# What values of $0^0$ would be consistent with the Laws of Exponents?

I am using the following fundamental properties of exponentiation on $N$ as as basis for this discussion:

(1) $0^1 = 0$

(2) $\forall x\in N (x\ne 0 \implies x^0 = 1)$

(3) $\forall x,y\in N (x^{y+1}=x^y\cdot x)$

Missing, of course, is a value for $0^0$. But only $0^0=0$ or $1$ are consistent with the Laws of Exponents:

(4) $\forall x,y,z\in N (x^{y+z}=x^y\cdot x^z)$

(5) $\forall x,y,z\in N (x^{y \space\cdot z}=(x^y)^z)$

EDIT:

From (5), we must have $(0^0)^2=0^{0\times 2}=0^0$. Therefore, $0^0= 0$ or $1$. Is this correct?

Is there any way to eliminate $0$ (or $1$) as a possible value, with reference to the fundamental properties or the laws of exponents?

• I am sure this has been discussed before, but I note, from my own efforts, that it is not easy to find the previous questions which have raised the point. – Mark Bennet Sep 5 '13 at 19:11
• Yes, this has been covered again and again, but there is a reason for that. We generally define $0^0=1$ on the natural numbers. This can be done in terms of set theory ($x^y$ being the size of $X^Y$ with $|X|=x,|Y|=y$.) It is also simpler - note how your (2) has a special case? Always unfortunate. The problem happens when you want $x^y$ to be continuous, but then you have a lot of problems - $x<0$, for example. – Thomas Andrews Sep 5 '13 at 19:14
• There was a discussion here – Old John Sep 5 '13 at 19:15
• I've been mystified by people trying to give a value to $0^0.$ It finally occurred to me that, in using summation signs for polynomials and power series, we use the shorthand $x^0$ to mean $1,$ and maybe people think that says something about $0^0.$ Hi, @OldJohn – Will Jagy Sep 5 '13 at 19:18
• There are indeed many postings on $0^0$ here -- maybe in every other math forum! -- but I am not asking for a value for $0^0$. There doesn't seem to be a consensus. But I have noticed that both $0^0=0$ and $0^0= 1$, at the very least, do seem to be consistent with the Laws of Exponents. I just wanted to confirm this fact. – Dan Christensen Sep 5 '13 at 19:28

If $c,d$ are cardinal numbers, then $c^d$ is the cardinal of the set of maps $d \to c$. This works for all cardinal numbers and the usual arithmetic laws hold. For $d=0$ we get $c^0=1$ since there is a unique map $\emptyset \to c$. This holds for all $c$, in particular $0^0=1$. So there is actually no debate what $0^0$ is or not, it is $0^0=1$ by the general definitions. No case distinctions are necessary. Forget about $0^0=0$, this is nonsense.

Ok, this may be glib, but following up on the comment of Will Jagy above, consider the following:

$$1 = 1^n = (1+0)^n = \sum_{i=0}^{n} \binom{n}{i} 1^i \cdot 0^{n-i}.$$

Do you see what I'm getting at? What sense do we want to make out of the last term, $1^n \cdot 0^0$?

Actually, I need to clarify why I posted this as an "answer." Since the OP is looking to define $0^0$ in a way that stays consistent with "fundamental properties of exponents," and I feel that the Binomial Theorem is fundamental enough to qualify as essential to arithmetic, the example above rules out the possibility of $0^0 = 0$ while reinforcing $0^0=1$.

• I think the proof of BT assumes $0^0=1$. – Dan Christensen Sep 5 '13 at 21:51
• Also, BT, as usually stated, assumes $0^0=1$. But it could easily be re-stated without this assumption, handling the zero-cases as exceptions similar to other exceptions for zero-denominators. Yes, it would be awkward. – Dan Christensen Sep 5 '13 at 22:39
• Of course, the PROOF of BT needs the assumption $0^0 = 1$, but here it is raised to an AXIOM which then actually implies the assumption; this means that $0^0=1$ is equivalent to BT. – Kofi Sep 6 '13 at 6:53

Dan, the proof of BT indeed assumes 0^0=1. The problem is that:

(a) there are many places where we assume 0^0 = 1, and (b) people do not acknowledge (a).

Calculus textbooks usually say that 0^0 is undefined, but at the same time, they contain dozens of formulas that assume 0^0=1.

Textbooks should honestly admits that it is standard practice to evaluate 0^0 to 1.

The argument for "undefined" is based on the mistaken belief that 0^0=1 leads to contradictions. It does not (if it did, we would have found out long ago, because the assumption 0^0=1 is used in subtle ways in lots of places).

But the real reason that I feel strongly that 0^0 must be 1 is because the "undefined" idea is contrary to a two deeply held convictions, namely (1) that the most convenient definition must be the best one, and (2) math is consistent, so given the fact that set theory tells us 0^0=1, it means that it is OK to use that everywhere.

• The problem in calculus is that choosing value for $0^0$ is choosing whether or not $0^x$ is continuous or $x^0$ is continuous, or seeing that $x^y$ cannot be continuous as a function of two variables. – Asaf Karagila Apr 21 '14 at 14:13
• Stewart (admittedly the only calculus book in easy reach) writes that $0^{0}$ is indeterminate (which is perfectly true), not "undefined". Can you supply specific citations of calculus authors who use "undefined" in this context, and/or do you dispute that $0^{0}$ is indeterminate? – Andrew D. Hwang Apr 21 '14 at 14:21
• I checked 4 calculus books, and they write that x^0 = 1 if x is not 0. They should write: x^0=1 for every x because that is what they actually use in several formulas. – Mark Apr 21 '14 at 15:56
• Asaf, even from the viewpoint of the function 0^x, the value 1 is better than the value 0. This function provides two values (0 and infinity) depending on whether x approaches 0 from the right or the left. Picking one of these two values is a arbitrary choice, and by symmetry a^(-x) = 1/a^x you can see that both choices must be wrong. The only value that is supported by a consistent argument is the value 1. – Mark Apr 21 '14 at 16:06
• If $0^0 =1$ then $0 = 0^{-2} \cdot 0^2 = 0^0=1$. – Kasper Mar 16 '18 at 9:15

We can make sense of $0^0=1$ in this way also : Expand $f(x)=a^x$ as Taylor series for real values of x and a is fixed positive real number and let a tends to zero. Its not rigorous but still supporting our assumption.

Follow-up

It can be shown that there exists not just 1 or 2 "exponent-like" functions on N as I suggested here, but an infinite number of them. And from any one of them, we can derive the usual Laws of Exponents. For formal proofs, see Oh, the ambiguity! at my blog.