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.

On squaring both sides we get,

p2 = 2q2

As 2q2 is even then p2 is also even and p is also even.

Let, p2 = 4a2

4a2 = 2q2

q2 = 2a2

As 2a2 is even then q2 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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<span lang="EN-USRS Aggarwal - Mathematics