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Alright, so the idea of an exponent, $x$, is that you are multiplying its base by itself $x$ number of times. With base $5$ and $x=3$, we have that $5^3$ = $5 \cdot 5 \cdot 5$

I understand that the logarithm with base $a$ of $x = c$,

tells us that

$$a^c = x$$

and for $c =$ positive; values for $x$ are greater than $1$,
and for $c =$ negative; values for $x$ are less than $1$,
and for $c = 0$, values for $x$ are...$1$.

So in short, I understand how, by means of observation of the graph of $f(x) = \log x$, we can see that $f(1) = 0$, BUT, I see no other way to understand why $x^0 = 1$, apart from the graph and everything around that very point.

I honestly cannot get my head around the idea, "$5$ times itself $0$ times... is one".

Is it that there is no fundamental answer for this but that we simply know by the graph? Or can I truly understand $x^0 = 1$ on its own?

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It's a definition, based mainly, perhaps, on some facts related to limits. It also makes a lot of sense within the context of series, in particular geometric ones. By the way, to avoid problems we usually require $\,x\neq 0\,$, otherwise things get a little foggy... – DonAntonio Nov 15 '12 at 23:23
The possibility of $x=0$ needs to be treated as a special case. – Ben Crowell Nov 16 '12 at 5:48
In all honestly, "5 times itself 0 times... is one" is not indisputably true, "5 to the power of 0 is one" is true on the other hand. The exponential function is conveniently defined as "a number times itself" for integer operands. Note: "5 times itself 0 times", how do you write that with the multiplication operand? You don't, and that's the problem :) – flindeberg Nov 16 '12 at 11:50
@BenCrowell: No, $x=0$ does not need special consideration. An empty product is an empty product, it cannot depend on what would have been in the product if it had not been empty. Anyone haveing qualms about indefinite forms should start to realize that $x^y$ for positive real $x$ just means $\exp(y\ln x)$, and the qualms will go away without having to undefine $0^0$. – Marc van Leeuwen Nov 16 '12 at 13:21
@FlybyNight: That is about the worst argument for not defining $0^0$ that I've seen. "The problem is similar to that with division by zero". No it isn't at all, there are plenty equations that perfectly well fix the value $0^0=1$ (like $X^0=1$ for the polynomial $X$; now substitute $X=0$); just the equation $0^1=0\times0^0$ does not fix it, so what? Also "$0$ to any positive power is $0$, so $0$ to the power $0$ should be $0$". Huh? Swallowing $n$ doses of cyanide is lethal for $n>0$, so it should be lethal for $n=0$. – Marc van Leeuwen Nov 17 '12 at 17:25

18 Answers 18

up vote 53 down vote accepted

$x^{n+1}=x\cdot x^n$ right?


$x^1=x \cdot x^0$ but $x=x^1$ so for that to hold true, $x^0$ must be $1$.


$\large x^{-n} = \frac{1}{x^n}$.

So $\large x^n \cdot x^{-n} = x^n \frac{1}{x^n} = 1$.

But $\large x^n \cdot x^{-n} = x^{n+(-n)} = x^0$, so once more, $x^0=1$.

There are really many reasons for that to hold, and all of them are just a consequence of some agreements we've made previously.

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Alright, appreciated. – khaverim Nov 15 '12 at 23:39
You need to assume that $x \neq 0.$ The expression $0^0$ is not well defined. – Fly by Night Nov 16 '12 at 20:01
It is implicitly assumed when I devide sides by $x=x^1$ since one cannot devide by 0. – Golob Nov 17 '12 at 10:44

Don't think of $x^n$ as $\overbrace{x\cdot x\cdots x}^{n\text{ copies}}$. Instead think of $x^n$ as representing the result of starting with $1$ and applying "${}\cdot x$" $n$ times: $1\overbrace{{}\cdot x\cdot x\cdot{}\cdots{}\cdot x}^{n\text{ copies}}$.

This also makes negative exponents a little nicer for beginners. $x^{-n}$ is just applying the opposite of ${}\cdot x$, which is ${}\div x$, $n$ times. $x^{-n}=1\overbrace{{}\div x\div x\div\cdots\,\div x}^{n\text{ copies}}$.

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This is better than "backward induction" from the cases $n=\ldots,2,1$, and it works painlessly if $x=0$. – Marc van Leeuwen Nov 16 '12 at 13:32

There are many answers about why $x^0 = 1$ for general $x$ so I'd like to address a different issue here, the way you think about exponentiation, which seems to be troubling you.

The definition you use for exponentiation is holds true for integers, and rationals if you define what $x^{1/n}$ means, but what about the irrationals? How can you times something by itself $\sqrt{2}$ times? The answer is - you can't! You need to come up with some sensible way to define exponentiation rigourously.

We define $$e^x = \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$$ which is absolutely convergent for any complex $x$, and only depends on us defining what $x^n$ means for the natural numbers and $0$. Considering the function restricted to the reals we can use this definition to show that the function is continuous, differentiable, has an inverse $\ln{x}$ and satisfies all the properties that we are used to with exponentiation. To then define, $a^x$ for $a \not= e$ we say $a^x = e^{a\ln(x)}$. This obeys all the properties that we want it to for rational $x$ and has the added bonus of being well defined on the irrationals too.

In fact, the definition of $e^x$ can even be seen as a motivation to define $x^0$ as $1$. $x^n$ for any natural $n$ is defined intuitively, but choosing any value for $x^0$ which is not $1$ will mean that this definition of $e^x$ will not longer obey the rules we have come to know and love e.g. $e^{a+b} = e^{a}e^{b}$. When you look at it like this, you really have no choice!

I hope you find this useful!

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A great perspective, although I'm slow to digest it all. We just reviewed that definition of $e^x$ in Calc II – khaverim Nov 15 '12 at 23:59
No problem, showing all the consequences of the definition that I stated would take a lecture or two and then some thought afterwards. I just wanted to convey the idea of how to think of exponentiation in a more formal, rigorous way, and the idea that our exponential function only works if $x^0 = 1$ just sort of fell out whilst doing it - so that was a nice surprise! – Tom Oldfield Nov 16 '12 at 0:04
To prevent bumping the post, I refrain from an edit, but there is an error here: $$a^x = e^{x\ln a}$$ and not the other way around. – AlexR May 8 '15 at 15:59

Well $5^{x} = 5^{0 + x} = 5^{0} 5^{x}$. Dividing by $5^{x}$ yields $1 = 5^{0}$. If you define at all, I don't think you will have a better choice.

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$\dfrac{a^b}{a^c} = a^{b-c}$ for $a \gt 0$. Now let $b=c$.

It can get more complicated for $a \lt 0$ and $b$ and $c$ non-integers, while for $a=0$ it is just a convenient definition.

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These are all good answers. Here's another way to think about it via the empty product. It may be the best way to see it "on its own". It also provides a nice way of thinking about other seemingly paradoxical operations involving zero like $0^0$ and $0!$

The empty product basically says that the product of no factors is 1. It is necessary if we want to recursively define a product of a set of commutative elements.

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Ah, great stuff. thanks! – khaverim Nov 15 '12 at 23:54

If you want to keep "the idea of an exponent, $x$, is that you are multiplying its base by itself $x$ number of times" then a simple way to think about is as follows:

$x^0 = 1 \times x^0 = 1x^0 = 1 \overbrace{\phantom{\small{x\text{ multiplied by itself 0 times}}}}^{x\text{ multiplied by itself 0 times}} = 1$

This works because $1$ is identity of multiplication ($\times$).

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I don't think you can "truly understand it on its own".

"5 times itself 0 times..." is just too naïve a way to think about it. It's true that we initially define powers by iterated multiplication, and in this context it is obvious that $x^{n+m} = x^n\times x^m$. Given this, there is an obvious generalisation for rational exponents: define $x^{\frac{p}{q}}$ to be the $q^{\mathrm{th}}$ root of $x^p$. The final step is also obvious: demand that $x^y$ be continuous in $y$.

I don't know a 'more fundamental' way to understand this.

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Although, I would interpret "5 times itself 0 times" as the empty product, which is defined to be 1. – Zach Langley Nov 15 '12 at 23:31
@ZachLangley: I'm not sure that helps. You've used the words 'interpret' and 'defined'... – Rhys Nov 15 '12 at 23:33
Your answer was quite helpful – khaverim Nov 15 '12 at 23:44
@Rhys It doesn't help. I was just suggesting that there is nothing wrong with thinking about $5^0$ as $5$ times itself $0$ times. – Zach Langley Nov 16 '12 at 3:42

What is $5$ added to itself $0$ times? The answer is $0$. Why? $0$ is the nothing of addition. It works the same way with $1$. You can change $5^0$ to $(5^0)\times1$. That is the thought behind it. Another way to look at it is a a nice and sleek definition I like for exponents. $x^1=x$ and $x^m\times x^n$=$x^{m+n}$. That means $x^n+x^0=x^{n+0}=x^n$. This goes back to the idea of $1$ being the identity of multiplication.

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I think it'd be a good idea, before you even answer questions, that you first learn how to properly type mathematics in this site with LaTeX. In the FAQ are some directions and it's pretty easy. – DonAntonio Nov 15 '12 at 23:36
I changed your formatting with LaTeX to make it a bit more readable, you may want to look over what I did to see how to make things show up more clearly in the future. Also, I think the idea of $1$ being the "nothing" of multiplication is misleading, since the nothing of multiplication could just as well mean zero. I think a better word is "identity" since that means that it "acts trivially" or "acts by not changing what it acts on" in a more general sense. – Tom Oldfield Nov 16 '12 at 0:10


$$\frac{x^m}{x^n} = x^{m-n}$$

thus if m=3 and n=3,

$$\frac{x^3}{x^3} = x^{3-3} = x^0$$


Anything devided by its self is 1, why?

$$\frac{10}{10} = 1,$$

because $\frac{y}{y} = y\left(\frac{1}{y}\right) = \left(\frac{y}{1}\right)\times \left(\frac{1}{y}\right)$ which can be displayed as

$$ \frac{y}{1} \times \frac{1}{y} = \frac{1}{1} = 1$$

here the $y$ values cancel out to become 1


so if anything devided by it's self is 1 and $x^0 = x^{m-n} = \frac{x^m}{x^n}$ then $X^0$ must = 1, as its simply just a way of saying that value devided by itself =?

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I see that this question already has seventeen answers, but I'm going to jump on the bandwagon because as a set theorist I feel very underrepresented by the current answers.

For $m,n \in \mathbb{N} = \{0,1,2,\dots\}$ we can define $m^n$ to be the number of functions from a set of size $n$ to a set of size $m$. For instance, $$3^5 = \text{the number of functions}\ \{1,2,3,4,5\} \to \{1,2,3\}$$ There's only one set of size $0$, namely the empty set $\varnothing$, and so for any $m \in \mathbb{N}$ $$m^0 = \text{the number of functions}\ \varnothing \to \{1,2,\dots,m\}$$ But for any value of $m$ there is only one such function, namely the empty function. (This is a function: its domain is empty so it satisfies the definition trivially.) Therefore $m^0 = 1$ for all $m \in \mathbb{N}$.

This definition of exponentiation extends naturally to $\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \cdots$ and also to infinite cardinal numbers.

More interestingly, according to this definition, we have $0^0=1$ (since there is exactly one function $\varnothing \to \varnothing$) without any fear of lack of definition. (It does mean that the function $x \mapsto 0^x$ is discontinuous at $0$, but who cares? It means that the power series definition of $\exp x$ is well-defined at $0$, which I think trumps this caveat!)

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Multiplying by $1$ is the same as not multiplying by anything at all, so it's the same as multiplying by $5$ zero times.

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Strictly speaking $x^0$ is defined to be 1 so that the power rules work out nicely for positive and negative powers.

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I think that $r^0=1$ mostly to make exponent arithmetics consistent. Since $r^a*r^b=r^{a+b}$ and since $r^a=\frac{1}{r^{-a}}$ we get: \begin{align} r^0 = r^{a-a}= r^a*r^{-a}=r^a*\frac{1}{r^{a}}=\frac{r^a}{r^{a}}=1 \end{align} So any other than 1 defenition to $r^0$ would be inconsistent with exponent arithmetics.

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So many have given so many answers

But on my behalf i want to put it in simple terms as below

Apply $\log$ on both sides so it becomes $$ \log x^0 = \log 1 $$ here $\log 1$ is zero and the rule of logarithms says that $\log a^b = b\cdot \log a$

so it would be $0\log x = 0$ So they are equal.

Simple right :)

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Are you assuming $x^0 = 1$? You should reverse your argument. – Zach Langley Nov 16 '12 at 13:27
yes i assumed that and at last i have proved it as well – Shiva Komuravelly Dec 3 '12 at 7:28

First of all, your identity doesn't hold true for $x = 0$. In another words, $x \in \mathbb R^+$. For the intuition,$$ a^0 = a^{x -x } = { {a^x} \over {a^x}} = 1$$

Another: $$a^{n -1} = {a^n \over a}$$Now,$$a^{1 - 1} = {a^1 \over a} = {a^1 \over a^1} = 1$$

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First of all you(not only you many) have to understand mathematics is also a language, another way of communication. Language is obviously based on grammar( lexical rules). Power is a operation which is based on multiplication. $2^2$ means its just a language which convey you that you have to multiply 2 number 2 times. $2^0$ means you should not use 2 at all. Automatically our mind fills the number 1 instead of void.

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Why should our mind fill in with 1 rather than any other number?? Mystical sounding hokum is not mathematics! – Simon Hayward Nov 16 '12 at 20:14
Why would our minds choose to fill the void with $1$? Why not $0$? Why not $\infty$? There has to be a better explanation than this. – chharvey Nov 17 '12 at 14:36

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