Sol purcell

CHAPTER

0

0.1 Concepts Review
1. rational numbers

Preliminaries
1 ⎡ 2 1 ⎛ 1 1 ⎞⎤
1
8. − ⎢ − ⎜ − ⎟ ⎥ = −
3 ⎣ 5 2 ⎝ 3 5 ⎠⎦
⎡ 2 1 ⎛ 5 3 ⎞⎤
3 ⎢ − ⎜ − ⎟⎥
⎣ 5 2 ⎝ 15 15 ⎠ ⎦

2. dense

1 ⎡ 2 1 ⎛ 2 ⎞⎤
1 ⎡2 1 ⎤
= − ⎢ − ⎜ ⎟⎥ = − ⎢ − ⎥
3 ⎣ 5 2 ⎝ 15 ⎠ ⎦
3 ⎣ 5 15 ⎦
1⎛ 6 1 ⎞
1⎛ 5 ⎞
1
=− ⎜ − ⎟=− ⎜ ⎟=−
3 ⎝ 15 15 ⎠
3 ⎝ 15 ⎠
9

3. If not Q then not P.
4. theorems

2

ProblemSet 0.1
1. 4 − 2(8 − 11) + 6 = 4 − 2(−3) + 6
= 4 + 6 + 6 = 16
2. 3 ⎡ 2 − 4 ( 7 − 12 ) ⎤ = 3[ 2 − 4(−5) ]
⎣
⎦
= 3[ 2 + 20] = 3(22) = 66
3.

–4[5(–3 + 12 – 4) + 2(13 – 7)]
= –4[5(5) + 2(6)] = –4[25 + 12]
= –4(37) = –148

4.

5 [ −1(7 + 12 − 16) + 4] + 2
= 5 [ −1(3) + 4] + 2 = 5 ( −3 + 4 ) + 2
= 5 (1) + 2 = 5 + 2 = 7

5.

6.

7.

5 1 65 7 58
–=
–=
7 13 91 91 91
3
313
31+ −=
+−
4 − 7 21 6 −3 21 6
42 6
7
43
=− +
−
=−
42 42 42
42
1 ⎡1 ⎛ 1 1 ⎞ 1⎤ 1 ⎡1 ⎛ 3 – 4 ⎞ 1⎤
=
⎜ – ⎟+
⎜
⎟+
3 ⎢ 2 ⎝ 4 3 ⎠ 6 ⎥ 3 ⎢ 2 ⎝ 12 ⎠ 6 ⎥
⎣
⎦
⎣
⎦
1 ⎡1 ⎛ 1 ⎞ 1⎤
= ⎢ ⎜– ⎟+ ⎥
3 ⎣ 2 ⎝ 12 ⎠ 6 ⎦
1⎡ 1
4⎤
= ⎢– + ⎥
3 ⎣ 24 24 ⎦
1⎛ 3 ⎞ 1
= ⎜ ⎟=
3 ⎝ 24 ⎠ 24

Instructor’s Resource Manual

2

2
14 ⎛ 2 ⎞
14 ⎛ 2 ⎞
14 6
⎜
⎟= ⎜ ⎟= ⎛ ⎞
9.
⎜⎟
21 ⎜ 5 − 1 ⎟
21 ⎜ 14 ⎟21 ⎝ 14 ⎠
3⎠
⎝
⎝3⎠
2

=

14 ⎛ 3 ⎞
2⎛ 9 ⎞ 6
⎜ ⎟ = ⎜ ⎟=
21 ⎝ 7 ⎠
3 ⎝ 49 ⎠ 49

⎛2
⎞ ⎛ 2 35 ⎞ ⎛ 33 ⎞
⎜ − 5⎟ ⎜ − ⎟ ⎜ − ⎟
7
⎠ = ⎝ 7 7 ⎠ = ⎝ 7 ⎠ = − 33 = − 11
10. ⎝
6
2
⎛ 1⎞ ⎛7 1⎞
⎛6⎞
⎜1 − ⎟ ⎜ − ⎟
⎜⎟
⎝ 7⎠ ⎝7 7⎠
⎝7⎠
7
11 – 12 11 – 4
7 21 = 7 7 = 7 = 7
11.
11 + 12 11 + 4 15 15
7 21 7 7
7
137 4675
−+
−+
5
12. 2 4 8 = 8 8 8 = 8 =
137 467 33
+−
+−
248 8888

13. 1 –1
1
2321
=1– =1– = – =
1
3
3333
1+ 2
2

14. 2 +

15.

(

3
5
1+
2

5+ 3

3
3
= 2+
25
7
−
22
2
6 14 6 20
= 2+ = + =
7777

= 2+

)(

) ( 5) – ( 3)

5– 3 =

2

2

=5–3= 2

Section 0.1

1

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16.

(

5− 3

) = ( 5)
2

2

−2

( 5)( 3) + ( 3)

2

27.

= 5 − 2 15 + 3 = 8 − 2 15

17. (3x − 4)( x + 1) = 3 x 2 + 3 x − 4 x − 4

= 3x2 − x − 4
18. (2 x − 3)2 = (2 x − 3)(2 x − 3)

12

4
2
+
x + 2x x x + 2
12
4( x + 2)
2x
=
+
+
x( x +2) x( x + 2) x( x + 2)
12 + 4 x + 8 + 2 x 6 x + 20
=
=
x( x + 2)
x( x + 2)
2(3 x + 10)
=
x( x + 2)
2

+

= 4 x2 − 6 x − 6 x + 9
= 4 x 2 − 12 x + 9

19.

28.

(3x – 9)(2 x + 1) = 6 x 2 + 3 x –18 x – 9

2
y
+
2(3 y − 1) (3 y + 1)(3 y − 1)
2(3 y + 1)
2y
=
+
2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)
=

2

= 6 x –15 x – 9

20. (4 x − 11)(3x − 7) = 12 x 2 − 28 x − 33x + 77

=

21. (3t 2 − t + 1) 2 = (3t 2 − t + 1)(3t 2 − t + 1)
3

2

3

2

2

= 9t − 3t + 3t − 3t + t − t + 3t − t + 1

6y + 2 + 2y
8y + 2
=
2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)

=

= 12 x 2 − 61x + 77

4

2
y
+
6 y − 2 9 y2 −1

2(4 y + 1)
4y +1
=
2(3 y + 1)(3 y − 1) (3 y + 1)(3 y − 1)

= 9t 4 − 6t 3 + 7t 2 − 2t + 1

0⋅0 = 0

b.

0
is undefined.
0c.

0
=0
17

d.

3
is undefined.
0

e.

05 = 0

f. 170 = 1

29. a.
22. (2t + 3)3 = (2t + 3)(2t + 3)(2t + 3)
= (4t 2 + 12t + 9)(2t + 3)
= 8t 3 + 12t 2 + 24t 2 + 36t + 18t + 27
= 8t 3 + 36t 2 + 54t + 27

23.

x 2 – 4 ( x – 2)( x + 2)
=
= x+2, x ≠ 2
x–2
x–2

24.

x 2 − x − 6 ( x − 3)( x + 2)
=
= x+2, x ≠3
x−3
( x − 3)

25.

t 2 – 4t – 21 (t + 3)(t – 7)
== t – 7 , t ≠ −3
t +3
t +3

26.

2

2x − 2x
3

2

2

x − 2x + x

=

2 x(1 − x)
2

x( x − 2 x + 1)
−2 x( x − 1)
=
x( x − 1)( x − 1)
2
=−
x −1

Section 0.1

0
= a , then 0 = 0 ⋅ a , but this is meaningless
0
because a could be any real number. No
0
single value satisfies = a .
0

30. If

31.

.083
12 1.000
96
40
36
4

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