Math 350
Advanced Calculus I
Fall 2009
Homework Solutions
Section 11
ORDERED FIELDS
11.3
Let
x
,
y
, and
z
be real numbers. Prove the following.
(a)
−
(
−
x
) =
x
Proof:
The element
−
(
−
x
) is by deFnition the additive inverse of
−
x
. To show
−
(
−
x
) =
x
,
we need to show that
x
is the additive inverse of
−
x
. By Axiom A5, the additive inverse of
any element is unique, so to show that
x
is the additive inverse of
−
x
, we only need to show
that
x
behaves like the additive inverse of
−
x
. That is, we need to show
(
−
x
) +
x
= 0
.
By Axiom A5, since
−
x
is the additive inverse of
x
, then
x
+ (
−
x
) = 0
.
By Axiom A2 Commutative Law of Addition,
x
+ (
−
x
) = (
−
x
) +
x.
Therefore
(
−
x
) +
x
= 0
,
as required. Thus
−
(
−
x
) =
x
.
(b)
(
−
x
)
·
y
=
−
(
xy
) and (
−
x
)
·
(
−
y
) =
xy
Proof:
The element
−
(
xy
) is by deFnition the additive inverse of
xy
.
So, to show that
(
−
x
)
·
y
=
−
(
xy
), we need to show that (
−
x
)
·
y
behaves like the additive inverse of
xy
. That
is, we need to show
xy
+ (
−
x
)
·
y
= 0
.
We have
xy
+ (
−
x
)
·
y
=
yx
+
y
·
(
−
x
)
=
y
[
x
+ (
−
x
)]
=
y
·
0
= 0
by M2 Commutative Law
by DL Distributive Law
by A5 Existence of Additive Inverses
by Theorem 11.1(b)
Alternatively, we can use the result of Theorem 11.1(c). We have
(
−
x
)
·
y
= [(
−
1)
·
x
]
·
y
= (
−
1)
·
(
x
·
y
)
=
−
(
xy
)
by Theorem 11.1(c)
by M3 Associative Law
by Theorem 11.1(c)
To prove (
−
x
)
·
(
−
y
) =
xy
, we can use the Frst result. We have
(
−
x
)
·
(
−
y
) =
−
[
x
·
(
−
y
)]
=
−
[(
−
y
)
·
x
]
=
−
[
−
(
yx
)]
=
−
[
−
(
xy
)]
=
xy
by the Frst part of (b)
by M2 Commutative Law
by the Frst part of (b)
by M2 Commutative Law
by part (a)
12
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Section 11
•
Ordered Fields
11.3
(continued)
(e)
If
x
6
= 0, then
x
2
>
0.
Proof:
Suppose
x
6
= 0. We need to show
x
2
>
0
.
By Axiom O1 Trichotomy Law, exactly one of the relations
x
= 0,
x >
0, or
x <
0 holds.
Since
x
6
= 0 by assumption, it follows that either
x >
0 or
x <
0. We need to prove
x
2
>
0 in
both of these cases.
Case 1.
Suppose
x >
0.
By Axiom O4, since
x >
0, when we multiply both sides of the
inequality
x >
0 by
x
, we preserve the inequality, so we obtain
x
2
=
x
·
x > x
·
0
.
By Theorem 11.1(b),
x
·
0 = 0, so this implies
x
2
>
0, as required.
Case 2.
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