# Proof that Square Root of Three is Irrational

Introduction

In The Intuitive Proof of the Infinitude of Primes, I showed you a proof strategy called proof by contradiction.  In this post, we use this strategy to prove that $\sqrt{3}$ is irrational.  In proof by contradiction, if the statement P is true, you have to assume the contrary, and then find a contradiction somewhere. Note that the proof in this post is very similar to the proof that $\sqrt{2}$ is irrational.

Theorem:  $\sqrt{3}$ is irrational.

Proof

We assume the contrary; that is, we assume that $\sqrt{3}$ is rational. Then

$\sqrt{3} = \displaystyle\frac{a}{b}$

where $a$ and $b$ are integers with no common factor except 1. This means that $\frac{a}{b}$ is in lowest terms.

Squaring both sides, we have

$3 = \displaystyle\frac{a^2}{b^2}$.

which implies that

$3b^2 = a^2$.

Since $3$ is a factor of $3b^2$, this means that $a^2$ is divisible by $3$. Now, if $a^2$ is divisible by $3$ then $a$ is also divisible by $3$ (we still have to prove this, but you can try) (1).

Now, since $a$ is divisible by $3$, $3$ is a factor of $a$. That is, $a = 3k$, where $k$ is some integer. Now,

$3b^2 = a^2 \implies 3b^2 = (3k)^2 \implies 3b^2 = 9k^2$

Dividing both sides by $3$, we have $b^2 = 3k^2$ which means that $b^2$ is divisible by $3$. This implies that $b$ is also divisible by $3$ (2).

Now, in (1), $a$ is divisible by $3$ and in (2) $b$ is also divisible by $3$ This contradicts the fact that $a$ and $b$ have no common factors except 1.

Therefore, the assumption is false. So, $\sqrt{3}$ is irrational. $\blacksquare$

By the way, David Feinberg has written a nice poem about the Square Root of Three.