This preview shows pages 1–2. Sign up to view the full content.
Due Thursday February 11
(1) In this exercise, we will show that the ordered ﬁeld
Q
is not complete.
(a) Suppose
q
∈
Q
and
q
2
<
2. Let
n
∈
N
and suppose that
n >
max
{
2

q

+1
2

q
2
,
1
}
.
Show that (
q
+
1
n
)
2
<
2.
If
n >
max
{
2

q

+1
2

q
2
,
1
}
then
n >
2

q

+1
2

q
2
, so 2

q
2
>
2

q

+1
n
and thus 2
> q
2
+
2

q

+1
n
.
We therefore have:
(
q
+
1
n
)
2
=
q
2
+
2
q
n
+
1
n
2
< q
2
+
2
q
n
+
1
n
because
n >
1
.
≤
q
2
+
2

q

+ 1
n
because
q
≤ 
q

.
<
2
.
(b) Suppose
r
∈
Q
and
r
2
>
2. Let
n
∈
N
and suppose
n >
2
r
r
2

2
. Show that
(
r

1
n
)
2
>
2.
Since
n >
2
r
r
2

2
, we have
r
2

2
>
2
r
n
and so
r
2

2
r
n
>
2.
We therefore have:
(
r

1
n
)
2
=
r
2

2
r
n
+
1
n
2
> r
2

2
r
n
>
2
.
(c) Using the results of (a) and (b), together with the fact that there is no
s
∈
Q
with
s
2
= 2 (do not prove this), show that
Q
is not complete. (Hint: show that
S
=
{
x
∈
Q
: 0
< x
2
<
2
}
is bounded above but has no least upper bound.)
It is clear that
S
is bounded above, for example by
b
= 2. We must show that
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 09/22/2010 for the course MATH 413 at Cornell University (Engineering School).
 '08
 PROTSAK

Click to edit the document details