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