Answer :

Let us suppose that √2 is an rational number and represent it in the form of where p and q are relatively prime and q≠0.

So,

On squaring both sides we get,

p^{2} = 2q^{2}

As 2q^{2} is even then p^{2} is also even and p is also even.

Let, p^{2} = 4a^{2}

4a^{2} = 2q^{2}

⇒ q^{2} = 2a^{2}

As 2a^{2} is even then q^{2} is also even and q is also even.

If p and q both are even, then they are further calculated which is not possible because of we assumed p and q are relatively prime number.

It means our assumption is wrong. So, by contradiction we can say that √2 is irrational number.

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