Contrapositive Proof , if n^2 is even , n is even.

[@xirenzhang9126]

I enjoyed this video today and it taught me something new. I like these sort of videos.

[@mathwiz4u]

It's really nice to hear that @xirenzhang9126, and I'll definitely post some more videos like this then. :)

[@gamerboy7224]

Wouldn't it be simpler to simply say that since 2 divides n, (given property of even numbers), then 4 divides n^2? which by extension means 2 divides n^2?

[@mathwiz4u]

That would be nicer but that would be proving it in the other direction; that would be proving if n is even then n^2 is even rather than our statement

[@gamerboy7224]

@mathwiz4u Yup my bad, although, im gonna double down and say that my proof works both ways. let p1^k1 * p2^k2 *... be the prime factorisation of n, then n^2 = p1^2k1 * p2^2k2 * ... if 2 divides n^2, then 4 must also divide n^2 (as all prime factors are squared, and 2 is prime), and hence 2 must also divide n originally ive only done like a year on number theory, so im very intermediate on rigorous proofs, but i feel this is pretty solid? fell free to give feedback

[@mathwiz4u]

@@gamerboy7224 Yeah that’s definitely a nice way to prove it; i’ve only taught myself proofs for 2 months from a book so i’m not exactly the expert

[@kfjw]

​​@@gamerboy7224 If I may, I'd like to bring something up: The prime factorization of any integer N is a multiset of prime integers. (I'll use the notation pX to denote the prime factors of an integer X) Given N and M, p(NM) is the multiset union of pN and pM. The multiset union operator cannot cause it to lose any elements it already has. If either N or M are even, that means it has 2 as a prime factor, and it can never lose a prime factor through multiplication. That means it can never stop being even through any kind of integer multiplication.

[@kfjw]

I should add something else. In order to prove it the other way, we can say that p(NM / M) = p(N). Therefore, p(N^2 /N) = p(N). Also, given any factor F from p(N^2), the number of copies of F is twice the number of copies of F in p(N). So we always have at least one copy left in p(N). This can't cause the complete loss of any element of the prime factor set.

[@jacobmerrill693]

Huh, I didn't think of it directly algebraicly like that. For the 'n even is n2 is even" in the thumbnail I thought "exponent is multiplication is repeat addition, addition of evens can only result in evens". To prove the otherway just had to add odds only add up to evens when in even sets, so odd × odd = odd. Which might be longer on the rigorous proof side but was super quick on the intuintive side

[@samar0029]

Nice, but you sound like a philosopher 😅

[@mathwiz4u]

@@samar0029 I cant tell if that’s a good thing or not 🥲