Proof that the square root of 2 is an irrational number

 Hello everyone. Today we will prove that $\sqrt{2}$ is an irrational number. We will use the method of contradiction as a proof method. This proof was first given by the Greek philosopher Aristotle. 

Theorem: $\sqrt{2}$ is an irrational number. 

Proof: 

Suppose that $\sqrt {2}$ is a rational number. Then $\sqrt {2}$ can be written in the form $\frac{p}{q}$ where $p$ and $q$ are coprime integers such that $q \neq 0$. Note that since $\frac{p}{q}$ is an irreducible fraction $p$ and $q$ cannot be even at the same time, otherwise the fraction would not be irreducible. 

From the equation $\sqrt{2} = \frac{p}{q}$ it follows that $2 = \frac{p^2}{q^2}$ or $p^2 = 2q^2$. So $p^2$ is an even number from which it follows that $p$ is also an even number so we can write the number $p$ in the form $p = 2k$ , where $k$ is an integer . 

So, we have that $(2k)^2 = 2q^2$ i.e. $4k^2 = 2q^2$, or $q^2 = 2k^2$. From the last equation we conclude that $q^2$ is an even number, so $q$ is an even number too. 

We have shown that $p$ and $q$ are even numbers, which contradicts the assumption that $\frac{p}{q}$ is an irreducible fraction. Thus, the initial assumption that $\sqrt{2}$ is a rational number is incorrect, which means that $\sqrt{2}$ is an irrational number. 

$\blacksquare$