**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 $latex \sqrt{3}$ 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 $latex \sqrt{2}$ is irrational.

**Theorem: ** $latex \sqrt{3}$ is irrational.

**Proof**

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

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

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

Squaring both sides, we have

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

which implies that

$latex 3b^2 = a^2$.

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

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

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

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

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

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

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

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