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# Basic Proof by Contradiction of the Zero Product Property

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Prove the Zero Producty Property in real numbers that:
If ab=0 then a=0 or b=0

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I'm going to call our assumption (ab=0) P, and the thing we want to prove (a=0 or b=0) Q. So we want to show that P => Q (P implies Q) and we can do this by contradiction: assume P and ~Q (the negation of Q) and show that there is a contradiction with this assumption.

I know what P is (ab=0) but what is ~Q? It is ~(a=0 or b=0). Now recall that ~(X or Y) is equivalent to (~X AND ~Y). So ~Q is (~a=0 and ~b=0). I'm going to rewrite ~Q as (a~=0 and b~=0) where ~= means "is not equal to".

So we assume ab=0 and a~=0 and b~-0 (that's P and ~Q). Now the second and third parts of the assumption mean that a is some nonzero number, and b is some nonzero number. Then we're supposed to believe that we multiply two nonzero numbers together and get ab=0? No, that's impossible. When we multiply nonzero numbers, we get a nonzero number. There are lots of ways you can make this more explicit. One way would be to form the product ab of the two nonzero numbers and call it k, and note that k~=0 because it's the product of nonzero numbers... and then we're assuming that ab=0 and ab=k~=0 which definitely is a contradiction. Choose whatever way feels most natural to you to describe why this is a contradiction.

Anyway, this all implies that ~Q (which is ~(a=0 or b=0)) must have been false. That means Q (a=0 or b=0) must be true, and our proof is done.

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