Here is the corrected version of your proof. Let A, B, and C be sets. If f(x) is a 1 - 1 and onto mapping from A --> B and g(x) is a 1-1 and onto mapping from B --> C, show g(f(x)) is 1-1 and onto. PF: Let x be an element of C. Then we know that since g(x) is 1 - 1 and onto then there is exact...
I just finished programming 3 an hour ago, but I don't like my brute force method. My running time varies from 3s to 500s depending on the given values, so I was lucky when I completed it. I want to know how other people have done.. Does someone want to describe their method or perhaps show me some ...
"n(n+1)/2 + (n+1)[(n+1)+1]/2" should be "n(n+1)/2 + (n+1)". Then continue: n(n+1)/2 + (n+1) = (n^2+3n+2)/2 = [(n+1)^2+(n+1)]/2 = (n+1)((n+1)+1)/2 I never liked induction in high school, but it isn't very difficult. I got inspired by your sum, so here is a sum for you. You know (f...
Very good ELorenz and Inferno96 . It's not allowed to divide by 0. Please tell me if it's too simple.. Here is the next "proof" (please give answers to both versions if you can): I suppose you know complex numbers. sqrt(a) means square root of a, sqrt(-1)=i. 1 = sqrt(1)*sqrt(1) = sqrt(1*1)...
I can make some math proofs which are (obviously) wrong. Your mission is to find out why the proofs ain't correct. When someone have submitted a good explanation, I will give you a new one.. Let's start simple. x = y Divide by (x-y) x/(x-y) = y/(x-y) Subtract y/(x-y) x/(x-y) - y/(x-y) = 0 Write on s...
It says "notice moo with !perm8", but moo doesn't respond. I've tried a lot of things, but none works: PRIVMSG moo :!perm8 PRIVMSG #perm8 :!perm8 And it doesn't work with a non-bot either: !perm8 /msg moo !perm8
I got two different solutions. 1: Alice is applied, Bob is pure, Charlie is insane, Dorothy is insane 2: Alice is applied, Bob is insane, Charlie is insane, Dorothy is applied