Proof: x is irrational => sqrt(x) is irrational

Ok so I am to prove: If x is irrational, then $$\sqrt{x}$$ is irrational. So I started by trying to prove the contrapositive: If $$\sqrt{x}$$ is rational, then x is rational.

So then $$\sqrt{x} = \frac{m}{n}$$ For integers m and n, n$$\neq$$0

Then square both sides. $$x = \frac{m^2}{n^2}$$

This is clearly rational because m^2 and n^2 are integers.

Now, is this a satisfactory proof? I am sure it is, it just seems as though it was too easy. Did my teacher ask it because it shows how proving the contrapositive can sometimes make life easy? Thanks.

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Yes, that's correct, and it is an easy proof. It boils down to this: Every square of a rational number is rational, and so these rational squares are the only numbers which have rational square roots.

Hurkyl
Staff Emeritus