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[2014-10-13. : 2:51 am]  O)FaRTy1billion[MM] -- Dem0nDem0n shouted: or do I check if it's NULL? Check if it's NULL.[2014-10-13. : 2:44 am]  jjf28 -- triggers only run if they're owned by players 1-8, force 1-4, or all players[2014-10-13. : 2:38 am] sigsaucy -- i put in some triggers owned by 'current player' and they didn't run? is current player triggers useless or what am i not getting?[2014-10-13. : 2:24 am]  Dem0n -- How can you check if memory allocation fails? Can you just do if(!malloc(5)) or something?[2014-10-13. : 12:39 am] trgk -- Contradiction works like that : "If that's true, then 1+1 = 3. So that should be false, or our math system should be inconsistent"[2014-10-13. : 12:27 am] jjf28 -- might help: http://www.personal.kent.edu/~rmuhamma/Philosophy/Logic/ProofTheory/proof_by_contradictionExamples.htm[2014-10-13. : 12:17 am] l)ark_ssj9kevin -- ZoanZoan shouted: assuming the sum of an irrational and a rational number is rational, you get a/b + irrational = c/d, and thus this irrational = c/d - a/b = (cb - ad)/bd, which means this irrational number is equal to an integer (cb - ad) over another integer (bd) worked fine and is a pretty discrete mathy solution[2014-10-12. : 9:58 pm] jjf28 -- nvm i must be on something, the hard logic just proved i used it wrong, gg me[2014-10-12. : 9:40 pm] jjf28 -- ok unified hard logic with the math methods I had forgotten, now i can  [2014-10-12. : 9:40 pm] Sand Wraith -- also happy early Columbus-was-a-fuckwit Day http://theoatmeal.com/comics/columbus_day[2014-10-12. : 9:40 pm] jjf28 -- If this assumption is false then by the square of opposition: every sum (r+i) is irrational[2014-10-12. : 9:38 pm] Sand Wraith -- If you use the sqrt(n) method, I generally see it being performed as a series of operations that go on infinitely.[2014-10-12. : 9:20 pm] jjf28 -- "some do not" certainly contradicts "all do" but doubts are still firing off because I remember proofs being harder xD[2014-10-12. : 9:12 pm] Zoan -- Well…. hrm.. I think maybe you still need to prove that for any given rational and irrational number their sum is irrational.[2014-10-12. : 9:03 pm] jjf28 -- did I get my NTS wrong? I think i'm allowed to go 0/1+root(2) == root(2) violating the assumption r+i1 != i2 but I feel like i cheated  [2014-10-12. : 8:45 pm] Zoan -- which is a contradiction, and therefore a rational plus an irrational must be irrational.[2014-10-12. : 8:45 pm] Zoan -- assuming the sum of an irrational and a rational number is rational, you get a/b + irrational = c/d, and thus this irrational = c/d - a/b = (cb - ad)/bd, which means this irrational number is equal to an integer (cb - ad) over another integer (bd)[2014-10-12. : 8:44 pm] jjf28 -- lol there's a certain number that's rational that makes showing that super easy[2014-10-12. : 8:19 pm] jjf28 -- assume it's true, follow the implications to an impossible consequence, then [2014-10-12. : 8:18 pm] Dem0n -- Prove by contradiction that the sum of a rational number and an irrational number is irrational.[2014-10-12. : 8:00 pm] MasterJohnny -- lol it depends on what you're proving...i kinda forgot alot of those basic proof kind of things...[2014-10-12. : 6:42 pm] Zerghuter666 -- is it my computer i have a school computer i am using and maybe i could use a different PC[2014-10-12. : 6:37 pm] Zerghuter666 -- here i will try to upload my map again and i will show you what happens[2014-10-12. : 6:35 pm] Zerghuter666 -- because the file is too big but i have it on Team Liquid tho here: http://www.teamliquid.net/forum/brood-war/468633-infested-custom-survival-map[2014-10-12. : 6:35 pm] jjf28 -- :sadface: , can you describe what's happening? dev seems to give those kinds of errors priority |
I dunno. I guess that works
it helps if you write out what the problem is saying explicitly.
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