Question

How do you proof that sqrt(2) is irrational?

I tried this,
p/q = sqrt(2)
p^2/q^2=2
p^2=2q^2
let p = q * sqrt(2) , (equation from top one)
(q*sqrt(2))^2=2q^2
2q^2=2q^2,
q=q

Is anything wrong or does this also prove that sqrt(2) is irrational, if so how?

Answer 1

Assume √2 is rational, so

√2 = p/q

for some relatively prime integers p,q - meaning we can't reduce p/q any further - and q ≠ 0.

Then

p = √2 q

and squaring both sides gives

p² = 2q²

This tells us that p² is even, so 2 divides p². But if 2 divides p², it must also divide p, since p is a square number. So p = 2r for some integer r, and we have

(2r)² = 2q²

4r² = 2q²

2r² = q²

We're in the same situation as before - 2 divides q², so 2 divides q. Both p and q are thus divisible by 2, but this contradicts our assumption that p and q are relatively prime, which means √2 cannot be expressed as a rational number p/q. Therefore √2 is irrational.

What you've written in your proof is technically correct, but it doesn't help in the irrationality proof.