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[2015-2-06. : 12:08 am]  jjf28 -- "Lol this is the most barebones production thread I have ever seen!"[2015-2-05. : 10:45 pm] Devourer -- Any set X and its complement X' together always form the universe, if I remember correctly (and this appears to be quite logic)[2015-2-05. : 10:31 pm]  MasterJohnny -- DevourerDevourer shouted: Is "I have a set A" implies that (what I called it) "A is the universe", then only consider the left case. So if A was the universe then A= (B union B complement) right?[2015-2-05. : 9:30 pm] jjf28 -- if SEN throws me many more SQL errors I might have to reverse engineer the database structure  [2015-2-05. : 8:57 pm] jjf28 -- NudeRaiderNudeRaider shouted: jjf28 Watching that, all I can concentrate on is how ugly the distortion is. ikr, but it was the only linkable place I found the quote[2015-2-05. : 8:48 pm] NudeRaider -- jjf28jjf28 shouted: http://youtu.be/M_jpGm_bN4k?t=15m54s Watching that, all I can concentrate on is how ugly the distortion is.[2015-2-05. : 8:08 pm] Lanthanide -- high larious: https://www.kickstarter.com/projects/strafegame/strafe/description[2015-2-05. : 7:00 pm] Devourer -- Is "I have a set A" implies that (what I called it) "A is the universe", then only consider the left case.[2015-2-05. : 6:50 pm] Devourer -- Further is C = {1, 2, 3, 5, 6, 7} not equal to B' = {4, 6, 7 ,8 , 9, 10}. Given that there were no restrictions to your 'rule', it should be able to be applied to my example as well. But it does not apply there, thus your 'rule' (I lack the english words) is false.[2015-2-05. : 6:42 pm] Devourer -- A is defined by B union C. Consider B = {1, 2, 3, 5} and C = {6, 7}. No assumption regarding the #ofElements was made, so we just assume the case that there are 10 elements, 1 to 10. So, B' would be B' = {4, 6, 7, 8, 9, 10}. However, A1 with A1 = B union C = {1, 2, 3, 5, 6, 7} and A2 with A2 = B union B' = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} are not equal. If we considered an infinite number of elements, the former assumption still is wrong, obviously.[2015-2-05. : 6:41 pm] jjf28 -- DevourerDevourer shouted: Zoan Are they? A = B U C therefore B, C ∈ A... but where on earth he gets ZoanZoan shouted: MasterJohnny only if B and C are subsets of A from idk[2015-2-05. : 6:39 pm] Devourer -- ZoanZoan shouted: Zoan wait ignore this - B and C are subsets of A Are they?[2015-2-05. : 6:37 pm] Zoan -- ZoanZoan shouted: MasterJohnny only if B and C are subsets of A wait ignore this - B and C are subsets of A[2015-2-05. : 6:30 pm] Zoan -- MasterJohnnyMasterJohnny shouted: If I have a set A=(B union C) and (B intersect C) is empty then is it true that C=(B complement)? only if B and C are subsets of A[2015-2-05. : 4:52 pm] BloodyZombie117 -- I don't remember doing any of this in school... I stopped at Algebra 2, because I can't do Calc or Trig... My brain is good at basic math, not that black magic stuff.[2015-2-05. : 4:50 pm] jjf28 -- though I suppose I could think of x3 and x2 as free variables and write x1, just to restate the solution set another way[2015-2-05. : 4:46 pm] jjf28 -- BloodyZombie117BloodyZombie117 shouted: x1 = 3, x2 = 0, x3 = 1. Solution solved. that's one of infinite solutions, not the set[2015-2-05. : 4:45 pm] BloodyZombie117 -- I'm just going off of basic principles... Unless you meant to make them exponents, not different x's.[2015-2-05. : 2:30 pm] Roy -- jjf28jjf28 shouted: got a job interview out of state, now only if I had a car... If they don't make travel arrangements, they aren't worth working for.[2015-2-05. : 8:55 am] Sand Wraith -- So it may be a proper subset of B complement, which is to say, not B complement, and so A would be B union C[2015-2-05. : 8:54 am] Sand Wraith -- C is only the component of B if it is composed of all elements not in B |
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