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If I'm asked to prove that an equation holds for all integers, how should I go about it?

My initial reaction would be to conduct two tests. The first one I plug in 2x into both sides of the equation and reduce and see if it comes out to the same thing on both sides. Then plug in 2x-1 into both sides and reduce and see if its the same. That would take care of both even and odd cases, but the more I think about this, the more incorrect it feels. I don't actually think that this proves anything since you are plugging the same formula into both sides. I don't know where I've seen this done before, but I know it is something that used to test something similar.

My next guess would be something like induction. Which is correct?

David G. Stork
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JosephTLyons
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  • Which equation? If it is $x/x = 1$, well, that's pretty easy! – David G. Stork Feb 16 '18 at 04:15
  • Without knowing a specific equation, there's no correct answer to this question. Some equations (e.g. some of those which have $(-1)^n$ terms) can be amenable to the former method; probably more are amenable to induction; but some won't readily yield to either. For some the equation holds for all integers but it's impossible to prove that it does. Without more context there can be no answer here. – Steven Stadnicki Feb 16 '18 at 04:16
  • $(-n) mod 2 = n mod 2$

    I do not want an answer, just verification if I can use either one of the proof techniques I mentioned. I would prefer to use the first, but I'm not sure how to see if it is ok or not.

     – JosephTLyons Feb 16 '18 at 04:22
  • Well, the first doesn't actually simplify the problem that much, and in this case induction is overkill. Here's a broad hint for you: $a\equiv b\pmod2$ is equivalent to saying $(a-b)\equiv 0\pmod 2$. What does that look like in your case? – Steven Stadnicki Feb 16 '18 at 04:39
  • Why would odd and even cases matter? And why wouldn't prime vs. non primes matter, or multiples of 7 vs numbers with remainders 1,2,3,4,5,6 not matter. It depends on the ewuation. – fleablood Feb 16 '18 at 05:41

2 Answers2

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A proof by induction would be appropriate in this case. Since I don't know the specifics of the equation you are dealing with, I'll give you a general outline.

1) Let k represent the integer in your equation. 2) Show that the equation holds for k=1. 3) Then assume that it holds for k=n and then show that it holds for k=n+1.

Without knowing the specific case I'm not sure how you would show it holds for the negative counterpart of each natural number, except to simply show that the value of the function when k is negative is the negative of the value of the function when k is positive.

TKoundak
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Proof by consideration of exhaustive subsets (evens and odds):

If $n$ is even, then $n\mod 2 = 0 = -n\mod 2$.

If $n$ is odd, then $n\mod 2 = 1 = -n\mod 2$.

$\blacksquare$

David G. Stork
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