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[2014-10-22. : 3:26 am] trgk -- Mini Moose 2707Mini Moose 2707 shouted: I summon Roy, trgk, and jjf to handle this one. And whoever else wants to come, I guess. Come back socrates. I'm not good at explaining things.[2014-10-22. : 3:22 am] trgk -- (p+q) * ((-p) + q + r) = p(-p) + pq + pr + q(-p) + qq + qr = 0 + pq + pr + (-p)q + q + qr[2014-10-22. : 3:22 am] trgk -- (p+q+r) * (p+q+(-r)) = (p+q)^2 + (r+(-r))(p+q) + r(-r) = (p+q) + (p+q) + 0 = p+q[2014-10-22. : 3:20 am]  Dem0n -- and it couldn't be that because the (r ^ ~r) becomes F, and if F is ANDed to anything, that whole thing is F, so I would just get (p v q) v F[2014-10-22. : 3:19 am] Dem0n -- my teacher went over this one, but I forget exactly what he did and I think I remember him factoring shit out again, so I'm not really sure if what I'm doing is right[2014-10-22. : 3:17 am] Dem0n -- but then that F would just get rid of (~p v q v r), which isn't right[2014-10-22. : 3:17 am]  Moose -- I summon Roy, trgk, and jjf to handle this one. And whoever else wants to come, I guess.[2014-10-22. : 3:10 am] Dem0n -- If you're proving something and you have p & q & F, can you just get rid of the F to have p & q?[2014-10-22. : 3:08 am] trgk -- Mini Moose 2707Mini Moose 2707 shouted: Ok, socratic method, bro, get ready @Like[2014-10-22. : 3:00 am] Moose -- That's probably more concise. Depends where you want to do the heavy lifting of the proof.[2014-10-22. : 3:00 am] Moose -- n * (n+1) * (n+2). If n is even, then for some integer k you can write the original expression as (2k)(2k+1)(2k+2), obvious factor of 2 up front. If n is odd, then express it as (2k+1)(2k+2)(2k+3), factor a 2 from the middle term and put it up front.[2014-10-22. : 2:55 am] Moose -- More difficult one? odd * odd = (2n + 1)(2m + 1) = 4nm + 2(m+n) + 1, 4nm and 2(m+n) are even, therefore their sum is even, therefore 4nm + 2(m+n) + 1 is odd[2014-10-22. : 2:53 am] Moose -- odd * even = (2n + 1) * 2m = 2 * [(2x + 1) * m], clearly divisible by 2[2014-10-22. : 2:51 am] Dem0n -- The thing is, I'm not sure if we're allowed to use those rules to justify it[2014-10-22. : 2:51 am] Moose -- So if your first integer is odd, you're multiplying odd * even * odd, if it's even, even * odd * even[2014-10-22. : 2:48 am] Moose -- Answer these three: odd times odd = odd or even? odd times even = odd or even? even times even = odd or even?[2014-10-22. : 2:47 am] Dem0n -- I have n, n+1, and n+2, each being equivalent to 2k, 2k+1, and 2k+2, respectively, but then I don't know what tod o[2014-10-22. : 2:47 am] Moose -- Dem0nDem0n shouted: how do you prove that the product of three consecutive integers is even? Delete facebook, lawyer up, hit the gym [2014-10-22. : 2:46 am] Dem0n -- how do you prove that the product of three consecutive integers is even? |
