Tim_Lou Posted July 11, 2004 Report Posted July 11, 2004 take out your calculator,put 0^0, it says error, but is it really an error? what does 0^0=?i heard it in a forum saying that in the graph n^0, the lim of 0 is 1,so it actually = 1, also, it says that the graph of 0^n is rather unimportant...what do you think?
Tim_Lou Posted July 12, 2004 Author Report Posted July 12, 2004 consider this, a^b/a^b = a^0 =1, so 0^x/0^x = 0/0.. after a little research on the internet, i found most of the time it is said that 0^0 is undefined. however:http://www.google.com/search?q=0%5E0&btnG=Google+Search http://us.metamath.org/mpegif/exp0.html these 2 sites say that it=1the 2nd one is totally... un-understandable, maybe somebody could explain it...?
TeleMad Posted July 18, 2004 Report Posted July 18, 2004 Tim_Lou: consider this, a^b/a^b = a^0 =1, so 0^x/0^x = 0/0.. Look up the definition of that exponential rule and see if it excludes 0 (that is, say a is not equal to 0). For example, one of the math books I have states: Quotient Rule for Exponentsa^m / a^n = a^(m - n), a != 0 another says The Quotient RuleFor any nonzero number a and any positive interges m and n, a^m / a^n = a^(m - n) So you can't use the quotient rule of exponents to try to show that 0^0 = 1. The reason for the restriction a != 0 is obvious (as you showed). Use a = 0 and say m = 5 and n = 3. That appears to give 0^5 / 0^3 = 0^(5 - 3) = 0^2 = 0. But actually, it doesn't. Note that in the denominator 0^3 = 0, and you can't divide by 0. So when you use a = 0, you actually get undefined.
Tim_Lou Posted July 25, 2004 Author Report Posted July 25, 2004 so, it is undefined... i think so too... wait a minute:a != 0 ??????? how is it possible???0! =11! =1.... in a combination or permutation, there is always at least 1 possibility....
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