It is not hard to show that [tex] \sqrt 2[/tex] cannot be represented by a rational number. Suppose [tex] \sqrt 2[/tex] is rational, then we must have
[tex] \sqrt 2 = \frac{p}{q}[/tex]
where [tex] p [/tex] and [tex] q [/tex] are both integers.

We further suppose that [tex] p[/tex] and [tex] q[/tex] have no common factors since we could cancel these factors out and replace [tex] p[/tex] and [tex] q[/tex] with [tex] p’[/tex] and [tex] q’[/tex] which are both integers and have no common factors.

We now square our equation to get
[tex] 2 = \frac{p^2}{ q^2}[/tex]
which implies [tex] p^2 = 2 q^2[/tex].

This means that [tex] p^2 [/tex]is divisible by [tex] 2[/tex]. The square of an odd number is always odd, so that [tex] p [/tex] can’t be odd. Hence [tex] p[/tex] must be even. Since [tex] p[/tex] is even, [tex] p = 2 k[/tex] for some integer [tex] k[/tex].

This gives
[tex] p^2 = (2 k)^2 = 4 k^2 = 2 q^2 [/tex]
and hence
[tex] q^2 = 2 k^2[/tex].
This means [tex] q^2[/tex] is divisible by [tex] 2[/tex], and by the same argument as before, [tex] q[/tex] must be even.

At the start we assumed [tex] \sqrt 2[/tex] was rational, so that [tex] \sqrt 2 = \frac{p}{q}[/tex] where [tex] p[/tex] and [tex] q[/tex] are integers with no common factors, but this leads us to conclude that both [tex] p[/tex] and [tex] q[/tex] are even, so they have the common factor [tex] 2[/tex].

We have a contradiction. The asumption that led us to the contradiciton is that [tex] \sqrt 2 [/tex] is rational. Either we accept the contradiction or we reject the assumption. As we can’t accept the contradiction, we must reject the assumption that [tex] \sqrt 2[/tex] is rational.