An
irrational number
is a real number that is not rational, that
is, cannot be expressed as a fraction (or
ratio
) of the form
p
/
q
,
where
p
and
q
are integers.
proofs that the square root of 2 is irrational
Pythagorean proof, as given by Euclid in his
Elements
proof by contradiction:

Assume that
2
is rational, that is, there exists integers
p
and
q
such
that
2 =
p
/
q
;
take the irreducible form of this fraction, so that
p
and
q
have no factors in common

square both sides, to give 2 =
p
^{
2
}
/
q
^{
2
}

rearrange, to give 2
q
^{
2
}
=
p
^{
2
}

hence
p
^{
2
}
is even

hence
p
is even (trivial proof left as an excercise for the
reader); write
p
= 2
m

substitute for
p
in (3), to give 2
q
^{
2
}
= (2
m
)
^{
2
}
= 4
m
^{
2
}

divide through by 2, to give
q
^{
2
}
= 2
m
^{
2
}

hence
q
^{
2
}
is even

hence
q
is even
(1) assumes that
p
and
q
have no factors in common; (5)
and (9) show they they both have 2 as a factor. This is a contradiction.
Hence the assumption (1) is false, and
2
is
not
rational.
proof based on unique factorisation
[I first saw this rather elegant little proof in Chaitin's
Meta
Math!
, p98. It uses the result of unique factorisation into
powers of primes (the
fundamental
theorem of arthimetic
).]

Assume that
2
is rational, that is, there exists integers
p
and
q
such
that
2 =
p
/
q
;
take the irreducible form of this fraction, so that
p
and
q
have no factors in common

square both sides, and rearrange, to give 2
q
^{
2
}
=
p
^{
2
}

factorise
p
and
q
into their
unique
prime
factoriastions,
p
= 2
^{
a
}
3
^{
b
}
5
^{
c
}
...,
q
= 2
^{
x
}
3
^{
y
}
5
^{
z
}
...

substitute into (2), and simplify, to give 2
^{
2
x
+1
}
3
^{
2
y
}
5
^{
2
z
}
...
= 2
^{
2
a
}
3
^{
2
b
}
5
^{
2
c
}
...
Since the LHS and the RHS refer to the
same
number, and since
such factorisation is unique, we have 2
x
+1 = 2
a
(or
odd
=
even
). This is a contradiction. Hence the assumption (1) is
false, and
2
is
not
rational.