Proof that $\sqrt{3}$ is irrational

This is a small proof that shows that $\sqrt{3}$ is irrational.

Proof:

Suppose that $\sqrt{3}$ is a rational number. If this is the case, then $\sqrt{3}$ can be written as a fraction in lowest form

$$ \sqrt{3} = \dfrac{a}{b}\quad (a,b \in \mathbb{N}). $$

If we square both sides to get $\frac{a^2}{b^2} = 3$, then we can rewrite it as $3b^2 = a^2$. If $a^2$ would be even, then $b^2$ would also be even, which is not possible because it is already in lowest form. Thus $a^2$ must be odd, which means that $a$ must also be odd. If you are not convinced of this, look at the following multiplications of even and odd numbers:

• $\mathrm{even} \times \mathrm{even}$ gives $(2n)^2 = 4n^2$ which is even.
• $\mathrm{odd} \times \mathrm{odd}$ gives $(2n+1)^2 = 4n^2$ which is odd.
• $\mathrm{even} \times \mathrm{odd}$ gives $(2n)(2n+1) = 4n^2+2n$ which is even.

This means that for $a^2$ to be odd, $a$ must also be odd. Any other case requires an even number. Using the assumption that $a$ is odd, we can replace $a$ with $2n + 1\ (n \in \mathbb{N})$, which gives 

$$ 3b^2 = (2n + 1)^2 $$

However, $3b^2$ must also be odd, from which follows that $b$ must be odd. Thus, we can rewrite $b$ as $2m + 1\ (m \in \mathbb{N})$. Substituting both and some algebra gives

$$\begin{align} 3(2m + 1)^2 &= (2n+1) \\ 12m^2 + 6m + 3 &= 4n^2 + 4n + 1 \\ 12m^2 + 6m + 2 &=  4n^2 + 4n \\ 6m^2 + 3m + 1 &= 2n^2 + 2n \\ 3(2m^2+m)+1 &= 2(n^2 + n). \end{align}$$

We now have that the left-side is an odd number, and the right side is an even number, which is not possible. This means that there is a contradiction in our assumption that $\sqrt{3}$ can be written as a fraction in lowest form. Therefore $\sqrt{3}$ is not rational, thus irrational. $\blacksquare$