Question

Prove that $\sqrt{2}$ is irrational.

Answer 1

Let us assume that $\sqrt{2}$ is rational. So, we can find integers $p$ and $q(\ne 0)$ such that,

$\sqrt{2}=\frac{p}{q}$
Suppose $p$ and $q$ have a common factor other than $1$.

Then, we divide by the common factor to get $\sqrt{2}=\frac{a}{b}$, where $a$ and $b$ are co-prime.
So, $b\sqrt{2}=a$
Squaring on both sides, we get
$2b^2=a^2$
Therefore, $2$ divides $a^2$
Now, we know that, if a prime number, $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.
Hence, $2$ divides $a$.

So, we can write $a=2c$ for some integer $c$.
Substituting for $a$, we get $2b^2=4c^2$

$\Rightarrow b^2=2c^2$
This means that $2$ divides $b^2$ and so, $2$ divides $b$.

Therefore, $a$ and $b$ have at least $2$ as a common factor.
But, this contradicts the fact that $a$ and $b$ have no common factors other than $1$.

This contradiction is because of our incorrect assumption that $\sqrt{2}$ is rational.

So, we can conclude that $\sqrt{2}$ is irrational. $\left[Henceproved\right]$