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I have been a witness to many a discussion about numbers to the power of zero, but I have never really been sold on any claims or explanations. This is a three part question, the parts are as follows...


1) Why does $n^{0}=1$ when $n\neq 0$? How does that get defined?

2) What is $0^{0}$? Is it undefined? If so, why does it not equal 1?

3) What is the equation that defines exponents? I can easily write a small program to do it (see below), but what about in equation format?


I just want a little discussion about numbers to the power of zero, for some clarification.


Code for Exponents: (pseudo-code/Ruby)

def int find_exp (int x, int n){     int total = 1;     n.times{total*=x}     return total; } 
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    Umm, I mean $n$ does not equal $0$, programming habit -_- And I am not good in TeX2012-11-11
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    Actually, $0^0=1$, too.2012-11-11
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    Really? I have always been told that it is undefined...2012-11-11
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    Think of $n^0 = 1$ as the base step of a recursive definition for $n^x$, much the same way as you have defined "total=1" in your pseudocode.2012-11-11
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    It is defined, but it is also not continuous. That means it is useless to define it that way in computer programs with real variables, for example, because floating point zero is just a number near zero2012-11-11
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    @ShaunAult Ah, true. But what if I set `n` to `n-1`, and set `total` to `x`? Wouldn't that throw your claim out the window?2012-11-11
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    @ThomasAndrews $0^0$ is an indeterminate form. As the limit of $x^x$ from the right, yes it limits to $1$. But as the limit of $0^x$, it tends toward 0.2012-11-11
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    Being an indeterminate form does not imply that something is undefined. $0^0$ is defined and equal to $1$. @ShaunAult2012-11-11
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    @ThomasAndrews Is there a mathematical way to prove so?2012-11-11
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    Sure, you can define your code that way. On the other hand, there are also many other good reasons to define $n^0 = 1$. Notably, it is consistent with $1 = n^p / n^p = n^{p-p} = n^0$.2012-11-11
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    No, by definition, being indeterminate means that this expression only makes sense when one takes a limit. By themselves, expressions such as $0/0$, $0 \cdot \infty$, and $0^0$ cannot consistently be defined as a particular real number.2012-11-11
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    @ShaunAult Then what kind of number is it? o.O2012-11-11
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    @CodeAdmiral: NaN. Not a real number, but a description of a potential limit. $\lim_{x \to 0^+} x^x$ and $\lim_{x \to 0^+} 0^x$ both have the indeterminate form $0^0$, and yet the former evaluates to $1$ while the latter to $0$.2012-11-11
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    @ShaunAult so which is commonly accepted? And what in the heck does "indeterminate form" mean? I am a highschool freshman ^^2012-11-11
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    @ShaunAult: That's basically a misunderstanding. It just mean that you cannot take the limit for $x\to 0$ by blindly substituting $0$ for $x$ into the expression. But if you could always do that, then there wouldn't be any point in having a concept of limit in the first place. The _arithmetic_ meaning of $0^0$ is or ought to be independent of any considerations of limits.2012-11-11
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    It's related to th concepts of limits (pre-calc and calculus material) If you're interested in learning more: http://en.wikipedia.org/wiki/Indeterminate_form2012-11-11
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    @ShaunAult Oh, trust me, I can tread water in calc. In church today, I was working the derivative of a circle equation.2012-11-11
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    @ShaunAult, the value of the expression $0^0$ is not necessarily the limit of a *different* expression $x^y$ as $x$ and $y$ tend to zero. The notion of "NaN" is a red herring in any case; it is a value defined by the IEEE 754 floating-point standard, not a concept in mathematics.2012-11-11
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    Duplicate of http://math.stackexchange.com/questions/11150/2012-11-11
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    @ThomasAndres Regarding $0^0$, two conventions are at odds. One is that $x^0$ should be $1$ for all $x$. The other is that $0^x$ should be $0$ for all $x$. Who wins? $0^0$ is just as much undefined as $\frac{0}{0}$ or $0\cdot\infty$. In fact, applying $\ln$ to $0^0$ and trying to apply logarithmic identities yields $0\cdot-\infty$.2012-11-16
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    This is a duplicate of two questions: $x^0$ $x \neq 0$: http://math.stackexchange.com/questions/9703/how-do-i-explain-2-to-the-power-of-zero-equals-1-to-a-child/564272#564272 $0^0$ not defined: http://math.stackexchange.com/questions/11150/2013-11-16
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    @ThomasAndrews $0^0\neq 1$. Because since $0=1-1$, we have $0^0=0^{1-1}=0\div 0=1$? But $0\div 0$ is strictly undefined! What you mean to say is... $$\Large\lim_{n\to 0}n^n=1.$$2018-10-07
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    See also [Why is $\ \large 2^0 = 1\ $?](https://math.stackexchange.com/q/6832/242)2018-10-07
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    Don't tell me what I mean to say, @user477343. There are elementary reasons $0^0=1.$ There are also convenience reasons to define it that way. Your argument is pointless, because $0^{2-1}=0/0,$ too. So is $0^1$ undefined?2018-10-07
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    @ThomasAndrews but... this means I have been taught wrong by every one of my math teachers thus far...2018-10-07
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    Then you have been taught wrong. $0^0$ is an indeterminate form, which is the equivalent of saying the function $f(x,y)=x^y$ cannot be made continuous at $(0,0),$ and some teachers confuse "indeterminate form" from "undefined." And people doing *numeric* calculations (like computer programming) will leave $0^0$ undefined because the inputs are actually approximations, and hence, because of the lack of continuity, it wouldn't make sense to define $0^0.$ But mathematically, $0^0=1.$ @user4773432018-10-08
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    @ThomasAndrews okay... thank you for that. I take back what I said :)2018-10-08
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    In lambda calculus, the exponentiation operator is remarkably trivial, and you get $0^0=1.$ In set theory, too. The only problem with the definition is a teaching problem - the notion that an indeterminate form is not the same as "undefined." But you get even more teaching problems when you get to complex exponentiation - exponentiation is tricky. @user4773432018-10-08
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    @ShaunAult: Your misconception is a common one. Please see [this post](https://math.stackexchange.com/a/2184974/21820) explaining the difference between $0^0$ and "$0^0$". One is equal to $1$ under any reasonable definition, while the other is a symbolic expression and not a number. This is completely in line with the comments by Thomas Andrews and Henning Makholm.2018-10-08

9 Answers 9

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It's basically just a matter of what you define the notation to mean. You can define things to mean whatever you want -- except that if you choose a definition that leads to different results than everyone else's definitions give, then you're responsible for any confusion brought about by your using a familiar notation to mean something nonstandard.

Most commonly we define $x^0$ to mean $1$ for any $x$. What you find in discussions elsewhere are argument that this is a useful definition, not arguments that it is correct. (Definitions are correct because we choose them, not for any other reason. That's why they are definitions).

Some people choose (for certain purposes) to explicitly refrain from defining $0^0$ to mean anything. That choice is (supposedly) useful because then the map $x,y\mapsto x^y$ is continuous in the entire subset of $\mathbb R\times\mathbb R$ it is defined on. But it's an equally valid choice to define $0^0$ to mean $1$ and then just remember that $x,y\mapsto x^y$ is not continuous at $(0,0)$.

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    I agree with most of your answer, *except* the definition we choose is not $x^0$ in this case, it is another definition which implies the value of $x^0$. See http://math.stackexchange.com/a/11211/23802012-11-12
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    But when considering the lone expression $0^0$ without the context of the function $x^y$, I've always seen the choice to refrain from defining $0^0$ to be firstly about two competing basics of arithmetic: that usually, $x^0=1$, and usually $0^x=0$. These two things are at odds with $0^0$.2012-11-16
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    @alex.jordan The two statements $P$: "$x^0 = 1$ for all $x$", and $Q$: "$0^x = 0$ for all $x$", are at odds with each other. However, statements can't be "at odds" with a number, so your last sentence makes no sense. Also, if $P$ and $Q$ contradict each other, all that that tells you is that at least one of them is false (why is it necessary to explain this on a math website?). And if you plug in $x=-1$ then it's pretty obvious which one that is.2017-02-24
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    @Mark. First, you just replied to a comment from 2012. Did you realize that? Second, I don't understand what you are saying. Your statement $Q$ is just plain false. $0^{-1}$ is not $0$, for example. Anyway, the upvotes on my comment indicate you may have misread or misunderstood something I said there (five years ago). People adopt the mathematical intuition that $0$ to a (positive) power should be $0$ and that a number to the $0$ power should be $1$. When they see $0^0$, these two competing intuitions clash (even though one of them shouldn't apply in the first place.)2017-02-24
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    @alex.jordan Sorry to reply to a very old comment. Your comment "These two things are at odds with $0^0$" says absolutely nothing about $0^0$ unless you make a basic error in logic. The only way the comment could be relevant to $0^0$ is if you assume that when two statements $P,Q$ are at odds, then they must be both wrong (even in situations where it is very obvious that $Q$ is wrong).2017-02-25
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    @Mark: It looks like you're simply trying to criticize, in a rather roundabout and combative way, the exact choice of _words_ Alex made more than 4 years ago, rather than the actual point he was making. Is life really long enough for that? What is your goal in this discussion?2017-02-25
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    My apologies for the time-frame and tone in my comment. I didn't intend to criticize the words, only the logic behind them.2017-02-25
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It is for various reasons convenient to define $0^0$ as being equal to $1$. For one thing, consider the Binomial Theorem, or power series. It is useful to be able to write $$(1+x)^n =\sum_{k=0}^n \binom{n}{k}x^k,$$ or $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}.$$ In each of these equations, if we want the expression on the right to give the correct answer when $x=0$, we need to set $0^0=1$.

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    Ah, that makes sense, thanks ^v^2012-11-11
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    Must a theorem be true? Or is it merely convenient for a theorem to be true?2017-02-17
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To 1): We define the exponents of a nonzero integer $a$ such that they satisfy the relation $a^ba^c=a^{b+c}$ for ay integers $b,c$, with $a^1=a$. In order for exponents to be well defined, we thus need $a^0=1$.

To 2): It depends on how you define it. If you define it via the limits $\lim_{x\rightarrow 0} x^0$ or $\lim_{x\rightarrow 0} x^x$, then $0^0=1$. If you define it as $\lim_{x\rightarrow 0} 0^x$, then $0^0=0$.

To 3): Exponents are defined simply by $a^n=\underbrace{a\cdot a\cdot \,...\, \cdot a}_{n}$.

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    For 1), so some mathematician once said that $a^{0} = 1$? For 2), Which is then correct? Or is it both? For 3), I can kinda wrap my head around that, but I have yet to come across that particular way of denoting a range...2012-11-11
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    No, we never define $0^0=0$. Just because it is a limit doesn't mean we define it that way. It is either $1$ or undefined.2012-11-11
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    1) is needed to make the notion of exponent well defined. For 2), there is no "correct" one. It is a matter of definition.2012-11-11
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    I thought the other way to think of 0 to the power 0 is like 0 / 0, and therefore is indeterminate2017-04-09
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From definition of division of powers with the same base we have that $$\frac{a^n}{a^m}=a^{n-m}$$ Assuming that $n=m$ from left side we get $$\frac{a^n}{a^n}=1$$ and from right side we get $$\frac{a^n}{a^n}=a^{n-n}=a^0$$ comparing the last two equations we have that $$a^0=1$$

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    Now let $a=0$. Does it hold that $0^0=1$?2018-10-07
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The invention of numbers was one of the biggest breakthroughs in the history of math. It marked the realization that this sack of pebbles $$\{ \blacktriangle\;\blacktriangle\;\blacktriangle\;\blacktriangle\;\blacktriangle \}$$ this string of knots $$-\bullet-\bullet-\bullet-\bullet-\bullet-$$ and this bone full of tally marks $$/\,/\,/\,/\,/$$ are all incarnations of a single thing, the abstract quantity five. That leap of abstraction has become so prosaic for us that it almost feels weird to do arithmetic by actually counting things. In some cases, though, it can be illuminating to go back to the basics—back to the days when we didn't have numbers, and we did all our arithmetic by counting things. Your question is one of those cases.

In what follows, I'll use a capital letter like $X$ to stand for a finite set of things, like a herd of goats or a pile of beads, and I'll use the symbol $|X|$ to stand for the number of things in the set.


Exponentiation is a tricky operation, as you've clearly noticed, so let's warm up with something simpler. If you have two piles of beads, $A$ and $B$, the simplest thing you can do with them is shove them together to make a bigger pile, which is often written $A \sqcup B$. You should easily be able to convince yourself that, on the level of numbers, $|A \sqcup B| = |A| + |B|$. In other words, the concrete operation of shoving two piles together corresponds to the abstract operation of adding two numbers. Addition of whole numbers is often defined in this way.


Here's a slightly tougher warm-up. If you have a bunch of shirts, $H$, and a bunch of skirts, $K$, you might wonder how many different outfits you can make by pairing a shirt with a skirt. The set of outfits is usually written $H \times K$. You should be able to convince yourself that $|H \times K| = |H| \cdot |K|$. In other words, the concrete operation of counting pairs corresponds to the abstract operation of multiplication. Multiplication of whole numbers is often defined in this way.


Now that we're warmed up, suppose you have a set of paints, $C$, and a bag of beads, $X$. You might wonder how many different ways there are to color each bead with one of the paints. The set of ways to color the beads is usually written $C^X$. If you try a few examples, you'll see that $|C^X| = |C|^{|X|}$. Exponentiation of whole numbers is often defined this way.

Finally, we can get to your question. Suppose you have a bunch of paints, but the bag of beads is empty. Is it possible for you to paint all the beads? Sure: you just don't do anything! In fact, not doing anything is the only way to paint all the beads in the bag, since there are no beads. So, when the set $C$ has a bunch of paints, but the bag $X$ is empty, $|C^X| = 1$. If you define exponentiation by counting colorings, that means $|C|^0 = 1$ for any positive number $|C|$.

To make matters worse, suppose you have no paints and no beads. Happily, you can still paint all the beads: once again, you just don't do anything. Like before, not doing anything is the only way to paint all the beads, so $|C^X| = 1$ even when both $C$ and $X$ are empty. If you define exponentiation by counting colorings, that means $0^0 = 1$.

On the other hand, suppose you don't have any paints, but you do have some beads. In this case, you can't paint all the beads, because you have no paints! There are just no ways to paint all the beads. In other words, when $C$ is empty but $X$ is not, $|C^X| = 0$. If you define exponentiation by counting colorings, that means $0^{|X|} = 0$ for any positive number $|X|$, just like you'd expect.


Here's a bonus. André Nicolas argued that $0^0$ should be $1$ in order to make the binomial theorem true. Even those weird-looking numbers $\binom{n}{k}$ can be defined using finite sets: if you have $N$ toys and $K$ kids, $\binom{N}{K}$ is the set of ways you can pick out enough toys to have one for each kid. (Note that you don't give each toy to a particular kid: you just want the numbers of kids and toys to be the same.) If you get out your set of paints $C$ and another set of paints $D$ and start painting various numbers of kids and handing out toys based on how many colors of kids there are, you should somehow be able to convince yourself that the binomial theorem is true, even when $C$ doesn't have any paints in it. That's why André Nicolas came up with the same rules for zero exponentials as we just did.

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    This is the best answer, in my opinion. $(+1)$2018-10-07
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(1) For intuition, if $k\ge0$ is an integer, take $x^k$ to mean "$1$ multiplied $k$ times by $x$", and $x^{-k}$ with $(x\neq 0)$ to mean "$1$ divided $k$ times by $x$." For integers $n\geq 0$, we may define $n!$ as the number of distinct ways to line up $n$ distinct objects--the only way to line up $0$ objects is to not line up any objects.

(2) We often define $0^0$ to be $1$, which accords with the intuitive definition above--if we multiply $1$ by $0$ not at all, then we still just have $1$. Now, sometimes we will not define $0^0$ at all, which I'll discuss further below.

(3) We can extend integer powers to rational powers as follows: We say $y=x^{\frac1m}$ for some integer $m>0$ if $x=y^m$. If $m$ is odd, there will be a unique solution $y$ to the equation $x=y^m$. If $m$ is even and $x<0$, there will be no real solution $y$; if $m$ is even and $x\ge0$, then there is at least one real solution $y$, and we will take $x^{\frac1m}$ to be the nonnegative solution. At that point, given integers $k,m$ with $m>0$ and $\frac k m$ in lowest terms, we define $x^{\frac k m}:=\left(x^{\frac1m}\right)^k$ for such $x$ as this is possible. Finally, for such $x$ that $x^{q}$ is defined for all rational $q,$ we can use continuity arguments to define $x^y$ for all real $y.$


In the manner described above, given real numbers $x$ and $y,$ we have defined a real number $x^y$ for all real $y$ when $x>0,$ for all nonnegative $y$ when $x=0,$ and for all rational $y$ with odd denominators when $x<0.$ Unfortunately, continuity arguments won't work to extend to any more $y$ when $x\le0,$ because the function behaves too erratically to extend continuously in such cases. In fact, the erratic behavior of the function $f(x,y)=x^y$ means that $f$ isn't even continuous at the origin! For example, we can approach the origin along the line $y=x$ in the first quadrant (that is, when $x$ and $y$ are positive), and find that $x^y$ approaches $1,$ which is what we would expect. However, if we try to approach it along the positive $y$ axis (that is, when $x=0$ and $y>0$), then we find that $x^y$ approaches $0,$ which is not at all what we want! This means that not only is $f(x,y)$ discontinuous at the origin, but that there is no way that we can define $f(0,0)$ to make it continuous there! Similarly, $f(x,y)$ is badly discontinuous when $x<0.$ For this reason, when trying to define a continuous real-valued exponential function, one cannot define $0^0$ at all, nor define $x^y$ when $x<0.$ This continuous function $g(x,y)=x^y$ is defined for all real $y$ when $x>0,$ defined for all positive $y$ when $x=0,$ and undefined otherwise. However, this doesn't alter the truth of $0^0=1,$ merely the domain of continuous definition.

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    @downvoter: I know it's probably futile to ask, but would you care to explain what you find unsatisfactory about my answer, so I can address any issues?2017-02-14
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    (1) is correct, but (2) is not, because to compute a function-value with limits, you need that function to be continuous at that point. It is self-contradictory to compute $0^0$ with a limit while also showing that $x^y$ is not continuous at the origin. It is true that some people use limits or continuity as a reason not to define $0^0$ but that does not make it a valid reason.2017-02-17
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    Despite my comment on (2), your answer is much better than a lot of other answers since you made it clear that one doesn't have to use limits. All I'm trying to emphasize is that using limits is wrong here, regardless of whether the conclusion is right or not.2017-02-17
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    @Mark: I wasn't attempting to compute the function value with limits. Rather, I was suggesting that one might attempt to define it by continuous extension of a function, and at the same time demonstrate briefly why that was problematic. Unfortunately, I phrased it very poorly, so it came out seeming as you describe. I have edited my answer. Hopefully it is more sensible now. Let me know! Thank you for your feedback.2017-02-22
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    All mathematical arguments against defining $0^0$ use invalid steps (= steps that are widely rejected in any other context). It is true what you wrote: "For this reason, many elect not to define $0^0$ at all". But this reason is invalid, i.e. widely rejected in any other context. If this reason were valid, it would also undefine floor($0$), or more generally, any $f(p)$ for any function $f$ with a discontinuity at $p$. It is misleading to list a reason without mentioning that this reason is invalid (i.e. would not be accepted in other contexts).2017-02-22
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    @Mark: I think I see what you mean, but I wouldn't say that the reason is invalid, just context-dependent. For example, if the context is defining continuous functions through real-analytic means, then it makes perfect sense to leave $\langle 0,0\rangle$ out of the domain of the real-valued exponential function. By the same token, in the context of real-valued functions on the reals, it wouldn't make sense to say that $i$ is a solution to the equation $x^2=-1,$ while it would make perfect sense in a more general context. I will alter my answer slightly to address context-dependence.2017-02-22
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    There is no proof that there is something wrong about $0^0 = 1$. Saying that a reason is context-dependent is the same as saying that it's not a proof.2017-02-22
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    The way to judge if the reasoning is valid or not is to apply the same argument in another situation. Suppose someone asks: does the fact that floor($x$) is not continuous at $x=0$ make floor($0$) undefined? With your reasoning, the answer would be: it's context-dependent, it is $0$ in some contexts, but undefined in the context where we insist that function-values must be computed with limits/continuity. I reject that because there is no rule that says that function-values must be computed with limits/continuity. It is meaningless to talk about "contexts" where this is a rule.2017-02-22
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    @Mark: Let's [move this discussion to chat](http://chat.stackexchange.com/rooms/54103/re-context-dependence), as this is growing too lengthy for a comment thread.2017-02-23
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To define $x^0$, we just cannot use the definition of repeated factors in multiplication. You have to understand how the laws of exponentiation work. We can define $x^0$ to be: $$x^0 = x^{n - n} = \frac{x^n}{x^n}.$$ Now, let us assume that $x^n = a$. It would then be simplified as $$\frac{x^n}{x^n} = \frac{a}{a} = 1.$$ So that's why $x^0 = 1$ for any number $x$.

Now, you were asking what does $0^0$ mean. Well, let us the example above: $$0^0 = 0^{n - n} = \frac{0}{0}.$$ Here is where it gets confusing. It is more likely to say that $\frac{0}{0}$ equals either $0$ or $1$, but it turns out that $\frac{0}{0}$ has infinitely many solutions. Therefore, it is indeterminate. Because we mathematicians want to define it as some exact value, which is not possible because there are many values, we just say that is undefined.

NOTE: $0^0$ still follows the rule of $x^0 = 1$. So it is correct to say that $x^0 = 1$ for ANY value of $x$.

I hope this clarify all your doubts.

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    Oh sorry, you probably don't know why 0/0 has infinitely many solutions. Well here's the thing: 0/0=x, where is x is allegadly undefined/unknown. Undo multiplication and you obtain the equivalent equation 0x=0. Every number (including complex numbers) make this equation true. So we can assume that 0/0 equal all numbers at the same time. This kind of values are called "indeterminate", meaning that there cannot be a specific value assigned for it. Call it 0^0=0/0 by the example shown above, 0^0 is just as indeterminate as 0/0.2013-11-16
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Another approach...

It can be shown that there exists infinitely many, what I call "exponent-like functions" that can be defined on the set of natural numbers $N$. By an exponent-like function $f$ on $N$, I mean $f$ such that:

  1. $f: N\times N\to N$

  2. $f(x,0)=1$ for $x\ne 0$

  3. $f(x,y+1)=f(x,y)\cdot x$

For all $x_0\in N$, there exists a unique exponent-like function $f$ such that $f(0,0)=x_0$.

It can be shown that, except for the value of $f(0,0)$, all exponent-like functions, as defined here, are identical.

From each such exponent-like function, we can derive the usual Laws of Exponents for non-zero bases corresponding to:

  1. $x^{y+z}=x^y\cdot x^z$

  2. $(x^y)^z= x^{y\cdot z}$

  3. $(x\cdot y)^z=x^z\cdot y^z$

If we define $0^0=1$, then these Laws of Exponents are true for all bases, including $0$. It might then be argued that we must have $0^0=1$. But the same is true for $0^0=0$ (but no other values), and we are no further ahead! If we are to look at exponentiation on $N$ as simply repeated multiplication, then $0^0$ is inherently ambiguous. We can formally define exponentiation on $N$ as follows:

  1. $\forall x,y\in N: x^y\in N$
  2. $\forall x\in N: (x\ne 0 \implies x^0 = 1)$
  3. $\forall x,y\in N: x^{y+1}=x^y\cdot x$

Here, $0^0$ is a natural number, but no value has been assigned to it.

See formal proofs, etc. in "Oh, the ambiguity!" at my math blog.