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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

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

1 Answer 1

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I prove the statement so you can see how this kind of proofs are done:

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?

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    $\begingroup$ 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? $\endgroup$
    – rogemuggle
    Nov 14, 2022 at 12:18
  • $\begingroup$ That's perfect! $\endgroup$ Nov 14, 2022 at 12:21

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