3 · 2 = 6.

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= 2√2.

Thus ( √

2 ) is the product of a nonzero rational number and an irrational number Now Problem 7 implies that√23is irrational.

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15 Suppose t is an irrational number Explain why at least one of t 2 and t 3 is irrational.

solution Suppose t 2 and t 3 are both rational Because

t = t3

t2, this implies that t is rational, which is a contradiction Thus at least one of t 2 and t 3 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.

− 238, − 3, 1, 13, 17, 23, 113, and 139.

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( 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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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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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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For Exercises 5–22, expand the given expression.

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7 ( 2x + 3 )

solution

( 2x + 3 ) = ( 2x ) + 2 · ( 2x ) · 3 + 32

= 4x2+ 12x + 9

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8 ( 3b + 5 )

solution

( 3b + 5 ) = ( 3b ) + 2 · ( 3b ) · 5 + 52

= 9b2+ 30b + 25

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9 ( 2c − 7 )

solution

( 2c − 7 ) = ( 2c ) − 2 · ( 2c ) · 7 + 72

= 4c2− 28c + 49

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10 ( 4a − 5 )

solution

( 4a − 5 ) = ( 4a ) − 2 · ( 4a ) · 5 + 52

= 16a2− 40a + 25

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= x2+ y2+ z2+ 2xy + 2xz + 2yz

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= x2+ 25y 2+9z 2 − 10xy − 6xz + 30yz

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