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human
ELI5: How do mathematical proofs work?

I don't understand how you could actually prove with absolute certainty that a concept or supposition will hold true in all possible circumstances.

I mean, for example, it seems obvious that all even numbers can be evenly divided by two. But how could you really PROVE that this is the case for some astronomically high even number that's so large that no one has ever even tried to divide it?
gpt
Well that's not really the right way to think about it because the *definition* of an even number is that it's a number of the form 2*n* where *n* is an integer. For any integer *n* the number 2*n* will *obviously* be divisible by two, because 2*n* ÷ 2 = *n*. There's your "proof," in a sense, but really it's just the definition of "even." It's tautological, which means it's a statement that's true *because it's true.*

But we can go on to prove something that's not tautological. For instance, consider this assertion: The product of two even numbers is even. Is that a true statement? Well, 6 is an even number and 4 is an even number, and 6 × 4 = 24, which is an even number … but that's just one case. Is it *necessarily* true that *for all* even numbers *a* and *b*, the product *a* × *b* must be even?

The answer is yes, but we don't have to just accept it on faith. We can *prove* it. Consider:

Start with some numbers *a,* *b* and *c* such that

*a* × *b* = *c*

where *a* and *b* are even numbers. Since *a* and *b* are even, then by the definition of "even" we can write

*a* = 2*a′*  
*b* = 2*b′*

and say that *a′* and *b′* are integers. This, again, is the *definition* of an even number. We can then put these expressions into our original equation and come out with this:

2*a′* × 2*b′* = *c*

We can factor this expression thusly:

2(*a′* × *b′*) = *c*

Now we know that the product of any two integers is an integer (not proved here, but you can work that out if you want), so since *a′* and *b′* are integers, *a′* × *b′* must also be an integer. In other words, there exists an integer *c′* = *a′* × *b′* such that

2*c′* = *c*

And by the definition of "even," that means *c* must necessarily be an even number.

That's how math proofs work in a nutshell. You start with something known to be true, then do things which are known to be valid (or which can be separately proved to be valid), and then you end up demonstrating that the thing you were trying to prove is *necessarily, unavoidably true.*

Or else you start by assuming the thing you wanted to prove is *false* and then, by working through steps, you show that some contradiction exists. This is called proof by contradiction and it's a very powerful technique. If you start by assuming a certain statement is true then follow the logical consequences of that statement and end up "proving" that 1+1=3 or something, then you know that either you made a mistake or else your initial assumption was wrong. And since a statement can generally only be true or false, if you were *wrong* to say it was *true,* then it must have been *false.*

I say "generally" there because if we go much further we're going to have to get into matters of decidability, and that's just too much for a short answer. Suffice to say that not every mathematical statement can be proved true or false. *Most of them* can, though, so the issue of undecidability rears its head very infrequently unless you happen to choose to study the question of decidability itself.