CHAPTER 1
INTRODUCTION
1
Chapter 1
Introduction
1.3
2.
(a) Put
p/
100
=
x
. Then the given expression becomes
a
+
ax
−
(a
+
ax)x
=
a(
1
−
x
2
)
, as required.
(b) $2000
·
1
.
05
·
0
.
95
=
$1995.
(c) The result is precisely the formula in (a).
(d) With the notation
used in the answer to (a),
a
−
+
(a
−
=
1
−
x
2
)
, which is the same expression as in (a).
4.
(a)
F
=
32 yields
C
=
0;
C
=
100 yields
F
=
212.
(b)
F
=
9
5
C
+
32
(c)
F
=
40 for
C
≈
4
.
4,
F
=
80 for
C
≈
26
.
7. The assertion is meaningless.
6.
x
=
R(
1
−
p)
−
S(
1
−
q)
(p
−
q)(
1
−
p
−
,
y
=
pS
−
qR
(p
−
1
−
p
−
if
(p
−
1
−
p
−
6=
0.
(Use, for example, (A.39) in Section A.9.)
1.4
2.
(a) Correct.
(b) Incorrect.
(c) Incorrect.
(d) Correct.
(e) Incorrect.
(f) Incorrect. (“Usually” the sum of two irrationals is irrational, but not always. For example,
π
and
−
π
are both irrationals, but
π
+
(
−
π)
=
0, which is rational.)
4.
(a)
y
≤
3
−
3
4
x
(b)
y>
3
2
z
(c)
y
≤
(m
−
px)/q
6.

2
·
0
−
3
=−
3
=
3,

2
·
1
2
−
3
2
2,

2
·
7
2
−
3
=
4
4
8.
(a) 3
−
2
x
=
5o
r3
−
2
x
=−
5, so
−
2
x
=
2o
r
−
8. Hence
x
1o
r
x
=
4.
(b)
−
2
≤
x
≤
2
(c) 1
≤
x
≤
3
(d)
−
1
/
4
≤
x
≤
1
(e)
x>
√
2or
x<
−
√
2
(f) 1
≤
x
2
≤
3, and so 1
≤
x
≤
√
3o
r
−
√
3
≤
x
≤−
1
1.5
2.
(a)
⇒
right,
⇐
wrong
(b)
⇒
wrong,
⇐
right
(c)
⇒
right,
⇐
wrong
(d)
⇒
and
⇐
both right (e)
⇒
wrong (0
·
5
=
0
·
4, but 5
4),
⇐
right
(f)
⇒
right,
⇐
wrong
4.
x
=
2. (
x
1, 0 or 1 make the equation meaningless. Multiplying each term by the common de
nominator
x(x
−
1
)(x
+
1
)
yields
(x
+
1
)
3
+
(x
−
1
)
3
−
2
x(
3
x
+
1
)
=
0. Expanding and simplifying,
2
x
3
−
6
x
2
+
4
x
=
0, or 2
2
−
3
x
+
2
)
=
0, or 2
−
1
)(x
−
2
)
=
0. Hence,
x
=
2 is the only
solution.)
6.
(a) No solutions. (Squaring each side yields
x
−
4
=
x
+
5
−
18
√
x
+
5
+