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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? $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ 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). $\endgroup$ 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} This inference is sound, but its assertion $(!!)$ begs the question, rendering the overall proof at best tautological (and informally fallacious).

Unfortunately, the overall proof is in fact invalid, because the contradiction that was subsequently reached was invalidly attributed to that explicit assumption "$\sqrt2$ is rational" instead of the actual assumption "assertion $(!!)$ and $\sqrt2$ is rational". If we fix this, then the main proof, in the context of $\mathbb R,$ just unhelpfully says this: if assertion $\boldsymbol{(!!)}$ holds, then $\sqrt2$ is irrational.

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  • $\begingroup$ If $\sqrt{2}$ were an integer, wouldn't that imply there was an integer $i$ s.t. $1 < i < 2$? $\endgroup$ 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. $\endgroup$ Commented Jun 16, 2022 at 3:53
  • $\begingroup$ @ryang No, you're right; I need to read these things more closely before commenting. $\endgroup$ Commented Jun 16, 2022 at 12:45

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