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Solutions for HWK 3
Based on the HWKs by my former student JINGYUAN LIU
October 10, 2009
Problem A
Part 1
Every number
r
∈
Q
can be expressed as
r
=
m
n
where
m
∈
Z
,
n
∈
N
, and
(
m,n
) = 1
. It is clear that
r
is
the root of the polynomial
f
(
x
) =
m

nx
, hence
r
∈
A
. So
Q
⊂
A
.
Part 2
The set
P
k
of polynomials of xed degree,
k
, with integer coe cients
z
0
,z
1
,...,z
k
∈
Z
consists of elements
(polynomials) of the form
z
0
x
0
+
z
1
x
1
+
···
+
z
k
x
k
where
z
k
6
= 0
. Each polynomial is uniquely identi ed by its
integer coe cients so there is a
1 : 1
correspondence between elements of this set, and the set
Z
k
×
(
Z
\{
0
}
)
.
We proved in class that
Z
is countable;
Z
\ {
0
}
)
is countable as a subset of
Z
. By a proposition proved in
the class the Cartesian product of a nite number of countable sets is countable, so the set
Z
k
×
(
Z
\{
0
}
)
is
countable, therefore the set
P
k
is countable, too.
Part 3
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