# Prove that number $\sqrt{2}$ is an irrational number using this theorem "if $a^2$ is even, $a$ must be even"

I have to prove the following:

Prove $$\sqrt{2}$$ is an irrational number using this theorem "if $$a^2$$ is even, $$a$$ must be even"

I made a proof by contradiction for the statement above, but I believed that there is a mistake in my proof.
Suppose $$\sqrt{2}$$ is a rational number,

let a = $$\sqrt{2}$$ \begin{align} a = \sqrt{2}\\ a^2 = 2 \end{align} Therefore, $$a^2$$ is even
Given that if $$a^2$$ is even, $$a$$ must be even,

but $$\sqrt{2}$$ is not even
Hence, we have a contradiction that if $$a^2$$ is even, $$a$$ must be even
Thus, $$\sqrt{2}$$ is an irrational number

Question: Would someone mind pointing out where my mistake is and why it is a mistake?
Thank you for your kind attention

• " is/is not even" means nothing for non integers like $\sqrt{2}$ or like your rational number $a.$ Nov 13, 2022 at 8:47
• Thanks. I get it now. Nov 13, 2022 at 8:52

For the sake of contradiction, suppose that $$\sqrt{2}$$ is rational. We can write $$\sqrt{2}=\frac{p}{q},$$ where $$p,q$$ are integers and the RHS fraction is irreducible. Squaring both sides and multiplying by $$q^2$$ we get $$2q^2=p^2.$$ Since, $$2q^2$$ is even then $$p^2$$ is even. Can you follow from here?
• Because "if $a^2$ is even, $a$ must be even", p is even. Since p is even, there is an integer $n$ that $p = 2n$. We can obtain $2q^2 = 4n^2$, then $q^2 = 2n^2$. Thus, q is even. But as the RHS fraction is irreducible, p and q are not both even. Hence, there is a contradiction. Is it reasonable? Nov 14, 2022 at 12:18