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Math 5615H: Introduction to Analysis I.
Fall 2011
Homework #2. Problems and Solutions.
#1.
Let
F
be a ﬁeld. Show that there exist not more that two diﬀerent solutions solutions of the equation
x
·
x
= 1. Is it possible that there is only one solution to this equation?
Solution.
If
x
·
x
= 1, then (
x

1)(
x
+ 1) =
x
2

1 = 0. Then
x
= 1 or
x
=

1. In the case
F
:=
{
0
,
1
}
with
1 + 1 = 0, we have

1 = 1. This is the only case when the given equation has exactly one solution.
#2.
Let
F
=
{
0
,
1
,
2
,
3
,
4
}
be a ﬁeld such that
1 + 1 = 2
,
1 + 2 = 3
,
1 + 3 = 4
,
1 + 4 = 0
.
Find the values of
xy
for all
x, y
∈
F
. Put the results into the table. You do not need to verify the axioms of a
ﬁeld.
Answer:
y
\
x
0
1
2
3
4
0
0
0
0
0
0
1
0
1
2
3
4
2
0
2
4
1
3
3
0
3
1
4
2
4
0
4
3
2
1
#3.
Show that any any two
rational
numbers
p
1
< p
2
there exists an
irrational
number
r
such that
p
1
< r < p
2
.
Solution.
We have
p
1
< r
:=
p
1
+(
p
2

p
1
)
/
√
2
< p
2
. If
r
is rational, then
√
2 = (
r

p
1
)
/
(
p
2

p
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This document was uploaded on 02/15/2012.
 Fall '09
 Math

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