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Is the following proof that $0^0 = 1$ correct?

Let $k = 0^0$ (assuming it has a meaningful value)

Observe that $k = 0^0 = 0^{n0} = (0^0)^n = k^n$, for all integers $n$.

$\implies k^n - k = 0$, for all $n \in \mathbb Z$.

$\implies k^n - k = 0$ and $k - k^{-n} = 0$, for all $n \in \mathbb N$.

$\implies k[k^{n-1} - 1] = 0$ and $\displaystyle\frac1{k^n}[k^{n+1} - 1] = 0$

$\implies [k = 0$ or $k =$ any $(n-1)$th root of unity$]$ and $[k =$ any $(n+1)$th root of unity$]$

All requirements are satisfied $\iff k = 1$ for any fixed $n$.

Therefore, $k = 0^0 = 1\quad\blacksquare$

Caleb Stanford
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A.Abbas
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    These steps show that if $0^0$ exists, it is 1. But many of these steps assumes the answer exists in the first place. – Paul Sep 09 '16 at 23:50
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    This does not qualify as a proof. You bring in the usual laws of exponents when it is convenient without justification even though there's no reason for them to apply, and you implicitly assume that $k$ is not $0$ when you take negative exponents, but give no reason why it can't be zero. – Matt Samuel Sep 09 '16 at 23:55
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    "Let $k=0^0$ (assuming it has a meaningful value)". It isn't. Whatever your proof continues to say no matter how logical or correct will not and can not be valid. – fleablood Sep 10 '16 at 00:08
  • Why can't $k=0$ by this argument? – Doug M Sep 10 '16 at 00:09
  • Similar reasoning: http://math.stackexchange.com/questions/1554929/zero-divided-by-zero-must-be-equal-to-zero – Simply Beautiful Art Sep 10 '16 at 00:09
  • Thanks to all that have commented. Your input is greatly appreciated. Originally, I thought that firstly, there is no reason for the usual exponential laws to forcibly not apply. But that includes negative exponentials as well. So my idea was that for the negative exponents to apply, this immediately eliminates k equalling zero. Of course, I never proposed that this proof is legitimate but was merely looking for reasons why it isn't. – A.Abbas Sep 10 '16 at 00:15
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    There is the ad hoc convergence theory which states that functions for values without direct meaning may be extended to allow general rules to work. We were lucky for $0^0$. In other cases, there is a cost and sometimes a bonus like for $i$ and the original $-2^{\frac12}$ –  Sep 10 '16 at 00:17
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    @Paul: Not even that. It shows that if $0^0$ exists and is invertible, it must be $1$; it uses $k^{-n}$ where $k$ is $0^0$. – celtschk Sep 10 '16 at 07:38
  • See Also: http://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1 – Caleb Stanford Sep 10 '16 at 16:22

2 Answers2

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Is my $0^0 = 1$ proof correct?

It's not a proof, really. You can never prove that something that does not have a meaning has a meaning. You can only define it to be so.

However, what you have written is an informal argument for the definition $0^0 = 1$, and it's a pretty good argument. Specifically, it's an argument that we should define $0^0 = 1$ IF we want some other properties to hold. What properties? Well, you only really employed one property, namely the law of exponents which says that if $a, b, c$ are integers and $a$ is not zero, then $$ (a^b)^c = a^{bc}. $$ Your "proof" is essentially an argument founded on the premise we should allow this to be true when $a = 0$ as well. Assuming that it's true when $a = b = 0$, you derive that $(0^0)^n = 0^0$ and $(0^0)^{-n} = 0^0$, and then go on to conclude that $0^0 = 1$. So what you've shown is that if we want the above law to be true when $a= b=0$, then we better define $0^0 = 1$.

To be sure, there are many objections to your argument:

  • You haven't really addressed the possibility that $k = 0^0 = 0$. When you take $(0^0)^{-n}$, you are implicitly not allowing $k$ to be $0$.

  • Similarly, you haven't really addressed the possibility that $k = \infty$.

  • The obvious objection: you've assumed the power law $(a^b)^c = a^{bc}$ should hold when $a = b = 0$.

  • Finally, you haven't really convinced readers who think that $0^0$ should be left undefined, because you've started out by assuming it has a value.

Fortunately, the first two bullets can be addressed by arguments similar to yours; for instance, if $0^0 = 0^{-0}$, then it should be its own reciprocal, so it doesn't make sense for it to be $0$ or $\infty$.

For the third bullet point, you could address it by considering other properties of exponents and extending them to $0$, and showing that in all or almost all cases $0^0 = 1$ seems to be the most natural. Of course, this won't convince someone who doesn't see any point in extending any of the properties to begin with.

The last bullet is more tricky. In analysis, $0^0$ is an indeterminate form meaning that we don't want to say that it is $1$ or any other particular number. And, it would actually be generally bad in analysis if we were to say that $0^0$ had a value; it would be wrong to assign it any value at all.

In summary, I have described what your argument is -- an argument, and not a proof, that $0^0 = 1$, and I think it's a decent argument. And I have discussed some possible objections to it.

In fact, note that $0^0 = 1$ is commonly taken to be the definition, used especially in abstract algebra and combinatorics. Even Google says it's $1$.

Caleb Stanford
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    You go google! Do them bad analysis things! – Simply Beautiful Art Sep 10 '16 at 00:56
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    Why would it be bad in analysis to say that $0^0=1$? After all, $x^y$ is discontinuous at $(0,0)$ no matter whether or how we define $0^0$. – celtschk Sep 10 '16 at 07:43
  • @celtschk Hold on, no. A function can't be discontinuous at a point where it doesn't exist. The "right" thing is to say $x^y$ is not defined at $(0,0)$. Anyway it's not bad as a notational convention (e.g. in writing out a Taylor series) but I don't think it's a good definition to take. – Caleb Stanford Sep 10 '16 at 16:13
  • And I claim the "right" thing to do it define $0^0=1$. And you still did not manage to demonstrate any harm done by that definition. – celtschk Sep 10 '16 at 17:02
  • @celtschk Here is the harm: $x^y$, for $x, y \ge 0$, becomes discontinuous at $(0,0)$. If that's not enough for you, then fine, no need to be a jerk about it on my answer. You will find that lots of other people agree with me that that is a good reason not to define it, including e.g. Arturo Magidin in the other thread. – Caleb Stanford Sep 11 '16 at 00:51
  • The power function is already necessarily discontinuous on a complete ray from $0$ (the branch cut), and definitely is not continuous at $0$ either way, so what would be the harm to define it at one more point connected to the already existing line of discontinuity? – celtschk Sep 11 '16 at 07:03
  • @celtschk I'm just talking about the real power function, $x^y$ for $x \ge 0$ and $y \in \mathbb{R}$. I agree with you that the complex power function, $e^{y \text{Log } x}$ has branch cut troubles either way. – Caleb Stanford Sep 11 '16 at 07:13
  • Ah, I see, you leave $(-1)^2$ undefined as well … – celtschk Sep 11 '16 at 07:30
  • @celtschk Yes, $(-1)^a$ is not usually defined in the context of a real number power function. – Caleb Stanford Sep 11 '16 at 07:33
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Just as an aside - a more 'proofy' way (if that even makes sense!) to show this is to calculate the (right) limit as $x \rightarrow 0$ of the function $x^x$ and to take this as the definition of $0^0$.

To do this simply note that $$x^x = e^{x\log x}. $$

Then standard limit rules give $$0^0 = \lim_{x \rightarrow 0^+} e^{x\log x}= e^0 = 1.$$

Zestylemonzi
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  • I don't think this is a good argument to conclude $0^0 = 1$. As you can see here, $\lim_{x,y \to 0} x^y$ does't exist, so it doesn't make sense to just pick $x = y$ and then take the limit. – Caleb Stanford Sep 10 '16 at 16:21
  • This is a completely different question and yes it does make sense to consider the case $x=y$ because this is the case when $x$ and $y$ vary in the same way when considering your limit. – Zestylemonzi Sep 10 '16 at 20:58