Ten points to the first person who can mathematically prove that women are evil.?
606037
Women explained mathematically\n\nFirst we state that women require time and money: \nWomen = Time X Money\n\n\nAnd as we all know “time is money”\n\nTime = Money\n\n\n\nTherefore by substituting Money for Time we get: \n\nWomen = Money X Money\n\nWomen = (Money)^2\n\n\n\nAnd because “money is the root of all evil” we therefore can state:\n\nMoney = (Evil)^1/2\n\nAnd Since\n\n(Money)^2 = Women \n\nThen (Money)^2 = Evil\n\n\n\nAnd we are forced to conclude by substituting "women" for "(money)^2" from above that:\n\nWomen = Evil
660726
Aryabhatta, the Indian mathematician who invented 0.
514566
G O d B L e s s u\n\n10 points to the genius please.
732707
Let x = points she still needs.\n235 + x > 462\nx > 227
9416
Vigintillion\n10 to the 63rd power\nor 9with 63 zeros and that is a real #
664410
First, we assume that all girls take time and money. \nGirls = Time x Money \nFurthermore, we take the statement, "Time is money", and we get: \nTime = Money \nGirls = Money x Money \nGirls = Money^2 \nMoreover, the statement, "Money is the root of all evil" means that: \nEvil = the square root of money \nGirls = (Evil)^1/2 (Evil)^1/2 \nwhich reduces to... \nGirls = Evil
486463
Quite clever you are. 10 points for being so smart!! Not cheating no since you are able to figure this out you get kudos for thinking creatively.
92274
Get solid proof to show her.
728920
mathematically speaking, incalculable arithmetic
375633
hummmmmmmmm....women are devious creatures that play games.......sorry
417753
This argument is called "reductio ad absurdum" in philosophy school, if I've got my Latin right.  Reduce to absurdity.  The idea being, if you want to test a principle (which presumably was your point, right?), you find where the extremes are, and use them to gage whether or not the principle can fit all cases.  So, for example, people use Hitler as a sort of icon for "bad guy."  And "rape" as a sort of icon for "bad act."\n\nAnd they are right to do so.  Having been there and suffered that, at the hands of a gang of 5, at the age of 13, and survived not only that but the humiliation of sitting in front of a grand jury and telling them what happened (at the age of 13, mind!), I can say that it is not unreasonable for a woman to think death is not so bad in comparison.\n\nThink about it, guys.  To have the one most precious thing she can give to someone she loves very dearly stolen from her and degraded and debased to less than an animal; and not just in philosophical or emotional, but intimately physical.  Sticking a knife in her chest would not be worse, under some circumstances.  And then to remember that as the first experience you ever had of something that can be pure magic when it is good?\n\nYou know how long it took me to find the magic?  Isn't that something that Steve should get sainthood for, at the very least?\n\nNo, a woman is not punished by God when she is raped; she is punished by brutality in a man.  She is blessed if she ever finds a man who can teach her about the good part, and be worthy of the devotion she will shower on him in gratitude.\n\nTHAT is God's Will!!!
651735
RELIGIONS DAMAGE ALL MANKIND'S INTELLIGENCE,\nNOT JUST WOMEN'S!
403535
Sounds like a repressive religion you belong to. A woman can be evil, but that doesn't mean that all women are evil.\nAnd how can a horse be evil? And a house?
377626
You asked the question in the Math. section. So:\n1) In letters: aeow = o awe\n   (not so different!)\n2) In numbers of letters: |3-5|=2\n3) In modern numerology:\n   man=10 - women=25\n   |10-25|=15\n   and 1+5=6
130447
ill debate you , you miserable fusk! look at all of your thought provoking questions. your a little lonely washed up headbanger that has nothing better to do than slam women. as for having broken down hostile witnesses, your sorry aas hasnt seen hostile. name it dipshite!
1074417
The answer is 10. \n\nIf you don't believe me, do the problem yourself and compare that with my answer.
1046884
To explain this matter, we have to use the Inclusion-Exclusion Principle. i think you probably aware of it. if don't, you can find about it easily on the internet and it is very simple and easy.\nBut the problem is that, in order to solve this analytically,the formulas used are too complicated and it is nearly impossible to write them down. so i'm gonna introduce the idea of how to solve this question.\n\nyou may  begin with following tasks\nFirst, to think of this, assign identity to men with 1,2,...,n numbers to distinguish from each other and let A_k denote kth man. (do the same to women as B_k). and let D2_n denote the number we are trying to find. (Dk-1_n for the number of possible arrangement of kth day)\n\nThere must be n! arrangements on the first day of dance. (for the first dance there are n! possibilities). for the second dance, each man has to choose as a partner a woman other than the one with who he first danced. the number of possibilities,D1_n, is the nth derangement number Dn. \n\nOn the third day, A_i cannot choose B_i1 and B_i2 whom he has already danced with. let the set T_k denote the way that A_k choose B_k1 or B_k2. So  you can easily find out that \n\nD2_n = |(T_1)c N (T_2)c N ... N (T_n)c|     ..............(1)\n\nwhere (T_k)c is the complemenary set of T_k and N stands for "Intersection" .\n\nto solve (1),  we need to use the Inclusion-Exclusion Principle and the way use it is too complicated.  you could derive the formula which equals Dk_n, but computing it with pen is nearly impossible,. you have to use computer to compute.\n\n\n\n\n\n\n\n\n(here goes additional answer to your question and this is the complete answer!<7:33pm 6.20>)\n\nI do some research with my computer and the correct answer is the mth day's possible arrangements can vary depend on the m-1th day's choice and various other people's choices.  if you take the number of people which is the initial value for this problem as n, it suddenly becomes extremely complicated. if you take the initial value more specific , let's say 10 (10 people), it is still impossible to say how many possible way to arrange people on mth (3<=m<=10) day. here goes an example for the reason why this is happening\n\ncase 1\nlet's set our initial value for n (there are n people)\non the first day and the second day, A1 danced with B1 and B2 respectively, and so does A2 with B2 and B1 respectively. on the third day, it is obvious that A1 and A2 cannot dance with two same women, so they must dance with the other women. so A1 has n-2 choices and can choose the women he likes. so let's say that A1 choose B3 as a dance partner, A2 must choose the women other than B1,B2 and B3 which leaves him n-3 choices. (the real situation may leave A1 and A2 choices far less because we are ignoring all the other people A3~An)\n\ncase2\non the first day and the second day, A1 danced with B1 and B2 respectively, and A2 danced with B3 and B4 respectively.on the third day, there are two choices which A1 can make.\n\ni). A1 choose B3 and B4.\nii). A1 choose the women other than B1,B2,B3 and B4.\n\nIf A1 takes i),this leaves A2 n-2 choices. If A1 takes ii),this leaves A2 n-3 choices\n\nyou can see that the situation on the mth day can be vary depend on whom you have already chosen for your dance partner on the m-1th day and other people's choice on the same day. so the correct asnwer for your question is that there is no established arrangement on the mth day. but there are still infinite amount of possibilities for you to figure this madness out, and when if you done, please let me know  :)