Suppose to the contrary that \(\sqrt{2}\) is rational. Then, there must exist two coprime integers \(p\) and \(q\) such that\(\sqrt{2} = \frac{p}{q}\). We can rearrange this equation as follows:

\[\sqrt{2} = \frac{p}{q}\] \[2q^2 = p^2\]

From the above it is clear that \(p^2\) is even. It follows that \(p\) is also even, since the square of any odd number is odd. Thus, we can substitute \(p\) with \(2k\), where \(k\) is any integer:

\[2q^2 = p^2\] \[2q^2 = 4k^2\] \[q^2 = 2k^2\]

By a similar argument, we can again see that both \(q^2\) and \(q\) are even. The fact that both \(p\) and \(q\) are even contradicts the initial statement that the two are coprime. Thus, we have reached a contradiction, and \(\sqrt{2}\) must be irrational.