Question

Prove that `sqrt2` is an irrational number.

Answer 1

Let `sqrt2` be rational

Then, its simplest form = `p/q`

Where p and q are integers having no common factor other than 1, and q ≠ 0

Now, `sqrt2=p/q`

On squaring both sides we get

`2=p^2/q^2`

`2q^2=p^2`   ...(i)

⇒ 2 divides p²

⇒ 2 divides p (∵ 2 is a prime and divides p² => 2 divides p)

Let p = 2r for some integer r

Putting p = 2r in (i) we get

`2q^2=4r^2`

⇒ `q^2=2r^2`

⇒ 2 divides p²    (∵ 2 divides 2r²)

⇒ 2 divides q     (∵ 2 is prime and divides q² => 2 divides q)

Thus, 2 is a common factor of p and q. But this contradicts the fact that p and q have no common factor other than 1.

Thus, contradiction arises by assuming `sqrt2` is rational.

Hence, `sqrt2` is irrational.