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Following up on my previous question at What values of $0^0$ would be consistent with the Laws of Exponents? I am still trying to get a handle on $0^0$. It now seems to me that not only is it undefined, but so is $0^1$. What mistake, if any, am I making here? (Yes, I know, you can simply define values for both and hope no contradictions arise, but I don't find such an approach to be very satisfying.)

In the natural numbers, exponents greater than 1 are naturally defined as follows:

$x^2=xx$

$x^3=xxx=x^2x$

$x^4=xxxx=x^3x$

$x^5=xxxxx=x^4x$

and so on.

Therefore $x^{n+1}=x^nx$ for $n\gt 1$.

What about $x^0$ and $x^1$?

Suppose $x^1x=x^2$. Then $x^1=x^2/x = x$ for $x\ne 0$.

Suppose $x^0x=x^1$. Then $x^0=x^1/x = 1$ for $x\ne 0$.

Thus $x^0$ and $x^1$ are both undefined for $x\ne 0$.

CONCLUSION:

For proof that $0^1=0$ assuming the usual Laws of Exponents, see comments by Henry Swanson below.

EDIT:

See FOLLOW-UP in my answer below.

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Dan Christensen
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    $0^1=0$ is surely defined in any mathematics. – PVAL-inactive Sep 16 '13 at 04:58
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    You have it backwards. $x^1 = x$ is the first definition and then for $x\neq 0$, $x^0 = 1$. Nonpositive powers do not make any sense a priori. – minimalrho Sep 16 '13 at 05:00
  • Yes, but how to justify it beyond it just seems to work. – Dan Christensen Sep 16 '13 at 05:00
  • How does "$x^1=x^2/x=x$ for $x\ne0$" suggest that $0^1$ is undefined? Why would we have to justify a definition? To compare: can you justify that the word "bird" refers to flying avian beings, or does it just seem to work that way? Why would one need to "hope no contradiction arises" in defining things like $0^0$ or $0^1$? Do you mean hope that all the laws (e.g. exponent laws) remain as universal as possible? – anon Sep 16 '13 at 05:01
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    Why would $0^1$ be anything other than $0$? – copper.hat Sep 16 '13 at 05:01
  • @anon It seems we can determine no value for $0^1$. – Dan Christensen Sep 16 '13 at 05:04
  • @Dan What makes you say that though? If you walk into a lit room and say to others "it seems no light is on" then we're going to ask why you think that. – anon Sep 16 '13 at 05:06
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    You can't determine a value from the equation $0 \cdot 0^1 = 0$, but that doesn't mean it doesn't have a value. – copper.hat Sep 16 '13 at 05:08
  • There is some element of perpetual motion about this. – copper.hat Sep 16 '13 at 05:12
  • Where does the set theory part comes into the question? – Asaf Karagila Sep 16 '13 at 05:26
  • @anon Re: Justifying definitions. Your definitions have to be consistent with one another. You could define $0^0=2$ but would soon run into contradictions. – Dan Christensen Sep 16 '13 at 05:31
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    But the definition of $x^1=x$ does not contradict anything with $x=0$ or any other value, obviously. You may as well ask if the distributive property contradicts the commutative property and study high-level model theory and formal logic for all that arbitrary skepticism. – anon Sep 16 '13 at 05:34
  • @anon I am not convinced that no contradiction would arise. Alternatively, we could leave $0^1 undefined as well. Yeah, it could get ugly. – Dan Christensen Sep 16 '13 at 05:37
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    Are you convinced that assuming $a+b=b+a$ and $a\cdot(b+c)=a\cdot b+a\cdot c$ does not lead to a contradiction? Why or why not? – anon Sep 16 '13 at 05:38
  • @anon You can prove these from first principles (e.g. Peano's axioms). – Dan Christensen Sep 16 '13 at 05:42
  • It seems odd to me that people only begin being skeptical about definitions like $x^1=x$ that do not depend on other things and therefore cannot contradict other things $-$ at most, other claims stated later on that build off of it (e.g. the universality of exponential laws or continuity of exponential expressions) could contradict it. By the way, there is a difference between defining what notation means, and then making claims using that notation (e.g. defining $x^n$ then saying $x^{n+m}=x^nx^m$ for all natural $n,m,x>0$), versus setting two numbers or expressions equal, like $1=2$. – anon Sep 16 '13 at 05:45
  • @anon Just like you can prove commutativity of addition from first principles, I think you should be able to prove $0^1=0$. If you can't, then leave it undefined. – Dan Christensen Sep 16 '13 at 05:51
  • Try proving "fsjkljf" exists or doesn't exist without first defining what that string of symbols means. Such a task doesn't make any sense. If we don't know what the words or expressions we use mean (i.e. their definitions) then we literally do not know what we're talking about! – anon Sep 16 '13 at 05:55
  • @anon At some point, you have to stop defining things, and start proving things. I think you can probably avoid specifying values for $0^0$ and $0^1$ and still have a workable definition of exponentiation. – Dan Christensen Sep 16 '13 at 06:08
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    Sure, but for $0^1$ and even $0^0$ that can be inconvenient, and there is no point in avoiding convenience all else equal. Funny you speak of "stop defining things," when in your question you define $x^n$ for $n=2,3,\cdots$ before you even define it for $n=1$ (you fail to give it any definition there, even though you would have to give $x^1$ some sort of meaning if you're going to "solve" for it), which is a strange order to do it in, because $1$ comes before $2,3,\dots$. You also don't seem to ask if $x^2=x\cdot x$ contradicts anything, even though that's the very first thing you defined. – anon Sep 16 '13 at 06:14
  • @anon You could start with $x^0=1$ for $x\ne 0$ if you like. Then $x^1=x$ for $x\ne 0$, etc. – Dan Christensen Sep 16 '13 at 06:21
  • possible duplicate of What values of $0^0$ would be consistent with the Laws of Exponents? – user91500 Dec 14 '13 at 17:35
  • Related: https://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1 – Henry Jan 08 '24 at 07:03

7 Answers7

9

In set theory and combinatorics, and for purposes of expansions in power series, $0^0$ is $1$.

It is $1$ because the number of factors multiplied is $0$. When you multiply no numbers, you get $1$, just as when you add no numbers at all you get $0$. Not multiplying by anything is the same as multiplying by $1$, just as adding no numbers at all is the same as adding $0$. Google the term "empty product" and read about it.

"Everybody knows" that $$ e^z = \sum_{n=0}^\infty \frac{z^n}{n!}. $$ But when $z=0$, then the first term is $\dfrac{0^0}{0!}$. But the first term must be $1$.

One the other hand $0^0$ is an indeterminate form in that if $f$ and $g$ approach $0$ as $x$ approaches something, then $f^g$ could approach any positive number at all or $0$ or $\infty$, depending on which functions $f$ and $g$ are. (But if $(f,g)$ approaches $(0,0)$ from within a sector bounded by two lines of positive slope, then the limit is $1$.)

Michael Hardy
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    Or if $f$ and $g$ are analytic with $f(x_0) = g(y_0) = 0$, then $\lim_{(x,y) \to (x_0,y_0)} f(x_0)^{g(y_0)}$ also exists. At least according to Wikipedia. – kahen Sep 16 '13 at 05:30
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    I've never cared for terminology like "$0^0$ is an indeterminate form". The function $(x,y) \mapsto x^y$ is discontinuous at (0,0), yes. But that doesn't really need to affect what we define $0^0$ to be. – Nate Eldredge Sep 16 '13 at 18:35
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Yes, for $x \ne 0$, $x^1 = x^2 / x$. Does that mean $0^1$ is undefined? No. We also know that for $x \ne 0$, $x^n = x^{n+1} / x$. So is $0^n$ undefined? No.

However, we must take care to define $0^1$ consistently. If you look at the function $x^y$ (or, if you don't know how to define exponents for real numbers, $e^{y \ln x}$) at the point $(0,1)$, you will see it behaves nicely.

But at $(0,0)$, it doesn't, so many people leave it undefined.

Henry Swanson
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  • I have $x^n$ defined above for $n\ge 2$. Only $n=0$ or $1$ are problematic. There doesn't seem to be any way to obtain values for $0^0$ or $0^1$ other than to simply define them. That is what I meant by them being "undefined." There may even be infinitely possible binary functions on N that are consistent with the usual Laws of Exponents (leaving out definitions of $0^0$ and $0^1$). – Dan Christensen Sep 16 '13 at 15:06
  • As I have defined exponentiation above, it can be shown to be unique except for possibly $0^0$ and $0^1$. In other words, if you where to make an infinite table of values for exponentiation as I have defined it, any values assigned to $0^0$ and $0^1$ would have no effect on the rest of the table. This doesn't mean you can assign arbitrary values. Values other than $0$ or $1$ for $0^0$, for example, would result in a contradiction. – Dan Christensen Sep 16 '13 at 16:56
  • Just playing around with a calculator, $|x|^y$ does seem to behave nicely around (0,1). So, there are probably only two realistic possibilities for the exponentiation function: one with $0^0=0$, the other with $0^0=1$. Can we obtain a contradiction from either $0^0=0$ or $0^0=1$? – Dan Christensen Sep 16 '13 at 18:10
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    Click the links in my post to see how x^y behaves. Also, I think I have a contradiction you'd like. We know $a^c b^c = (ab)^c$. Thus, $a^1 0^1 = 0^1$. So, we have some $x$ such that $a x = x$ for all $a$. Can you prove this $x$ must be zero? (this doesn't work for $0^0$, because anything non-zero to the zero is one, so we get $1x=x$, which does not force a value on $x$). – Henry Swanson Sep 16 '13 at 18:53
  • Say $a=2$. Then we must have $2\times 0^1=0^1$. Therefore $0^1=0$. Thanks! – Dan Christensen Sep 16 '13 at 22:00
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    You leave $0^0$ undefined. I define it by $0^0=1$ and have many good reasons for doing so. So, please, don't use “we”. – egreg Sep 17 '13 at 17:46
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    Their is no reason why all definition must behave nicely. Many define $0^0$ to be $1$. – Baby Dragon Sep 17 '13 at 17:53
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    I just use 'we' because that's the pronoun I've always seen for math/science writing. But if you insist... – Henry Swanson Sep 17 '13 at 19:59
  • I will point to the point that many people say that $0^0=1$ because in set theory $x^y$ is the cardinal of the set of functions from a set of cardinal $y$ to a set of cardinal $x$ and the set of functions from the empty set to the empty set has one element: the empty function. Nevertheless, the better is to defined it according to the best convention in the context we are working in. – Josué Tonelli-Cueto Sep 18 '13 at 16:17
  • Yeah, whenever it's from a discrete context, I've never seen it become anything but $1$. Maybe it's good to think of them as two different functions that happen to agree on the natural numbers? (Or just accept that $x^y$ is discontinuous at the origin, which seems like a better idea, the more I read about complex analysis and branch cuts and all that fun stuff) – Henry Swanson Sep 19 '13 at 19:52
2

$0^0$ can be defined as $1$ for the following reason:

  1. The alternating sum of binomial coefficients from the $n$-th row of Pascal's triangle is what you obtain by expanding $(1-1)^n$ using the binomial theorem, i.e., $0^n$. But the alternating sum of the entries of every row except the top row is $0$, since $0^k=0$, $\forall k > 1$. But the top row of Pascal's triangle contains a single $1$, so its alternating sum is $1$, which supports the notion that $(1-1)^0=0^0$, if it were defined, should be 1.
Anthony Peter
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    That's wrong. You should study the function $\displaystyle \lim_{(x,y) \to (0,0)}x^y$ instead. – user66733 Sep 16 '13 at 05:05
  • @some1.new4u If we simply view the graph we see that it approaches a value of one... – Anthony Peter Sep 16 '13 at 05:07
  • But you're looking at the wrong graph. The graph that you should look at is $f(x,y)=x^y$, not $f(x)=x^x$. – user66733 Sep 16 '13 at 05:07
  • http://www.wolframalpha.com/input/?i=x%5Ex – Anthony Peter Sep 16 '13 at 05:10
  • Again, you're studying $f(x)=x^x$. That's a single variable function. The real function is $f(x,y)=x^y$ which is a multi-variable function with two inputs, and this is the function you should look at because exponentiation is a binary operation, not a unary operation. – user66733 Sep 16 '13 at 05:14
  • @some1.new4u but for sake of simplicity to the original proposer, one can simple view it in this way – Anthony Peter Sep 16 '13 at 05:17
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    the first point is the worst possible argument you can give for defining $0^0$. The two occurrences of $0$ are in fundamentally different roles, and there is no reason at all to consider the equality of these values as anything but a accidental situation. – Marc van Leeuwen Sep 16 '13 at 05:17
  • No, you CAN'T. It's like you project a photo on a line and then you say that we have lost no details about it. – user66733 Sep 16 '13 at 05:18
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You can't conclude that $x^1$ is undefined for $x \neq 0$ just because division by $0$ is undefined. The reason that $x^0$ is undefined for $x=0$ is rather more subtle. Moreover, I'm failing to see why we can't conclude that $x^n$ is undefined for $x=0$ by your logic? Can't we say $\displaystyle x^n = \frac{x^{n+1}}{x}$ for $x \neq 0$ ?

To know that why $0^0$ is undefined you can look at the behavior of the function$x^y$ near $(0,0)$ to see that $x^y$ has serious complications around $(0,0)$. That's why $x^y$ is left undefined for $(x,y)=(0,0)$, because we can't define a particular value for it. It's like that you define a binary operation $x*y$ by $\displaystyle x*y=\frac{\sin(x)}{y}$ and you insist on defining $0*0$.

See also here on wikipedia. Cauchy was the first one who objected to $0^0=1$.

user66733
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  • Cauchy did not write that. He simply warned about limits of $f(x)^g(x)$ as $f(x),g(x)$ converge to $0$. But this has no impact on the value of $0^0$ itself. – Mark Feb 13 '17 at 21:48
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$$x^{n-m}=\underbrace{x^{-1}\ldots\cdot x^{-1}\cdot x^{-1}}_\text{m times} \cdot 1\cdot \underbrace{x\ldots\cdot x\cdot x}_\text{n times}$$

Nathaniel Bubis
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0

$0^1$ is surely defined, and it's zero (0)

as for $0^0$, it is undefined, becuase if x=0 we'll get:

$0^{x-1} = 0^{-1} = \frac{1}{0}$ which is really undefined expression.

BUT, there is a convention I read about over the net that $0^0 = 1$, without further reasoning

DanielY
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  • I had been assuming $0^1=0$ as well, but now I am wondering if it is justified. – Dan Christensen Sep 16 '13 at 05:07
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    How the hell does $0^{x-1}$ relate to undefinedness of $0^0$? $0^0$ is undefined in analytical context is because the function $x^y$ has no limit as $(x,y)\to (0,0)$. – tomasz Nov 29 '13 at 04:13
-1

In the natural numbers, exponents greater than 1 are naturally defined as follows:

$x^2=xx$

$x^3=xxx=x^2x$

$x^4=xxxx=x^3x$

$x^5=xxxxx=x^4x$

and so on.

More formally, exponentiation can be defined as a binary function on $N$ such that

(1) $\forall x\in N:x^2=xx$

(2) $\forall x,y\in N:x^{y+1}=x^y x$

It can then be shown that

(1) $\forall x\in N:(x\ne 0 \implies x^1=x)$

(2) $\forall x\in N:(x\ne 0 \implies x^0=1)$

(3) $\forall x,y,z\in N:(x^{y+z}=x^y x^z) \implies 0^1=0 \land (0^0=0 \lor 0^0=1)$

Formal proof in DC Proof format (126 lines) at http://dcproof.com/Ambiguity-of-0-to-the-0.htm

FOLLOW-UP

Down vote quite justified! My thinking on this topic has evolved considerably in the weeks since I posted the above, although I still have exponentiation being defined as a binary function on $N$ such that

(1) $\forall x\in N: (x\ne 0 \implies x^0=1)$, or equivalently $\forall x\in N:x^2=xx$

(2) $\forall x,y\in N:x^{y+1}=x^y x$

$0^0$ is assumed to be a natural number, but no is actual value is assigned to it. Contrary to my previous claim, however, it can be shown that we must have $0^1=0.$

See details and formal proofs at "Oh, the Ambiguity!" at my math blog.

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Dan Christensen
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