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24 replies to this topic

#1 jremoletesilerio

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Posted 19 February 2008 - 03:00 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?
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#2 Aatif

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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?


Indeterminate
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#3 unreality

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Posted 19 February 2008 - 10:16 PM

0^0 = 0/0

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 :D)

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
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#4 bonanova

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Posted 19 February 2008 - 10:29 PM

The function f(x,y) = x^y has a discontinuity at the origin: the limit along x=0 is 0, and the limit along y=0 is 1.

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.
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#5 unreality

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Posted 19 February 2008 - 10:34 PM

x0 = xa / xa

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)

so:

x0 = x1 / x1

x0 = x / x

00 = 0 / 0

it's simple exponents.

50 = 51 / 51 = 5 / 5 = 1
or
50 = 52 / 52 =25 / 25 = 1

etc

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
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#6 bonanova

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Posted 20 February 2008 - 10:35 AM

x^y is well defined everywhere except the origin.
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:

For example,
[any_value]x=0 -> [any_value]=0/x. let x->0. [any_value]=0/0.

The 0^0 case is more well behaved.

[1] [limit x->0] x^0 = 1.
[2] [limit y->0] 0^y = 0.

Look more closely at case [1] 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.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#7 carlosn27

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Posted 21 February 2008 - 01:33 AM

0/0 is not 1, it's indeterminate because 0 technically isnt a real number, it's more of a lack of number.
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#8 unreality

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Posted 21 February 2008 - 02:11 AM

0 is a number too :D 0 real and 0 imaginary on the complex plane (in other words, the origin point of the complex plane)

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)

lol
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#9 RoseOnPlayer

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Posted 21 February 2008 - 03:44 PM

It is true that x/x = 1 and will always be equal to one but there should be a limit that x <> 0 (x must not be equal to zero). If the problem given is (2x+y)/(x-1), then it should always be understood that x must not be equal to positive 1 to avoid undetermined answer.

I hope this simple info helps.

Edited by RoseOnPlayer, 21 February 2008 - 03:46 PM.

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#10 unreality

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Posted 21 February 2008 - 11:37 PM

Exactly. 0/0 isnt always 1 (though it can be). 0^0 is equal to 0/0, so the same applies. Bonanova was saying that 0^0 is a special case of 0/0 only if the current meaning of 0/0 was 1, however I am disputing that. He is using the function created by x^y as evidence, but a wedding ring doesn't make you married. Being married makes you have the wedding ring. The laws of exponents are the laws of exponents, nothing else as far as I know. 0^0 = 0/0. Though maybe bonanova has some more proof showing that 0^0 is only a special case of 0/0 and somehow set apart differently, I'd like to see it :D
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