I've been trying to find a proof by contradiction that $$\forall n \in \mathbb{Z}(n^2-n \in \{\text{evens} \})$$ Every time I assume $n^2-n$ is odd, the only way I can seem to prove it is just going case by case (assume $n$ is even and show it's true, same with $n$ being odd). Is there a proof by contradiction of this theorem, such as assuming $n^2-n=2m+1$ then showing that $m=\frac{n^2-n-1}{2} \not\in \mathbb{Z}$?
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5Hint: factor $n^2-n$ – Karl Oct 20 '24 at 11:24
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Are you looking for a proof that doesn't use cases? – Zoe Allen Oct 20 '24 at 11:25
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To avoid cases you could say it's equivalent to $n^2+n$ being even, and prove $\binom{n}{2} = \frac{n^2+n}{2}$ so it's an integer. But that's quite involved. – Zoe Allen Oct 20 '24 at 11:29
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1See "Proving $n^2+n$ is even for any integer $n$", "Proof by induction $n^2+n$ is even", "$n^2+n$ is even", among others. (Answers provide various approaches.) – Blue Oct 20 '24 at 11:30
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1Induction proofs can be viewed as proofs by contradiction. See https://en.wikipedia.org/wiki/Proof_by_infinite_descent – GEdgar Oct 20 '24 at 11:31
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2Let $f(n)=n^2-n$. Show that if $f(n)$ is odd then so is $f(n-1)$. Show that $f(1)$ is even. Note contradiction. – Gerry Myerson Oct 20 '24 at 12:18
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Why do you want a proof by contradiction? A typical example (as seen in the answers here) seems to be: (1) assume that $n^2 - n$ is odd. (2) argue that in fact $n^2-n$ must be even. (3) This is a contradiction. Just cross out steps (1) and (3) and get a shorter and more direct proof. – John Palmieri Oct 20 '24 at 18:24
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@John That's why "proof by contradiction" is usually meaningless in elementary contexts, since we can naively turn any (constructive) proof $P$ into contradictive form by first assuming "not $P$" then proving $P$ (which is how constructivity is destroyed in many modern reformulations of Euclid's proof of infinitely many primes). – Bill Dubuque Oct 20 '24 at 19:02
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@GEdgar But any proof can be viewed as a "proof by contradiction" - see the prior comment. – Bill Dubuque Oct 20 '24 at 19:05
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For some integer $n$ suppose $n^2-n$ is odd
$$n^2-n \equiv 1 \pmod2\iff n(n-1)\equiv1\pmod2$$
Let $m=n-1$. One of $m,n $ must be even, WLOG assume $m$ is even then $m=2k,k\in\mathbb{Z}$
$$n(n-1)=mn=2kn\equiv0\pmod2$$ Which contradicts the supposing argument so $n^2-n$ is even for $n\in\mathbb{Z}$
Antony Theo.
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Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Oct 20 '24 at 19:03