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I have just taught the classic proof by contradiction that $\sqrt 2$ is irrational, and one of my students came up with the following proof:

Assume that $\sqrt 2$ is rational, so $\sqrt 2=\frac{a}{b}$ where $a$ and $b$ are integers such that $\frac{a}{b}$ is irreducible.

$2=\frac{a^2}{b^2}$

$b^2=\frac{a^2}{2}$

$b^2$ is a square number, and $a^2$ is a square number, but a square number divided by 2 cannot equal a square number, so there is a contradiction.

To justify this claim:

If a number $a$ is even, then $a=2n$, so $a^2=4n^2$.

$\frac{a^2}{2}=2n^2$ and the square root of $2n^2$ is $\sqrt{2}n$, which is clearly not an integer, therefore $2n^2$ is not a square number.

If a number $a$ is odd, then $a=2n+1$, so $a^2=4n^2+4n+1$.

$\frac{a^2}{2}=2n^2+2n+\frac{1}{2}$ which is an integer add a half, so it is not an integer. Therefore it is not a square number.

So, since a square number divided by 2 is not a square number, the contradiction is reached in the line:

$b^2=\frac{a^2}{2}$

Therefore $\sqrt 2$ is irrational

Is this sound?

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    $\begingroup$ Why is $\sqrt 2\, n$ "clearly" not an integer? Isn't that what you are trying to prove? $\endgroup$
    – lulu
    Commented Jun 15, 2022 at 12:49
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    $\begingroup$ I take the offence with the sentence: “... and the square root of $2n^2$ is $\sqrt{2}n$, which is clearly not an integer.” This is a circular argument. $\endgroup$
    – user700480
    Commented Jun 15, 2022 at 12:49
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    $\begingroup$ It's exactly a rewording of what you're trying to prove. $\endgroup$ Commented Jun 15, 2022 at 12:52
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    $\begingroup$ It’s just a rewording. “$\sqrt{2}n \not \in \mathbb{Z}$ for all $n \neq 0$“ is true if and only if “for all $\sqrt{2} \neq \frac{m}{n}$ for all $m,n \in \mathbb{Z}$, $n \neq 0$“ is true. The statement on the right hand side is exactly that $\sqrt{2}$ is irrational. $\endgroup$
    – Joe
    Commented Jun 15, 2022 at 12:54
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    $\begingroup$ On the other hand, there are proofs other than the classic one. You may enjoy these (which of course also include the classic proof). $\endgroup$
    – J.G.
    Commented Jun 15, 2022 at 13:21

1 Answer 1

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$\sqrt{2}n$, which is clearly not an integer

This step appears to be implicitly making the inference \begin{gather}n\ne0\implies\Big(\sqrt2\not\in\dfrac{\mathbb Z}n \implies \sqrt{2}n\not\in\mathbb Z\Big)\tag✓\\\text{and}\quad\color{red}{\sqrt2\not\in\dfrac{\mathbb Z}n};{\tag!}\\\text{therefore,}\quad{\big(n\ne0\implies\sqrt{2}n\not\in\mathbb Z\big)}.\end{gather} Although this inference per se is sound, by begging the question, it invalidates the proof containing it. Therefore, the proof is fallacious.

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  • $\begingroup$ If $\sqrt{2}$ were an integer, wouldn't that imply there was an integer $i$ s.t. $1 < i < 2$? $\endgroup$
    – Spencer
    Commented Jun 15, 2022 at 23:26
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    $\begingroup$ @Spencer Yes, but the fact that $\sqrt{2}$ is not an integer does not automatically imply that for all positive integers $n$, $\sqrt{2} \ n$ is not an integer. $\endgroup$ Commented Jun 15, 2022 at 23:38
  • $\begingroup$ @Spencer (If I've misunderstood your remark, apologies.) $1.5$ isn't an integer yet equals the quotient of two integers; in any case, while the red portion is actually true, the point is that the student was assuming it in order to prove it, thereby committing circular reasoning. $\endgroup$
    – ryang
    Commented Jun 16, 2022 at 3:53
  • $\begingroup$ @ryang No, you're right; I need to read these things more closely before commenting. $\endgroup$
    – Spencer
    Commented Jun 16, 2022 at 12:45

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