x^0 = 1??

If i do X^0 it returns 1...

exactly WHERE does the 1 come from in that? If I multiply X by itself 0 times, I'd get 0, I'm not seeing where the 1 comes from.

:confused:

Code:

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This proof uses the laws of exponents. One of the laws of exponents is:

  n^x
  --- = n^(x-y)
  n^y 

for all n, x, and y. So for example,
  3^4
  --- = 3^(4-2) = 3^2
  3^2


  3^4
  --- = 3^(4-3) = 3^1
  3^3
 

Now suppose we have the fraction:
    3^4
    ---
    3^4

This fraction equals 1, because the numerator and the denominator are the same. If we apply the law of exponents, we get:
      3^4
  1 = --- = 3^(4-4) = 3^0
      3^4

So 3^0 = 1.
We can plug in any in number in the place of three, and that number raised to the zero power will still be 1. In fact, the whole proof works if we just plug in x for 3:

                  x^4
  x^0 = x^(4-4) = --- = 1
                  x^4

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Makes sense...I knew that rule (I'm in Algebra II) but never really thought of it that way :)

BTW:

0! = 1

AND
there IS a (-0.5)!, though i can't remember its exact amount (a surd of some sort)

0! is 1???:confused:

Yes. I think it is just to make some sums possible.
Like Combinations. 3-C-0 = 3!/(3!*0!) = 1 => 0! = 1