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angela210793
 3 years ago
Best ResponseYou've already chosen the best response.0every no in exponent 0 is =1.....idk how to prove tht though :(...my teacher has never asked me tht and she has never proved tht to me O.o...sorry :/

ragnarok23
 3 years ago
Best ResponseYou've already chosen the best response.0There's a combinatorial argument. Consider trying to arrange 0 objects in a line  this is 0^0. There's exactly one way to do this: do nothing.

101Ryan101
 3 years ago
Best ResponseYou've already chosen the best response.0Is there an algebraic proof involving an equation?

101Ryan101
 3 years ago
Best ResponseYou've already chosen the best response.0It seems more intuitive that the answer would be 0 to me, that's all.

joemath314159
 3 years ago
Best ResponseYou've already chosen the best response.3you could use the binomial theorem, and the fact that 0! = 1 \[(x+y)^n = \left(\begin{matrix}n \\ 0\end{matrix}\right)x^n+\left(\begin{matrix}n \\ 1\end{matrix}\right)x^{n1}y+\ldots+\left(\begin{matrix}n \\ n\end{matrix}\right)y^n\] \[=\sum_{i=0}^{n}\left(\begin{matrix}n \\ i\end{matrix}\right)x^{ni}y^i\] Let n = 0, and y = x, we end up with: \[(xx)^0 = \left(\begin{matrix}0 \\ 0\end{matrix}\right)x^0(x)^0 \iff 0^0 = 1\]

jimmyrep
 3 years ago
Best ResponseYou've already chosen the best response.13^8 / 3^8 = 1 = 3^(88) = 3^0

101Ryan101
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks Joe, I don't understand that proof but it helps to know that it's beyond my level at this point.

joemath314159
 3 years ago
Best ResponseYou've already chosen the best response.3>.< im sry, is there a certain part of it you dont understand? i only have 15 mins till my next class, but im down to try and explain.

101Ryan101
 3 years ago
Best ResponseYou've already chosen the best response.0no, it's fine.. I'm serious.. thax... Just wanted to look into this and get some feedback is all..

101Ryan101
 3 years ago
Best ResponseYou've already chosen the best response.0Jimmyrep! nice! that's a good one for me to grasp..
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