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in reply to More Fun with Zero!
Okay
Now you got me thinking. I have always been taught (as most of us have) that n^0 is 1. Now, why, do you suppose, that n multiplied by itself zero times, would it equal one? It isnt the same as n*0, I realize that, but if you think about n^1, you get n, right? meaning n^1=n. n^2=n*n, and n^0=? you dont have anything multiplied by anything else, so shouldnt this be undefined as well? I realize, also, that this has probably been defined by some dead mathematician who used this case to prove some wildy fantastic theorem, but still.
Am I totally off base with this?
thanks!
E-Bitch
p.s.:
my 2 cents about the 0^0 thing: 0^1 = 0 right? nothing multiplied by itself one time, should still equal nothing. so, if we were to think of this in terms of, oh, say, atoms, if you have no atoms muliplied by no atoms one time, you get no atoms (SURPRIZE!!!). now, if you have no atoms, multipled by no atoms, 0 times, you get 1 atom!
what?
we just created matter!
call the pentagon!
whoohoo!
note: energy to matter conversions do not apply in this simple example |
Re: Re: More Fun with Zero!
by spudzeppelin (Pilgrim) on Jul 23, 2001 at 21:13 UTC
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Now you got me thinking. I have always been taught (as most of us have) that n^0 is 1. Now, why, do you suppose, that n multiplied by itself zero times, would it equal one?
It's a matter of consistency: remember, when you take two values which are exponents of the same base and multiply them together, you get that base, to the sum of the exponents. So, what is
((n)^1) * ((n)^-1) ?
It is n * (1/n) = (n/n) = 1.
This behaves politely for all real n != 0. But, it still remains 1 for all n->0 from each side, so it's what our friend in analysis call (and what tilly alluded to above) a removable singularity. That is, there is definitely a "hole" there, but for all-intents-and-purposes we can "plug" it when doing calculus.
Spud Zeppelin * spud@spudzeppelin.com
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The essence of the difficulty in this case is that defining a "consistent" value of the function f(x,y)=x^y isn't possible. By definition, x^y is defined as e^(y log x). (Which, incidentally, is itself not well-defined; one has to make a branch cut for log to have it make sense. If you don't know what that means, just assume that we've done it, as it doesn't really impact this discussion very much.) Note that the limit as x->0 (with y <> 0 fixed) is 0 and the limit as y->0 (with x <> 0 fixed) is 1. So we can't fill in a value at (0,0) that is going to make the function continuous there.
However, as others have noted, one can make a sensible convention 0^0=1 -- sensible in the sense that it makes doing certain things easier. There are other contexts in mathematics where one makes similar conventions (without sacrificing rigor, of course) in order to simplify notation or statements of results and so on. (E.g. in measure theory one might use the convention that 0 * infinity = 0.)
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