$\sqrt{3}$ is an irrational number.

Proof:

Suppose it is a rational number and let $\sqrt{3} = \frac{p}{q}$ where $p$ and $q$ are integers and $q$ is not zero. Furthermore, assume that the fraction is in its simplist form. Now we can square each side and simplify

$$ \begin{align*} (\sqrt{3})^2 &= \big(\frac{p}{q}\big)^2 \\ 3 &= \frac{p^2}{q^2} \\ p^2 &= 3q^2. \\ \end{align*} $$

From this we can conclude that $p^2$ is divisible by 3. But by the fundamental theorem of arithmetic $a$ is also divisble by 3. So let $p = 3k$ where $k$ is an integer. Let’s plug in $2k$ back in the main equation and simplify

$$ \begin{align*} p^2 &= 3q^2 \\ (3k)^2 &= 3q^2 \\ 9k^2 &= 3q^2 \\ 3k^2 &= q^2 \\ \end{align*} $$

But now $q^2$ is divisible by 3 which means that $q$ is divisible by 3 but this means that $q$ and $p$ must have 3 as a common factor. However, we said earlier that $p^2/q^2$ is in its simplest form. Therefore, this is a contradiction and $\sqrt{3}$ can not be a rational number.