# How to prove square root of 2 is irrational?

A rational number is a number that can be expressed in a fraction with integers as numerators and denominators.

Some examples of rational numbers are 1/3, 0, -1/2, etc. Now, we know that $\sqrt{2}\approx 1.41421\cdots$.

Is the square root of 2 rational? Or is it irrational (the opposite of rational)? How do we prove it? It turns out we can prove that the square root of two is irrational using a technique called proof by contradiction. (One of the earlier posts on this blog also used proof by contradiction to show that there are infinitely many prime numbers.)

First, we suppose that $\displaystyle\sqrt{2}=\frac{p}{q}$, where $\displaystyle\frac{p}{q}$ is a fraction in its lowest terms.

Next, we square both sides to get $\displaystyle 2=\frac{p^2}{q^2}$.

Hence, $2q^2=p^2$. We can conclude that $p^2$ is even since it is a multiple of 2. Thus, $p$ itself is also even. (the square of an odd number is odd).

Thus, we can write $p=2k$ for some integer k. Substituting this back into $2q^2=p^2$, we get $2q^2=4k^2$, which can be simplified to $q^2=2k^2$.

Hence, $q^2$ is also even, and hence $q$ is also even!

But if both $p$ and $q$ are even, then $\displaystyle\frac{p}{q}$ is not in the lowest terms! (we could divide them by two). This contradicts our initial hypothesis!

Thus, the only possible conclusion is that the square root of two is not a rational number to begin with!

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