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I was asked to prove that the difference of two odd numbers is even but, given that I might be the most distracted person in the world, I proved the sum. However, I think it should still be correct but I'll first show you my proof.


Let $n$ and $m$ be odd integers. Then $\exists k,l\in\mathbb Z:m=2k+1\ \land n=2l+1$ . Then, $$m+n=(2k+1)+(2l+1)$$ $$=\ 2k+2l+2$$ $$=2(k+l+1)$$ Since $k+l+1$ is an integer, then $2(k+l+1)$ is even.


I told my professor that this proof still worked for the difference of odd numbers because if you make $n$ a negative integer, then that is basically subtraction. He says that that would've been correct had I specified that $n$ was negative. But I find that redundant because by definition an integer can be either positive or negative. Then this proof should work for the four cases in which a) both $m$ and $n$ are positive, b) $m$ is positive and $n$ is negative, c) $m$ is negative and $n$ is positive, and d) both $m$ and $n$ are negative.


I didn't feel it was right to with him, given he's a professor and I'm just an undergrad, but I just can't really understand the difference. I hope you can help me understand and correct me if I'm wrong.

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    $\begingroup$ You succeeded in showing that given arbitrary odd integers $m,n$ that $m\color{red}{+}n$ is even. You did not actually specify anything about subtraction in your argument, but you could reuse all of your earlier work to show that $m\color{red}{-}n$ is also even by noting $m\color{red}{-}n=m+(\color{red}{-}n)$ and that $n$ odd implies that $-n$ is also odd, showing that the difference of two odd integers can be represented instead as the sum of two odd integers allowing you to use your previous work to complete the proof. $\endgroup$ – JMoravitz Sep 8 '17 at 19:34
  • $\begingroup$ If you had explicitly stated that the proof also works for subtraction using $\,m-n = m+(-n)\,$ and you also proved $\,n\,$ odd $\,\Rightarrow\, -n\,$ odd, then there could be no objection. Did you do so? $\ $ $\endgroup$ – Bill Dubuque Sep 10 '17 at 16:43
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Let $n$ and $m$ be odd integers. Then $\exists k,l\in\mathbb Z:m=2k+1\ \land n=2l+1$ . Then, $$m+n=(2k+1)+(2l+1)$$ $$=\ 2k+2l+2$$ $$=2(k+l+1)$$ Since $k+l+1$ is an integer, then $2(k+l+1)$ is even.

Your argument proves that "the sum of two odd integers is even", which indeed implies that the difference of two integers is also even. But this argument is not identical to a proof of want you want.

I told my professor that this proof still worked for the difference of odd numbers because if you make $n$ a negative integer, then that is basically subtraction. (No. It is not. In your proof, n and m are only assumed to be odd integers. There is no assumption about whether they are positive or negative. When $n$ is negative, $m+n$ means a subtraction between $m$ and $(-n)$, not a subtraction between $m$ and $n$.)

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  • $\begingroup$ You said, "No. It is not. In your proof, n and m are only assumed to be odd integers. There is no assumption about whether they are positive or negative." But there is; namely, the assumption is made by saying "integer". Otherwise, what's the difference between an odd integer and an odd natural number? By saying integer I'm assuming the number is either positive or negative. Am I not? @BillDubuque I'm defining $m$ and $n$ to be odd integers. Namely, $\exists k,l\in\mathbb Z:m=2k+1\ \land n=2l+1$. Since $k$ and $l$ are also integers, they are either positive or negative. Are they not? $\endgroup$ – Alex D Sep 12 '17 at 5:54
  • $\begingroup$ @AlexD.: Again: "Your argument proves that "the sum of two odd integers is even", which indeed implies that the difference of two integers is also even. But this argument is not identical to a proof of want you want." $\endgroup$ – Jack Sep 12 '17 at 11:58
  • $\begingroup$ @AlexD. Let $m=5$ and $n=3$. How does your argument show that $5-3$ is even? $\endgroup$ – Jack Sep 12 '17 at 13:06

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