This preview shows page 1. Sign up to view the full content.
Math 370
Solutions to Assignment 1
5, 8.1, 8.2, 18, 22, 25, 28.
5.
Let
A
=
f
n
2
N
:
n >
3
g
:
Then
A
has no maximal element.
Proof.
Suppose that
A
has a greatest element
m:
Then
m
2
A
so
3
< m < m
+1
2
A:
This contradicts
the property that
m
is maximmal in
A:
Thus our supposition is false so
A
has no maximal element.
8.1
Let
F
x
2
F:
Then
x >
0
i/
x <
0
:
Proof.
Suppose
x >
0
:
Then by
O
2
,
x
= 0 + (
x
)
< x
+ (
x
) = 0
:
Conversely, suppose
x <
0
:
Then
0 =
x
+
x <
0 +
x
=
x:
8.2
Let
e
F:
Then
e >
0
:
Proof.
We are given that
e
6
= 0
:
By the trichotomy law there remain only two mutually exclusive
possiblities:
0
< e
and
e <
0
:
Suppose that
e <
0
:
Adding
e
, we get
0
<
e:
Thus
0
<
(
e
) (
e
) =
e:
This contradiction shows that
0
< e:
18
Let
S
F
. The
b
= inf
S
i/ the following conditions hold
(
i
)
b
±
x
for all
x
2
S
and
(
ii
)
" >
0
)
b
+
" > x
for some
x
2
S:
Proof.
(=
)
)
Suppose
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '09
 Math

Click to edit the document details