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101Ryan101 Group Title

Medals for ones who prove: 0^0 = 1 (I wrote the last one wrong) :/

  • 3 years ago
  • 3 years ago

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  1. angela210793 Group Title
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    every 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 :/

    • 3 years ago
  2. ragnarok23 Group Title
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    There'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.

    • 3 years ago
  3. 101Ryan101 Group Title
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    Is there an algebraic proof involving an equation?

    • 3 years ago
  4. 101Ryan101 Group Title
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    It seems more intuitive that the answer would be 0 to me, that's all.

    • 3 years ago
  5. joemath314159 Group Title
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    you could use the binomial theorem, and the fact that 0! = 1 \[(x+y)^n = \left(\begin{matrix}n \\ 0\end{matrix}\right)x^n+\left(\begin{matrix}n \\ 1\end{matrix}\right)x^{n-1}y+\ldots+\left(\begin{matrix}n \\ n\end{matrix}\right)y^n\] \[=\sum_{i=0}^{n}\left(\begin{matrix}n \\ i\end{matrix}\right)x^{n-i}y^i\] Let n = 0, and y = -x, we end up with: \[(x-x)^0 = \left(\begin{matrix}0 \\ 0\end{matrix}\right)x^0(-x)^0 \iff 0^0 = 1\]

    • 3 years ago
  6. jimmyrep Group Title
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    3^8 / 3^8 = 1 = 3^(8-8) = 3^0

    • 3 years ago
  7. 101Ryan101 Group Title
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    Thanks Joe, I don't understand that proof but it helps to know that it's beyond my level at this point.

    • 3 years ago
  8. joemath314159 Group Title
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    >.< 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.

    • 3 years ago
  9. 101Ryan101 Group Title
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    no, it's fine.. I'm serious.. thax... Just wanted to look into this and get some feedback is all..

    • 3 years ago
  10. 101Ryan101 Group Title
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    Jimmyrep! nice! that's a good one for me to grasp..

    • 3 years ago
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