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Is my proof correct? Asking because it differs from solution in by book. Thanks in advance.

Lemma 1 given: "If the sum of two integers is even then they have the same parity"

Proof by contradiction.
Assume that an odd number is odd.

Case 1: 1 number is odd:
WLOG let $a + b$ be odd. Thenn by lemma 1 $a$ and $b$ have different parity. WLOG let $a$ be even. Since $a$ is even $c$ is also even, since $c$ is even $b$ is also even, which is a contradiction.

Case 2: all three numbers are odd: WLOG let $a$ be even, since $a$ is even $c$ is odd, since $c$ is odd $b$ is even, which results in both $a$ and $b$ being even which is a contradiction.

$\blacksquare$

a a
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    You could make it clearer by distinguishing between numbers ($a,b,c$) and sums ($a+b,b+c,c+a$). "Assume that an odd number is odd" becomes "Assume that an odd number of sums is odd". Also, note that the sum of the three sums is even; so you really only have to show that if the sum of any three numbers is even, then an even number of them must be odd. – TonyK Nov 13 '21 at 23:26
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    More generally it's easy to show that the parity of a sum is the parity of the number of odd summands. Since your $3$ terms have $\rm\color{#c00}{even}$ sum, there are an $\rm\color{#c00}{even}$ number of odd terms. – Bill Dubuque Nov 14 '21 at 00:26
  • This answer in the dupe has a simple proof of the claim in my prior comment, using arithmetic $!\bmod 2.,$ Notice how using modular arithmetic makes it completely obvious. – Bill Dubuque Nov 14 '21 at 01:10

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Yes, this should be correct. Your application of WLOG looks correct to me. However, I would recommend explaining why you say "since $a$ is even $c$ is also even" and "since $c$ is even $b$ is also even" - I know it's because $a+c$ and $b+c$ are even due to your assumption that $a+b$ is the odd one, but it would make it more thorough to state this.

There will usually be many ways to prove anything so don't worry that it differs from the book. It's a bit more complicated than needed though - if you assume that an odd number of $a+b,b+c,a+c$ are odd, then $(a+b)+(b+c)+(a+c)=2(a+b+c)$ is odd which is clearly impossible - but yours is still a valid proof.

hyper-neutrino
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