The Google calculator says that $\left(\frac00\right)^0=1$. Is this true?

Question

According to Google, $\left(\frac00\right)^0=1$. Is this true? Why or why not?

Answer 1

This certainly isn't literally true: $\frac{0}{0}$ is undefined, and hence $\left(\frac{0}{0}\right)^0$ is also undefined.

You might ask whether $\left(\frac{0}{0}\right)^0=1$ is true in the limiting sense that $\lim_{(x,y,z)\to(0,0,0)}\left(\frac{x}{y}\right)^z=1$. However, this is also false. For instance, let $x=t^2$, $y=t$, and $z=1/\log t$ for $t\in(0,1)$. As $t\to 0$, $x$, $y$, and $z$ each go to $0$. But $$\left(\frac{x}{y}\right)^z=t^{1/\log t}=e$$ for all $t$, so the limit as $t\to 0$ is $e$. Since you can find other parametrizations along which $(x,y,z)\to (0,0,0)$ for which the limit of $\left(\frac{x}{y}\right)^z$ is something else (for instance, if $x=y=z=t$, $\left(\frac{x}{y}\right)^z=1^t=1$ for all $t$), it follows that the limit $\lim_{(x,y,z)\to(0,0,0)}\left(\frac{x}{y}\right)^z$ does not exist.

(Strictly speaking, when we write $\lim_{(x,y,z)\to(0,0,0)}\left(\frac{x}{y}\right)^z$, we should be taking the limit on the domain where $\frac{x}{y}>0$, since otherwise $\left(\frac{x}{y}\right)^z$ itself may be undefined.)

Answer 2

It depends whether you are adopting a computational vs. a symbolic-analytical perspective.

From a computational perspective, the parentheses indicate that the number crunching inside the parentheses is to be completed first. However, dividing by zero is undefined for integer math, or for floating point divide, you get a result NaN (meaning not a number) and this result cannot go forward into exponentiation, so computationally, the expression is non-computable into a usable result. And, since questions of T vs. F are predicates in computation, and you haven't specified your question in the form of a computable predicate, the question you've asked can't be answered computationally.

From a symbolic-analytic perspective, using limits you could prove that x/x is a stable result equal to 1.0 as x approaches 0. You could separately show that 1^n power gives a stable result equal to 1 as n approaches 0. You could combine the two limits results to claim that (x/x)^n yields a stable value of 1 as both x and n approach 0.

That's not the same as just saying that the equation (0/0)^0 = 1 is true, because by itself, the equation appeals to a singularity exception in both the divide and exponentiation functions. I think the most you can say is that in the limit as the operands approach zero, the equation is true.

Answer 3

No. Under the usual conventions, this expression is "ill-typed". You don't write $0/0$.

On the other hand, under the conventions that I describe here, we have $$\left(\frac{0}{0}\right)^0 = (\mathbb{R})^0 = 1$$

So "yes", if you want to define things that way.

Perhaps this is evidence that my evil twin is working for Google. I'll find you eventually Darkgoblin.