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This is a proof in Discrete Mathematics.

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    $\begingroup$ Please include the proof in the text to make it accessible to more people. Also, what are your thoughts? $\endgroup$ – Carsten S Jul 27 '17 at 12:52
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    $\begingroup$ @CarstenS He already included his thoughts: "I can't find what it is wrong". $\endgroup$ – Peyton Jul 27 '17 at 12:53
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    $\begingroup$ The problem is with the $\leftarrow$. That you would need $a^2$ to be the double of an even number plus $1$. Therefore that implication is still missing proving that if $a^2$ is odd, it should have that form. $\endgroup$ – Peyton Jul 27 '17 at 12:54
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    $\begingroup$ After the third $\leftrightarrow$, it says $4k^4 + 4\color{red}{K} + 1$. Should be $4k^2 + 4\color{red}{k} + 1$. Nailed it! $\endgroup$ – tilper Jul 27 '17 at 12:55
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    $\begingroup$ @tilper LOL!!!! $\endgroup$ – Peyton Jul 27 '17 at 12:56
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The problem lies in the spurious use of biimplications. If we read the left-to-right implications, we get a correct proof of the statement

If $a$ is odd, then $a^2$ is odd

However, the reverse implication is not correct. We cannot conclude that if $a^2$ is odd, then $a^2 = 2(2k^2+2k)+1$ for some $k \in \mathbb{N}$. All we know is that then $a^2 = 2K+1$ for some $K \in \mathbb{N}$.

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  • $\begingroup$ Well, yes, you can conclude that, although one could argue that there are steps missing in the proof. $\endgroup$ – Dirk Jul 27 '17 at 13:09
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The problem is when you go back from

$a^2$ is odd

To

$a^2 = 2(2k^2+2k)+1$

That does not (immediately) follow!

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To be correct, you have to write

Let $n\in \Bbb N $.

$n $ is odd $\implies \exists k\in \Bbb N \;\; n=2k+1$

$$\implies \exists k\in \Bbb N \;: n^2=2 (2k^2+2k)+1$$

$$\implies \exists K=2 (2k^2+2k)\in \Bbb N \; :$$ $$ n^2=2K+1$$ $\implies n^2$ is odd.

Conversly,

$n$ is even $\implies \exists k\in \Bbb N \; : n=2k $

$$\implies \exists k\in\Bbb N \; : n^2=4k^2$$

$$\implies \exists K=2k^2\in \Bbb N \;: n^2=2K $$ $\implies n^2$ is even.

we conclude that $n $ is odd $\iff n^2$ is odd.

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It should be: $a$ is odd $\iff a=2k-1, k\in N.$

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Alberto Debernardi Jul 27 '17 at 14:19
  • $\begingroup$ @AlbertoDebernardi, thank you for advice. I think this is the seriuos flaw in the proof, since it leaves out the odd number $1$. $\endgroup$ – farruhota Jul 27 '17 at 15:27

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