# Is $0^0=1$ postulate independent of all other axioms of complex numbers?

This question is inspired by the other question which asked for a proof that $i^i$ is a real number.

Many calculators when asked for $0^0$ return 1. I asked a mathematician how to prove that but he replied that it is impossible and that one only can postulate this by a convention.

So I wonder whether he is right? Is $0^0=1$ axiom really independent of all other axioms defining standard complex numbers and exponentiation?

UPDATE

It seems that no response so far tried to provide an answer to my clear question, that is whether $0^0=1$ is an independent axiom or not. Most answers are trying to defend particular values for $0^0$ which the authors prefer with some indirect or abstract argumentation, mostly involving taking limits or citing practical purpose.

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Is there a standard axiomatization of "standard complex numbers and exponentiation"? – Jason DeVito Oct 27 '12 at 2:39
Many calculators return 1? Really? I have a TI-60X which I have had for around 15+ years and it returns an error. And, my TI-84 which is several years newer still returns error. So, I am guessing TI calculators probably all return errors. Can you give an example of one that returns 1? – Graphth Oct 27 '12 at 15:34
@Graphth Windows calculator, for example. – Anixx Oct 27 '12 at 15:58
This is, at most, a definition. – Mariano Suárez-Alvarez Oct 27 '12 at 19:56
@Graphth Testing on all my calculators: results. – Navin Oct 28 '12 at 5:20

In complex analysis, exponentiation is defined by first choosing a branch of the complex logarithm and then defining $a^b = \exp(b \log a)$, where $\exp$ is the complex exponential (defined in terms of a series) and $\log$ is the chosen branch of the complex logarithm. Because there is no branch of the complex logarithm defined at $z = 0$, the expression "$0^0$" is not defined for complex exponentiation.

For natural number exponentation, the identity $0^0 =1$ is indeed a postulate, which is useful in many settings, and it follows from the definition $0^0$ as an empty product. But the definition of complex exponentiation does not verify this identity, because it is not defined in terms of "products", it is defined in terms of the complex function $\exp(z)$ and a branch of the complex logarithm.

In the complex case, it would make no real sense to try to define $0^0 = 1$; the logarithm has a singularity at $z = 0$ and there is no benefit of trying to make the function defined there, because it will still not be analytic there.

The thing that is often confusing is that there are several different definitions of exponentiation, and the fact that they agree at various points is a theorem, not something that "has to be". The theorem breaks down in many cases when we look at $0^0$; this expression is given a value by some definitions of exponentiation but not others. It is somewhat analogous to the fact that $5/2$ is undefined in the natural numbers but defined in the complex numbers, except opposite: $0^0$ is defined for natural number exponentiation but not for complex exponentiation.

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Are not natural numbers a subset of complex numbers? Can $0^0=1$ be independently postulated for complex numbers as an axiom? – Anixx Oct 27 '12 at 3:25
Of course $\mathbb{N}$ is a subset of $\mathbb{C}$; the issue is that there is more than one "take the power" function. Unfortunately, since we don't use a symbol for these functions, it's hard to see visually that there are different ones. If we look at the "repeated multiplication" definition, thinking of the exponent as natural, then $0^0$ will be $1$. But complex multiplication is not defined that way, and it is defined the same way for all inputs, not by cases. But since the complex power $0^0$ is undefined, you can pick any value for it you want, if you only want to assign it a value. – Carl Mummert Oct 27 '12 at 3:29
so $0^0=1$ is an independent axiom BOTH for natural exponentiation and for complex exponentiation? – Anixx Oct 27 '12 at 3:33
It depends on how you define natural exponentiation; when it is defined in terms of repeated products, the empty product is $1$ and so you get $x^0 = 1$ from that. – Carl Mummert Oct 27 '12 at 3:35
that empty product is 1 is an axiom or provable? – Anixx Oct 27 '12 at 3:37

I don't think it's just a convention. $0^0$ can be considered an empty product, and when you don't multiply by anything, it's the same as multiplying by $1$. The identity $$e^z= \sum_{n=0}^\infty \frac{z^n}{n!}$$ does not hold when $z=0$ unless $0^0=1$, since the first term is $\dfrac{0^0}{0!}$.

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Well I see but this does not answer the question directly. – Anixx Oct 27 '12 at 2:53
This is one case where it is $1$ but if you evaluate the limit $$\lim_{x\to0, y\to0}x^y$$ in some directions you will not get 1 – Navin Oct 27 '12 at 3:03
The issue is that the $z^n$ in the Taylor series is a different operation than the complex exponential $z^w = \exp(w \ln z)$, even though they are both written in the same way. The $z^n$ is defined as repeated multiplication, but the complex $z^w$ isn't. – Carl Mummert Oct 27 '12 at 3:07
@Anixx: look at $x = \exp(-1/t)$ and $y = at$. Then both $x$ and $y$ tend to $0$ as $t \to 0^+$ but $x^y$ tends to $\exp(-a)$ -- and $a$ was arbitrary. – Carl Mummert Oct 27 '12 at 3:32
@Carl Mummert oh I see I confused it with another limit, $\lim x^x$, in that case you always get 1 at zero unless you are approaching zero by a spiral. – Anixx Oct 27 '12 at 3:36

The most basic definition of addition is: $a+b$ is the cardinality of a set that is a disjoint union of two sets of cardinalities $a$ and $b$. This matches the intended interpretation of addition for natural numbers and extends readily to $\mathbb Z$, $\mathbb Q$, $\mathbb R$, $\mathbb C$ by the usual constructions.

The most basic definition of multiplication is: $a\cdot b$ is the cardinality of the cartesian product of two sets of cardinalities $a$ and $b$. This matches the intended interpretation for natural numbers and extends readily to $\mathbb Z$, $\mathbb Q$, $\mathbb R$, $\mathbb C$ by the usual constructions.

The most basic definition of exponentiation is: $a^b$ is the cardinality of the set of maps from a set of cardinality $b$ to a set of cardinality $a$. Since there is exactly one map from $\emptyset$ to any other set (namley the inclusion map), we have $a^0=1$ for all cardinalities $a$, including $a=0$. On th other hand, for a non-empty set, there are no maps to the empty set, hence $0^b=0$ for all $b\ne 0$. The basic rules of exponentiation are valid for this definition (as provable by simple set-theoretic arguments), e.g. $a^{b+c}=a^b\cdot a^c$, $(a\cdot b)^c=a^c\cdot b^c$, $a^{b\cdot c}=(a^b)^c$. This is all very nice and pretty and consistant and thus give s definition of exponentiation in $\mathbb N_0$. You have problems already when trying to extend the definition to $\mathbb Z$ (what is $0^{-1}$?). You have more problems when extending to $\mathbb Q$ (watch out for the error in $-2 = (-8)^{\frac13}=(-8)^{\frac 26}=((-8)^2)^{\frac16}={64}^{\frac16}=2$). Problems in the negative continue when trying to extend the definition to $\mathbb R$ by continuity (because $\frac{2n}{2n+1}\to 1$, we should have $(-1)^{\frac{2n}{2n+1}}\to -1$, in fact already the rational extension is not continuous at $(0,0)$), not to mention $(-1)^{\frac12}=?$). All this is the reason why the existence of a nice function $\exp\colon \mathbb C\to \mathbb C$ with the fundamental properties $\exp(x+y)=\exp(x)\cdot\exp(y)$ and $\exp(0)=1$ that resemble the above laws of exponentiation so well is taken as a way to define exponentiation in $\mathbb C$ with its help (but still with restrictions due to branching) - even the suggestive notation $e^z$ for $\exp(z)$ is so common! However, this definition forces one to leave certain gaps in order to keep continuity. While one might choose some unusual branch cut other than the negative reals, $0^0$ will always be sacrificed. In the end, what we want is persistance (that is: as few exceptions to theorems as possible). Therefore, in the $\mathbb C$ and probably already in $\mathbb Q$, it is preferable to leave $0^0$ undefined. But when working e.g. with integers $0^0=1$ is preferable.

Note also the following example: One writes polynomials e.g. as $P(X) = \sum_{k=0}^n a_k X^k$. Here it is conventional to use $X^0=1$, even if one computes $P(0)$. This is much more convenient than needing to write $P(X)=a_0 +\sum_{k=1}^n a_k X^k$. Some might argue that $X\ne 0$ as element of the ring $\mathbb Z[X]$, therefore $X^0=1$ does not reference the problem whether $0^0$ is defined or not.

UPDATE: For all mathematical operations, it is relevant how they are defined. If I take ZFC as "the" basis of most of mathematics, there is not a single axiom of it dealing with exponentiation at all. In fact, thee are not even axioms about addition or multiplication. Thus for example $2+2=4$ is not an axiom or postulate, it is simply a theorem (based on the definitions of "+" and of course of "2" and "4"). I have described above, how addition and multiplication are defined for elements of suitable sets of numbers. Contrary to these two operations, exponentiation as a few problems: There are several ways of defining it, essentially one is coming from exponentiation of ordinal numbers (and gives $0^0=1$) and one coming from the exponential and logarithm function and their defining power series (plus some conventions about how to treat multivaluedness of the logarithm in the complex case) which has difficulty at assigning a value to $0^0$. Thus while the first definition does not apply at all to general real or complex numbers and does assign e.g. the value $-1$ to the expression $e^{\pi i}$, the second definition does not assign a value to $0^0$. Thus under closer inspection there are diffeent operations that are known under the name of exponentiation (and the same name is justified and even motivated by the fact that both definitions agree where they both define a value). Thus it may depend on the context whether or not you assign a value to the expression $0^0$ (or maybe even which value you assign?). Either way $0^0=1$ is not an axiom, instead it is either part of a definition or a theorem follwoing from the specific definition of exponentiation.

Then again, you may consoider all definitions as extensions to a theory by introducing new symbols and axioms describing them. If this is how you look at it, then $0^0=1$ can be viewed as an independent axiom iff it is not part/consequence of the basic definition of exponentiation- see above.

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Isn't saying that there exists a function $f:\emptyset \rightarrow A$ also a convention? I don't disagree with it (it ultimately boils down to the principle of vacuous truth), but it seems like circular reasoning to use it as an argument here. – asmeurer Oct 27 '12 at 20:27
@asmeurer: I don't see why that should be "just a convention". (The graph of) such a function $f$ is a subset of $\emptyset\times A$ such that (vacuous truth), hence $f=\emptyset$ (the only subset of $\emptyset\times A$) is the one and only function $\emptyset\to A$. The same works with funcitonal predicates. There is no definitional loophole (compared with what I encoountered in another discussion today: Is a topological space disconnected if it has more than one component? or if the number of components is $\ne1$?) – Hagen von Eitzen Oct 27 '12 at 20:36

$0^0$ is indeterminate. We define it to be either 1 or 0 depending on the circumstances. There are various proofs which prove either case. However, $0^0 = 1$ is seen more often because it is required to define some Taylor series, it has set-theoretic interpretation, combinatorial interpretation, etc. It is pretty much the same as saying that $0! = 1$, it is just a convention, which also has a set-theoretic interpretation and a combinatorial interpretation. It is most definitely not an axiom. Neither is $0^0 = 1$. By the way, if $0^0 = 0$ then there are exactly four Munchhausen numbers, otherwise there are only two if you count the trivial case $1^1$.

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Well if there are proofs that prove either of the contradicting cases, then the set of axioms is inconsistent. -1 – Anixx Oct 27 '12 at 3:18
@Anixx - the point is that the value of $0^0$ (if any) you adopt depends on the context i.e. the set of axioms you are working with, and the definition of exponentiation you have chosen. It isn't that the axioms are inconsistent, but that the definition of the exponential function is different - and that arises because there isn't a definition which does everything you want it to - like $x^y$ being continuous at $(0,0)$ – Mark Bennet Oct 27 '12 at 3:41
@MarkBennet the answer says "There are various proofs which prove either case." and also "It is most definitely not an axiom.". So I want to see the proofs that the answer claims to exist. – Anixx Oct 27 '12 at 3:43
@Anixx Well try defining $0^0$ by continuity from $0^r$ for positive real $r$. That gets you to zero. (Try defining $0^r$ for negative real $r$ to see why there might be a problem at $r=0$). – Mark Bennet Oct 27 '12 at 3:54
@BrianM.Scott: I call $a\circ b$ indeterminate, if $a_n\circ b_n$ may show different behaviour (that is: not determined by $a$ and $b$ alone) for different sequences $a_n\to a$ and $b_n\to b$. This is in general the case where $(a,b)$ is a boundary point of the realm where $\circ$ is continuous and no continuous extension to $a\circ b$ is possible. In other words, we use indeterminate mainly in the context of $\lim a_n\circ b_n \stackrel ?= \lim a_n\circ \lim b_n$, which relies on continuity. Thus I personally consider $0^0$ both indeterminate and (via $|\emptyset^\emptyset|$) defined. – Hagen von Eitzen Oct 27 '12 at 16:15

$0^0$ is your first introduction to ambiguity of notation. There are several things one could mean by exponentiation, and the particular case of $0^0$ is the most prominent manner in which they conflict.

The usual real^real exponentiation, it is clearly undefined due to the discontinuity. The multi-valued complex^complex exponentiation is undefined for a similar reason.

Another example is a limit form, which is more of a formal description of the behavior of a limit $\lim f^g$ rather than an operation. (although it is closely related to real number exponentiation)

However, in (ring element)^(natural number) exponentiation, it is equal to $1$.

But the most relevant case is actually something different: it is the evaluation of a power series (or polynomial) at zero -- specifically, the power series $x^0 \; (= 1)$. So $0^0$ is not actually exponentiation at all, but it is substituting $x=0$ into $x^0$. Of course, this can be computed using the (ring element)^(natural number) operator.

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The standard definition of exponentiation of positive real numbers is to use the standard definition for positive exponents, extend this to negative and rational exponents (there is a theorem giving uniqueness of the integer root of any positive real number that is important here). We then define all other exponents by taking limits, i.e., $a^x = \lim_{x_n\rightarrow x} a^{x_n}$, where $x_n\in \mathbb{Q}$. The important thing here is that the definition is well-defined, that is, any sequence $x_n\in \mathbb{Q}$ that converges to $x$ gives the same value in the formula.

For complex exponentiation, we might construct things in a similar way. The important thing again, though, is that we get the same value no matter how we take the limit. For example, if $x_n \rightarrow 0, x_n > 0$, then $\lim_{n\rightarrow \infty}0^{x_n}=0$, since each term in the sequence is $0$. But on the other hand, $\lim_{n \rightarrow \infty}x_n^0=1$, since each term in the sequence is $1$. Therefore, as far as analytic definitions go, $0^0$ is not well-defined.

$0^0=1$ is well-defined from other perspectives, such as ones stemming from number theory or combinatorics, but analytically, it doesn't make sense. That is why we say that this is only a convention. We tend to think of definitions that make sense analytically as the "right" ones, and any defining anything that can't make analytical sense as a convention. $1/0=\infty$ is another example of a convention.

By the way, another way to see this is that there is a discontinuity of $x^y$ at $x=y=0$. You can see this from the graph.

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You wrote: "For example, if $x_n \rightarrow 0, x_n \neq 0$, then $\lim_{n\rightarrow \infty}0^{x_n}=0$, since each term in the sequence is $0$." - This is NOT true. Actually, this limit does not exist. – Anixx Oct 27 '12 at 7:41
@Anixx: On the contrary, it certainly is true: under the stated hypotheses the sequence $\langle 0^{x_n}:n\in\Bbb N\rangle$ is the constant sequence $\langle 0,0,0,\dots\rangle$, which does indeed converge to $0$. – Brian M. Scott Oct 27 '12 at 8:58
@Anixx: No, it is not wrong. $0^x=0$ for all non-zero $x$. – Brian M. Scott Oct 27 '12 at 10:03
Does someone really think $0^{-1} = 0$ ??? – GEdgar Oct 27 '12 at 16:42
I fixed the answer. $x_n$ should be positive as pointed out. This does not change the argument, though. – asmeurer Oct 27 '12 at 19:08

It seems that no response so far tried to provide an answer to my clear question, that is whether $0^0=1$ is an independent axiom or not

I am interpreting this question as "is $0^0=1$ an {independent axiom} or {non-independent axiom}".

No one has given a yes/no answer to whether it is independent because it is not an axiom! To summarize all the other answers:

In natural number exponentiation, " $0^0=1$ " is a true statement because it is an empty product. Also see Hyperoperation for a recursive definition of addition in terms of succession, multiplication in terms of addition, exponentiation in terms of multiplication, tetration in terms of exponentiation, ...ad infinitum. If you use this definition, $0^0$ is the base case of the recursion and evaluates to $1$ because it is the identity.

In complex exponentiation, " $0^0=1$ " is not a well formed statement because $\log{0}$ is not defined in any branch ($x^y~\stackrel{def}{=}~e^{y\log{x}}$) and the limit $$\lim_{\mathop{x\to0,}{y\to 0}}x^y$$ is not the same in all directions

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"true statement because it is an empty product" that empty product is equal 1 an axiom? If not, how to prove it? – Anixx Oct 29 '12 at 0:17
@Anixx If you use the hyper operator definition, it is defined in the recursion. See the $n\ge 3, b=0$ case – Navin Oct 29 '12 at 0:19
"is not a well formed statement because the limit..." How on earth existence or the value of the limit makes the statement about value of a function in a point well formed or not? – Anixx Oct 29 '12 at 0:19
@Anixx because usually these edge cases where the original definition may not apply are defined as their limits. – Navin Oct 29 '12 at 0:21
Re the link: as I see the definitions on that page are in fact axioms or equivalent to axioms. – Anixx Oct 29 '12 at 0:21