Solutions to Homework Assignment 4
Math 74, Fall 2006
October 6, 2006
1.
Definition 1.
A natural number
n
is a
perfect square
if there exists an integer
k
such that
n
=
k
2
.
Theorem 2.
If
n
∈
N
then
√
n
∈
Q
if and only if
n
is a perfect square.
Proof.
Clearly if
n
∈
N
is a perfect square, then
√
n
∈
N
⊂
Q
. Suppose, on the other
hand, that
n
∈
N
is not a perfect square. Define
B
=
k
∈
N

k
√
n
∈
Z
.
We will show that
B
is nonempty. Proceeding by way of contradiction, suppose that
B
is not empty. By the WellOrdering Principle,
B
has a least element. Let
b
= min
B
.
Then
b
√
n
∈
Z
since
b
∈
B
. Now choose an integer
m
such that
m <
√
n
≤
m
+ 1
.
By the WellOrdering Principle, it is obvious that such an integer
m
exists, and that
it is unique. Also, since
n
is not a perfect square,
√
n
=
m
+ 1, and so
√
n < m
+ 1.
Now define
k
=
b
(
√
n

m
)
.
Notice that
k
is an integer, since
k
=
b
√
n

bm
is the difference of two integers. Also,
our choice of
b
guarantees that 0
<
√
n

m <
1. So we conclude that
0
< k < b.
Since
k
√
n
=
bn

m
(
b
√
n
) is the difference of two integers, and is therefore an integer,
we deduce that
k
∈
B
. But
k < b
contradicts the fact that
k
= min
B
.
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 Winter '06
 doolittle
 Mathematical Induction, Perfect square, Natural number, Prime number, Proposition 8

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