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Just wondering if this is the correct way to write this proof. Thank you!

Assume $a$ and $b$ have opposite parity. We’ll consider two cases: $a$ is even, $b$ is odd or $a$ is odd, and $b$ is even. WLOG, suppose $a$ is even, and $b$ is odd. By definition, $a=2m$ and $b=2n+1$ for some integers $m,n$. Then $a-b=2m-2n+1$ and therefore $a-b=2(m-n)+1$. Since $m-n$ is an integer, then $a-b$ is odd.

Wng427
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    For one, it seems you have proved it the wrong way (you showed $a,b$ have opposite parity $\implies a-b$ is odd). Also, since it is $a-b$ and not $b-a$, you have to show that you can suppose $a$ is even without loss of generality. – healynr May 06 '20 at 20:19
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    If $P$ is the statement $a-b$ is odd, and $Q$ is the statement $a$ and $b$ have opposite parity, the problem asked you to prove that $P\implies Q$. Here however, your proof is for $Q\implies P$ which is not the same thing. "If $x$ is a healthy normal horse then $x$ has four legs" is not the same thing as "If $x$ has four legs then $x$ is a healthy normal horse." My desk has four legs but clearly is not a horse. – JMoravitz May 06 '20 at 20:20

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You have proven the statement: "If a and b have opposite parity, then a-b is odd" instead of "If a-b is odd, then a and b have opposite parity".

In particular, if you want to show that a and b have opposite parity, you cannot start your proof with "Assume a and b have opposite parity".

You either have to start from your assumption, "Assume that a-b is odd", or from contraposition: "Asssume that a and b do not have opposite parity", and from there conclude a contradiction with your assumption, i.e. show "if a and b do not have opposite parity, then a-b is not odd".

Hetebrij
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Let $a-b$ be odd. Then $a-b = 2n+1$ for some integer $n$. If we take both sides $\mod 2$, then we have $a-b \equiv 1 \mod 2$. Therefore $a,b$ have opposite parity since their difference $\not \equiv 0 \mod 2$.

healynr
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  • You're just saying that a-b is odd (since $a-b\equiv 1 \text{ mod } 2 $) then a and b have opposite parity. You're basically assuming the result without actually showing it clearly. – Anonmath101 May 06 '20 at 22:39
  • @Anon I am taking the definition of being odd (ie. equaling $2n+1$ for some $n$) and showing it is equivalent: My definition of opposite parity is incongruence modulo 2, which is what I showed. (I don't know what else you would want to see) I am not assuming the result, I am showing the result follows with very little manipulation from the given. $a-b \equiv 1 \mod 2$ is not, as far as I know, the definition of being odd, but rather a consequence of it. – healynr May 06 '20 at 22:49
  • What you mean is that if a and b had the same parity then a-b=0 mod 2. Which is what the proof is. – Anonmath101 May 07 '20 at 16:43