Assume that \(\sqrt{2}\) is a rational number, and therefore can be expressed as \(a/b\), where \(a\) and \(b\) are integers and \(b \ne 0\).

We additionally assume that \(a/b\) is fully simplified, and therefore one or both must be odd. If both were even, then it can be simplified further.

\[\sqrt{2}=a/b\]\[2=a^2/b^2\]\[2b^2=a^2\]
\(a^2\) must be even, therefore \(a\) must also be even, because even \(×\) even is even. Even numbers can be represented as \(2n\), so we substitute \(a\) for this.

\[2=a^2/b^2\]\[2=(2n)^2/b^2\]\[2=4n^2/b^2\]\[2b^2=4n^2\]\[b^2=2n^2\]
\(b^2\) must be even, therefore \(b\) must also be even.

We have come to the conclusion that both \(a\) and \(b\) must be even, which contradicts the assumption we made at the start, which is that \(a\) and \(b\) can't both be even.

Assume that there is a finite number of prime numbers, which can be listed as:

\[p_1, p_2, p_3, ..., p_n\]
Consider the number which is one more than the product of all existing prime numbers:

\[N=p_1×p_2×p_3×...×p_n+1\]
When you divide \(N\) by any of the prime numbers \(p_1, p_2, p_3, ..., p_n\), you get a remainder of 1, therefore none of the prime numbers \(p_1, p_2, p_3, ..., p_n\) is a factor of \(N\).

So \(N\) must either be prime or have a prime factor which is not in the list of all prime numbers.

Both of these contradict the assumption made at the start that there is a finite list of prime numbers, therefore there is an infinite number of prime numbers.

(This is Euclid's proof.)