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Posted 19 February 2008 - 03:00 AM
Posted 19 February 2008 - 03:31 AM
If zero to the nth power is zero, and n to the zeroth power is one, then what is zero to the zeroth power?
Posted 19 February 2008 - 10:16 PM
so it's whatever you think 0/0 is. This isnt really a riddle...
yes there are laws like the ones you stated in your OP as 'quick laws' to help people out such as yourself who don't understand what setting something to the zeroeth power actually means (sorry lol, but you should look it up or type it into your calculator )
there are other 'quick laws'... all with exceptions. If they didn't, 0/0 would be 1, judging by the 'quick laws'. And maybe it is. But as Aatif said, it is indeterminate
Posted 19 February 2008 - 10:29 PM
There is some benefit if we define the value to be 1: for the binomial theorem valid everywhere, x^0 must = 1 for all x.
There is no comparable benefit for making the function 0^x well behaved at x=0, so much of modern thought favors defining the value as 1.
But others argue it should remain indeterminate, because x^y is in fact discontinuous at the origin.
- Bertrand Russell
Posted 19 February 2008 - 10:34 PM
where a is any number, in the case of 0 it doesn't matter, you might as well use 1 for a (you just cant use 0)
x0 = x1 / x1
x0 = x / x
00 = 0 / 0
it's simple exponents.
50 = 51 / 51 = 5 / 5 = 1
50 = 52 / 52 =25 / 25 = 1
it's simple exponents. The number that is the actual exponent isn't a real number
53 is just 5*5*5
we only base the "quick-rule" property of 0^0 based on the fact that ALL NUMBERS EXCEPT 0 are 1 when divided by themselves
I'm not gonna speak for programming languages and stuff, they obviously are gonna make it whatever is best for programming with that language. Though if you're asking what it is at its core, it's 0/0
Posted 20 February 2008 - 10:35 AM
That is, [limit as x,y -> 0] x^y is not well defined - it depends on how x and y approach 0, or which one gets there first.
But that does not mean 0^0 can have any value whatsoever, the way 0/0 can:
[any_value]x=0 -> [any_value]=0/x. let x->0. [any_value]=0/0.
The 0^0 case is more well behaved.
 [limit x->0] x^0 = 1.
 [limit y->0] 0^y = 0.
Look more closely at case  and invoke the equation x^0 = x/x:
[limit x->0] x^0 = [limit x->0] x/x = [limit x->0] 1 = 1.
That is, x/x is well behaved [continuous] at 0; even tho it's 0/0 there, its value is unambiguously 1.
- Bertrand Russell
Posted 21 February 2008 - 01:33 AM
Posted 21 February 2008 - 02:11 AM
bonanova: I could also use limits to prove, with x/y, that as y approaches 0, the answer approaches infinity as it gets bigger and bigger (and thus disprove the any value agreement). So in both cases 'limits' and 'algebra' are saying different things. In both cases, they're undefined. 0/0 is undefined. 0^0 is undefined. The function of x^y is just following the trend of exponents, of which x^0 is simply expressed by the laws of exponents as x/x. Since 0/0 is not always 1, neither is 0^0. Then again, it could be blue cows ;D
Rule #1: There's an exception to every rule (including Rule #1)
Posted 21 February 2008 - 03:44 PM
I hope this simple info helps.
Edited by RoseOnPlayer, 21 February 2008 - 03:46 PM.
Posted 21 February 2008 - 11:37 PM
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