Question

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

Answer 1

Let $\sqrt{2}$ is a rational number.

So, we can find integers $p$ and $q(\ne 0)$ such that $\sqrt{2}=\frac{p}{q}$.

Let $p$ 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 coprime numbers.

So, $b\sqrt{2}=a$.

Square both the sides, $2b^2=a^2$.

Therefore, $2$ divides $a^2$. Now, by theorem, it follows that $2$ divides $a$.

So, we can write $a=2c$ for some integer $c$.

Therefore, $2b^2=4c^2$

and $b^2=2c^2$.

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 has arisen because of our incorrect assumption that $\sqrt{2}$ is a rational number.

So, we can conclude that $\sqrt{2}$ is an irrational number.

Therefore, hence proved.