We prove that #\sqrt{2}# is irrational. We use a proof by contradiction. That means we assume #\sqrt{2}# is rational, then we will find a contradiction, and then we conclude that it cannot be true that #\sqrt{2}# is rational.

If we assume that #\sqrt{2}# is rational, we can write it as a fraction #\tfrac{\blue p}{\orange q}# for which #\blue p# and #\orange q# are integers. The integers #\blue p# and #\orange q# have no common divisors because if they did, we could further simplify the fraction.

Therefore:

\[\sqrt{2}=\frac{\blue p}{\orange q}\]

When squaring both sides, we find:

\[2=\frac{\blue p^2}{\orange q^2}\]

By multiplying both sides by #\orange q^2#, we find:

\[2\orange q^2=\blue p^2\]

The left-hand side is divisible by #2#, so the right-hand side should be too. Therefore, #\blue p^2# is divisible by #2# and therefore, so is #\blue p#.

Next, we note that if #\blue p# is divisible by #2#, we can write #\blue{p}# as #2 \cdot a#, for which #a# is an integer. This means:

\[2 \orange q^2=(2a)^2=4a^2\]

If we divide both sides by #2#, we find:

\[\orange q^2=2a^2\]

This means that #\orange q^2# is divisible by #2#, and therefore #\orange q# is also divisible by #2#.

So we see that both #\blue p# and #\orange q# are divisible by #2#. This contradicts the fact that #\blue{p}# and #\orange{q}# have no common divisors. Therefore, our assumption was wrong and we cannot write #\sqrt{2}# as a fraction. #\sqrt{2}# is therefore irrational.