| Off Topic > Off Topic |
| Impossible Math |
| << < (7/12) > >> |
| Inv3rted:
I see your logic, but it's an issue of semantics as to whether the problem is broken or not. When you divide by the same number on both sides of the equation, even if it is undefined, the equality remains true. |
| Riot:
--- Quote from: Drak on July 28, 2009, 10:47:02 PM ---Exponents. x2 --- End quote --- aah, great I know what you ment but i never learned that sign |
| Rughugger:
I find this topic hilarious, you're asking kids who haven't even gotten to basic algebra about exponents and things like that. Oh you slay me, mighty troll. |
| Jsk2003:
--- Quote from: Lord Pie on July 28, 2009, 10:39:56 PM ---x = y x^2 = xy x^2-y^2 = xy-y^2 (x+y)(x-y) = y(x-y) x+y = y 2y = y 2 = 1 1 = 0 But, the only number that x can equal is clearly 0. Damn cancelling of variables. It's clearly broken. --- End quote --- You divided by (x-y) and x=y, thus you divided by zero. So you're wrong. |
| Muffinmix:
Guys. Zeros don't vanish. (x+y)(x-y) = y(x-y) assuming y=x (x+x)(x-x) = x(x-x) (x+x)(0) = x(0) 0 = 0 What's the problem? Also, if you divide both sides by (x-y) (x+y) = y This still works if x = y = 0, or if unlike in this problem, x =/= y and x = 0 |
| Navigation |
| Message Index |
| Next page |
| Previous page |