If $0^0 = 1$, then is it true that $0/0 = 1$?

Question

By Knuth, Concrete Mathematics (2nd ed.) page 162, it is convenient that

$$0^0 = 1$$

Then, is it true that

$$0^1/0^1 = 0^{1-1}= 0^0 = 1$$ and we are free to exclude indeterminate statement of $0/0$ ?

Long answer short: no, it is not correct to say that $0/0=1$ even with the definition of $0^0 = 1$. — Ben Grossmann, Mar 02 '20 at 08:36
You are doing it wrong! It should be $\frac{0}{0}=\frac{12\cdot 0^1}{0^1}=12\cdot 0^0=12$. — , Mar 02 '20 at 08:40
Or $\frac{233}{0} - \frac{233}{0} = 1 \rightarrow \frac{233}{0} = 1 + \infty$, which is actually right :) But, not sure about $\infty - \infty = 1$ :) — Petro Kolosov, Mar 02 '20 at 08:42
@PKK It is important to note that $0^0$ is still an "indeterminate form" in some contexts. In particular, if $\lim_{x \to 0}f(x) = \lim_{x \to 0}g(x) = 0$, then we can't conclude that $\lim_{x\to 0}f(x)^{g(x)} = 0^0 = 1$. — Ben Grossmann, Mar 02 '20 at 08:46
Here's another silly one: $$ 0 = \log(1) = \log(0^0) = 0 \log(0) = 0 \cdot (-\infty) $$ — Ben Grossmann, Mar 02 '20 at 08:48
@KiênP.S. If you're being serious, note that this would require a redefinition of "=" to some extent. For instance, taking your statement literally implies that $1 = 0/0$ and $0/0 = 2$, so by the transitive property of equality we have $1 = 2$. — Ben Grossmann, Mar 02 '20 at 08:51
@KiênP.S. If you're interested in seeing a system in which $0/0$ is "defined", I would recommend that you look into wheel theory — Ben Grossmann, Mar 02 '20 at 08:58

score 4 · Accepted Answer · answered Mar 02 '20 at 09:05

There's a good reason why it makes more sense to define $0^0$ as $1$ than it does to define $0/0$ as $1$. Notice that $0=\alpha\cdot 0$ for any real number $\alpha$. Substituting this in $\frac00=1$ would give $1=\frac00=\frac{\alpha\cdot 0}{0}=\alpha$, so this convention would not be consistent.

Conversely, the convention that $0^0=1$ is consistent, at least by this measure, because $0^{\alpha\cdot 0}=(0^0)^\alpha=1^\alpha=1$, and $(\alpha\cdot 0)^0=\alpha^0\cdot0^0=1\cdot 1=1$.

score 1 · Answer 2 · answered Mar 02 '20 at 08:59

1

No, the argument doesn't pass through.

You cannot define $0^{-1}$, so using $0^{1-1}=0^1\cdot0^{-1}$ is invalid.

answered Mar 02 '20 at 08:59

egreg

238,574

Christian Prince · Answer 3 · 2020-03-02T09:56:39.617

Assuming that $0^0=1$ is wrong because, if you take natural logarithms both sides, this will mean $$0=\ln (1) =\ln(0^0)=0 \cdot \ln (0)=0 \cdot \lim_{x \to 0^{+}} \ln (x)=0 \cdot (-\infty) $$ which is a contradiction, because $0 \cdot (-\infty)$ is an indeterminate (an operation that is meaningless.)

Similarly, suppose $\frac{0}{0}=1$ as evaluated by yourself, then that will mean $$\infty -\infty = \ln \left( \frac{0}{0}\right)=\lim_{x \to 0} \ln (x) -\ln(x)=\ln(1)=0$$ which is also a contradiction.

Let me supply an example, we know that $0.0000001 \approx 0 $ and $0.01 \approx 0$, then $$\frac{0}{0} \approx \frac{0.01}{0.0000001}=100000>>1$$

Let me give another example, since the symbol $\infty$ means a biggest number, I can consider the number 120 to be my $\infty$ just like in a T-distribution table so that $$\infty-\infty=125-120=5 >0$$ This is the reason why these operations are called indeterminate.

First, you must understand that writing $\frac{0}{0}$ on its on is a problem, but we still write it. I am calculating limits to take care of the domain problems and to show that $\frac{0}{0}$ is not only undefined but it is also undetermined. I hope I have answered your question @N.F.Taussig — Christian Prince, Mar 17 '20 at 11:42

If $0^0 = 1$, then is it true that $0/0 = 1$?

3 Answers3