# 0^0 = 1?

by The Rev
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 Sci Advisor HW Helper PF Gold P: 12,016 BoTemp was right in critizing me, shmoe; I hadn't made the proper restriction on x.
 Sci Advisor HW Helper P: 1,994 How you stated it is fine to me, you were talking about a right hand limit. There's no reason for anyone to assume (or even care) if your function was defined at 0.
 P: 41 i agree that 0^0= 1 a^x= 1/(a^(-x)) if a=x=0 then there is no way that 0^0= 0 because we'll have 0=infinite(1/0)
Emeritus
PF Gold
P: 16,091
 if a=x=0 then there is no way that 0^0= 0 because we'll have 0=infinite(1/0)
No, there's no way that 0^0 = 0 because 0^0 is undefined.

If you like heuristic reasoning from identities, what about 0^x = 0?
 P: 223 I don't think of 0 as being in the same number system as anything else really. It's more of a concept like infinity. So you really can't do all of the same math with 0 as you can with other numbers. $$0^0$$ makes no sense. Nor would $$log(0)$$. LOL... that's so funny... where I listed ['tex'] 0^0 [/'tex'] it put "infinity". EDIT: And now it's back to 0^0. Hmmm....
Emeritus
PF Gold
P: 16,091
 So you really can't do all of the same math with 0 as you can with other numbers.
You meant arithmetic. And that comes directly from the definitions -- division is defined for any nonzero denominator.

Incidentally, though, for each operation which is undefined at zero (such as 1/x), there's a corresponding operation which is undefined at one. (such as 1/(x-1)) So, in a very real sense, you can do exactly as much with zero as you can do with any other real number.
PF Gold
P: 1,236
 Quote by Hurkyl No, there's no way that 0^0 = 0 because 0^0 is undefined. If you like heuristic reasoning from identities, what about 0^x = 0?

0^x = 0

defined: for all x > 0
undefined: for all x < 0, x = 0
 P: 223 Wait! Did someone already do this? $$0 \approx (1/\infty)$$ Not quite; but approximately. $$1/(1/\infty) \longrightarrow (1/1)/(1/\infty) \longrightarrow (\infty/1) * (1/1) \longrightarrow \infty*1 = \infty$$ Though $$0$$ is a little less than $$1/\infty$$. So what value is it really?
 P: 4 As far as math teachers are concerned, you may safely assume $$0^0 = 1$$ (/End deliberate hand-waving mode). As a precaution, most documents or proofs that require its use (most that I've seen, anyway) will still explicitly state it as a useful interpretation before applying it.
 Quote by Rahmuss Wait! Did someone already do this? $$0 \approx (1/\infty)$$ Not quite; but approximately. $$1/(1/\infty) \longrightarrow (1/1)/(1/\infty) \longrightarrow (\infty/1) * (1/1) \longrightarrow \infty*1 = \infty$$ Though $$0$$ is a little less than $$1/\infty$$. So what value is it really?
You are wrong $$\frac{1}{\infty}$$ does not make sense for real numbers as $$\infty$$ is not an element of the real number set. As for sets in which it does exist, you need to learn their axioms and not assume that they are the same as the real numbers (otherwise they would be the real numbers).