THREE PROBLEMS
PROBLEM 1
:
Prove that
√
2 is irrational.
Proof
:
Assume to the contrary that
√
2 is rational, that is
√
2 =
a
b
,
where
a
and
b
are integers
and
b
6
= 0
.
Moreover, let
a
and
b
have no common divisor
>
1
. Then
2 =
a
2
b
2
⇒
2
b
2
=
a
2
.
(1)
Since 2
b
2
is even, it follows that
a
2
is even. Then
a
is also even
(in fact, if
a
is odd, then
a
2
is
odd). This means that there exists
q
∈
Z
such that
a
= 2
q.
Substituting this into (1), we get
2
b
2
= (2
q
)
2
⇒
b
2
= 2
q
2
.
Since 2
q
2
is even, it follows that
b
2
is even. Then
b
is also even
by
the arguments above. This is a contradiction.
±
THEOREM
(DIVISION ALGORITHM):
For any integers
a
and
b
with
a
6
= 0
there exist
unique integers
q
and
r
such that
b
=
aq
+
r,
where
0
≤
r <

a

.
The integers
q
and
r
are called the
quotient
and the
reminder
respectively.
PROBLEM 2
:
Prove that
√
3 is irrational.
Proof
:
Assume to the contrary that
√
3 is rational, that is
√
3 =
a
b
,
where
a
and
b
are integers
and
b
6
= 0
.
Moreover, let
a
and
b
have no common divisor
>
1
. Then
3 =
a
2
b
2
⇒
3
b
2
=
a
2
.
(2)
Since 3
b
2
is divisible by 3, it follows that
a
2
is divisible by 3. Then
a
is also divisible by 3.
In fact, if
a
is not divisible by 3, then by the Division Algorithm there exists
q
∈
Z
such that
a
= 3
q
+ 1
or
a
= 3
q
+ 2
.
Suppose
a
= 3
q
+ 1
,
then
a
2
= (3
q
+ 1)
2
= 9
q
2
+ 6
q
+ 1 = 3(3
q
2
+ 2
q

{z
}
q
0
) + 1 = 3
q
0
+ 1
,
which is not divisible by 3. We get a contradiction. Similarly, if
a
= 3
q
+ 2
,
then
a
2
= (3
q
+ 2)
2
= 9
q
2
+ 12
q
+ 4 = 3(3
q
2
+ 4
q
+ 1

{z
}
q
00
) + 1 = 3
q
00
+ 1
,
which is not divisible by 3. We get a contradiction again.
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 Fall '10
 GIESEKER,D.
 Algebra, Division, Integers, Prime number, Greatest common divisor, Divisor, common divisor

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