Math 523H Homework #1
Section 1.1
1.1 #1. Prove that if
x
and
y
are real numbers, then

xy

=

x

y

.
Proof.
Suppose that
x
,
y
∈
R
.
•
Case 1:
x <
0
,y <
0
.
Then
xy >
0, so

xy

=
xy
. But

x

=

x
and

y

=

y
, so

x

y

= (

x
)(

y
) =
xy
by Prop 1.1.1(c). Thus,

xy

=
xy
=

x

y

.
•
Case 2:
x <
0
,y >
0
.
Then
xy <
0, so

xy

=

(
xy
). But

x

=

x
and

y

=
y
, so

x

y

= (

x
)
y
=

(
xy
) by Prop 1.1.1(c). Thus,

xy

=

(
xy
) =

x

y

.
•
Case 3:
x >
0
,y <
0
.
Then
xy <
0, so

xy

=

(
xy
). But

x

=
x
and

y

=

y
, so

x

y

=
x
(

y
) =

(
xy
) by Prop 1.1.1(c). Thus,

xy

=

(
xy
) =

x

y

.
•
Case 4:
x >
0
,y >
0
.
Then
xy >
0, so

xy

=
xy
. But

x

=
x
and

y

=
y
, so

x

y

=
xy
.
Thus,

xy

=
xy
=

x

y

.
•
Case 5:
x
= 0. Then

x

= 0.

xy

=

0
·
y

=

0

= 0 = 0
· 
y

=

x

y

. Similarly if
y
= 0,
since we can simply reverse the roles of x and y.
Therefore,
∀
x,y
∈
R
,

xy

=

x

y

.
±
1.1 #3. Prove that

(

x
) =
x
for all real numbers
x
.
Proof.
Let
x
∈
R
. By (P4), we know that there exists

x
such that
x
+ (

x
) = 0. By (P1),
x
+ (

x
) = (

x
) +
x
, so (

x
) +
x
= 0. By (P4), we also know that there exists

(

x
) such that
(

x
) + (

(

x
)) = 0. By Prop 1.1.1(b), additive inverses are unique, so we have

(

x
) =
x
.
±
1.1 #4. Prove that all real numbers
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 Fall '10
 Nahmod
 Math, Real Numbers, Addition, Archimedian, field., xy .

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