SOME IRRATIONAL NUMBERS
PETE L. CLARK
Proposition 1.
The square root of
2
is irrational.
Proof.
Suppose not:
then there exist integers
a
and
b
= 0 such that
√
2 =
a
b
,
meaning that 2 =
a
2
b
2
. We may assume that
a
and
b
have no common divisor – if
they do, divide it out – and in particular that
a
and
b
are not both even.
Now clear denominators:
a
2
= 2
b
2
.
So 2

a
2
. It follows that 2

a
. Notice that this is a direct consequence of Euclid’s
Lemma – if
p

a
2
,
p

a
or
p

a
.
On the other hand, we can simply prove the
contrapositive: if
a
is odd, then
a
2
is odd.
By the Division Theorem, a number
is odd iff we can represent it as
a
= 2
k
+ 1, and then we just check: (2
k
+ 1)
2
=
4
k
2
+ 4
k
+ 1 = 2(2
k
2
+ 2
k
) + 1 is indeed again odd. So
a
= 2
A
, say. Plugging this
into the equation we get
(2
A
)
2
= 4
A
2
= 2
b
2
, b
2
= 2
A
2
,
so 2

b
2
and, as above, 2

b
. Thus 2 divides both
a
and
b
: contradiction.
Comment: This is a truly “classical” proof.
In G.H. Hardy’s
A Mathematician’s
Apology
, an extended rumination on the nature and beauty of pure mathematics,
he gives just two examples of theorems: this theorem, and Euclid’s proof of the
infinitude of primes. As he says, this is inevitably a proof by contradiction (unlike
Euclid’s proof, which constructs new primes in a perfectly explicit way). The orig
inal statement is logically more complicated than what we actually prove in that
it takes for granted that there is some
real
number
√
2 – characterized by being
positive and having square equal to 2 – and then shows a “property” of this real
number, namely it not being a fraction.
But the essence of the matter is that a
certain mathematical object
does not
exist – namely a rational number
a
b
such that
(
a
b
)
2
= 2. This was the first “impossibility proof” in mathematics.
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 Spring '11
 Staff
 Number Theory, Integers, Prime number, Rational number, pete l. clark

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