Is there any good reason not to define $0^0=1$ , such as contradictions in algebra or arithmetic?

Math people:

The title is the question: Is there any good reason not to define $0^0=1$ , such as contradictions in algebra or arithmetic?

I searched for similar questions before I posted this question, and couldn't find any. After I posted it, I got some comments citing similar questions. There is a similar question at What values of $0^0$ would be consistent with the Laws of Exponents? . I checked some of the other questions in the links in the comments and the links posted with those links. There is a closer match at How to define the $0^0$? That question was closed as a duplicate, but the older, duplicated question was not identified. I could not find a convincing answer anywhere to my essential question: "does defining $0^0=1$ lead to contradictions in algebra and arithmetic?" I'll leave it up to others to decide whether my question is a duplicate. If it is, maybe you can close this question and give a better (in my opinion) answer to one of the older questions.

Let me get one thing out of the way up front: yes, I know "$0^0$" is an indeterminate form. That is, if $f$ and $g$ are real-valued functions with $f(t) \to 0^+$ as $t \to 0$ and $g(t)\to 0$ as $t \to 0$, then you don't know what $\lim_{t\to 0}f(t)^{g(t)}$ is, or even whether exists, without more information. I don't consider this a good reason to declare that $0^0$ itself must be considered undefined. I know many people will disagree with me here. I expect at least one answer and some comments arguing why this is a good reason for $0^0$ to be considered undefined. Everyone is entitled to their opinion, and you are free to leave such an answer. I will not attempt to change your mind, beyond what I write in this question.

If you define $f(x,y) = x^y$, then $f$ cannot be continuous on $[0,\infty) \times \mathbb{R}$ no matter what value, including $1$, you assign to $f(0,0)$. But why should every function have to be continuous?

If the mathematical community ever does come to the consensus that $0^0=1$, and I were teaching calculus students about limits involving indeterminate forms, I probably would not even mention the question of whether $0^0$ itself had a value, because it probably just confuse the students. They probably wouldn't even notice the omission.

To me, a "good reason" not to define $0^0=1$ would be if this definition resulted in a contradiction, when used in expressions involving multiplication and exponentiation of real numbers and the rules used to simplify such expressions. Here is an attempt to produce such a contradiction: assuming $0^0=1$, $(0^0)^2=1^2=1$, and $(0^0)^2=0^{(0*2)}=0^0=1$. No contradiction. In constrast, if you define $0/0 = 1$ and you want the associative property to hold (a reasonable expectation), then you can derive the contradiction $1=0/0=(2*0)/0=2*(0/0)=2*1=2$.

It just occurred to me that there is another good reason for not declaring officially that $0^0$ must always equal $1$: if defining $0^0=0$ does not lead to contradictions in algebra or arithmetic, either.

I am not claiming $0^0 = 0/0$. Of course you can never divide by zero, or raise zero to a negative power.

Of course, when people use power series, they use $0^0=1$ all the time, and no one complains. I have read that "$0^0=1$" is used often in combinatorics, but I don't know much about combinatorics.

Based on what I have seen in the older questions, their answers, and the answers and comments to this question, it seems that no one has discovered any way in which defining $0^0$ to be $1$ leads to contradictions when using the usual rules of multiplication and exponentiation. It also seems that defining $0^0$ to be $0$ does not lead to such a contradiction. So I'm guessing it is impossible to produce such a contradiction. But I have never heard of anyone wanting to define $0^0$ as $0$.

Solution 1:

At the very least, the answer $0^0 = 1$ is consistent with cardinal arithmetic on the set $\{0, 1, 2, \ldots\}$. Under this interpretation, the number $m^n$ is defined to be the number of functions from an $n$-element set to an $m$-element set. There is exactly one function from a $0$-element set to a $0$-element set, so in this interpretation, $0^0 = 1$. The laws $a^m a^n = a^{m+n}$ and $(a^m)^n = a^{mn}$ can be proven with that definition. Thus, exponentiation as repeated multiplication can be recovered. (In one of your links, Matt N. gives this same idea as an answer.)

The first criticism I heard involved the subtraction law $\frac{a^x}{a^y} = a^{x-y}$. The idea is that $\frac{0^x}{0^x} = 0^0$, yet $\frac{0^x}{0^x} = \frac{0}{0}$, so $0^0$ is undefined. The problem here isn't with $0^0$, it's with saying $\frac{0^x}{0^x}$ is equal to anything, when it is undefined. We could use the same reasoning to say that $0^2$ is undefined, since $0^2 = \frac{0^7}{0^5} = \frac{0}{0}$. At worst, the subtraction law has to be modified to say that it only applies when $a^y \neq 0$. We make such modifications for all of our other laws involving division, so why should this be any different?

Solution 2:

In many settings, the point-set version of the notion of function is a useful fiction, but is ultimately wrong.

For example, in real analysis, the value of a function at a point is pretty much wholly irrelevant -- it is only the "bulk" behavior of a function in a regino that matters. The value of an integrand at individual points already has no bearing whatsoever to a Riemann integral. Most operations are implicitly modified to append "... and then take the continuous extension of the result" to their usual meaning. The definition of "derivative" eventually gets refined to the point where discontinuities get washed out. One even works with objects like the dirac delta function that can't even be fully described by specifying its 'values' at all points.

Definition: The cvalue of a function $f$ at the point $a$ is defined to be, if the it exists and is unique, the value $g(a)$ where $g$ is a function continuous at $a$ and differs from $f$ at only finitely many points. This value will be notated $f[a]$.

(replace "finitely many points" with "a set of measure zero" to get a better definition, if you know what that means)

The name and notation is made up for this post -- I think analysis would simply call this the "value" of $f$ at $a$ and assume that their colleagues know what they mean.

When limits exist, we can compute them simply by plugging in cvalues:

$$ \lim_{x \to a} f(x) = f[a] $$

It's also not too hard to see, for example, that

$$ \int_a^b f(x) \, dx = \int_a^b f[x] \, dx $$

so we can already see that cvalues are already good enough for computing integrals.

Also, recall that the major property of the dirac delta function is that if $f$ is continuous at zero, we have

$$ \int_{-\infty}^{+\infty} f(x) \delta(x) \, dx = f(0) $$

This generalizes to

$$ \int_{-\infty}^{+\infty} f(x) \delta(x) \, dx = f[0]$$

when $f$ is discontinuous at zero.

Another nice example is that the partial function $f(x) = \frac{x}{x}$ and the function $g(x) = 1$ satisfy $f[x] = g[x]$ everywhere.

I hope these give some of the flavor of why cvalues have importance.

The point of bringing all of this up is that in the continuous context, putting emphasis on values at individual points is the wrong way to think about things. Setting $0^0 = 1$ washes out, because the cvalue of $x^y$ doesn't exist at $0^0$. However, continuous functions are very special because they are much easier to manipulate and work with and have loads of nice properties. Modifying exponentiation to be discontinuous makes it awkward to take advantage of the nice features that exponentiation would otherwise have from its status as being continuous on the region defined by $x > 0$ and on $x = 0 \wedge y > 0$.