# Is Zero Raised to the Power of Zero Equal to One?

• The Rev
In summary, any number raised to the zeroeth power is equal to 1, including zero raised to the zeroeth power. There is some debate about the value of 0^0, but it is usually defined as 1 for notational convenience. However, in some cases, it may be defined as 0 or left undefined. Overall, 0^0 is considered an indeterminate form and may have different values depending on how the numbers are approaching zero. In polynomials and power series, 0^0 is defined as 1, but in other contexts, it may have a different value.
The Rev
I've been told that any number raised to the zeroeth power is equal to 1. What about zero raised to the zeroeth power? Is that 1 or 0?

$$0^0=1?$$

The Rev

undefined!

The Rev said:
I've been told that any number raised to the zeroeth power is equal to 1. What about zero raised to the zeroeth power? Is that 1 or 0?

$$0^0=1?$$

The Rev
...and indeterminate. In some cases when a generalization is desired, it is profitable to define it as 1, in others, as 0.

In series, it is always taken to be 1, so that you can write a series expansion compactly:

$$f(x)=\sum_{n=0}^{\infty}a_nx^n$$
so that $f(0)=a_0$

For example, the geometric series:

$$\sum_{n=0}^{\infty}x^n =\frac{1}{1-x}$$
is true at x=0 only if we define 0^0=1.

I don't think that it's undefined; I think that it's 1. 0^0 is an empty product, and empty products are necessarily equal to 1. As Galileo points out above, we need 0^0=1 for series to have compact formulas.

Everyone agrees that $$x^0=1$$ for $$x\neq0$$, but there's no reason to think that it should be different at 0 -- $$0^x$$ is only 0 for x > 0, since it's not defined for negative x.

There's no problem accepting 0^0 as 1, and there are many good reasons to think it shouldn't be undefined or 0.

Well, I think that the formula a^0=1 appears when you try to divide a^m by itself:

(a^m)/(a^m) = a^(m-m) = a^0

Since the first part of this equation equals 1, we have a^0=1

But if a=0 we can't do the division (0^m)/(0^m)

CRGreathouse said:
I don't think that it's undefined; I think that it's 1. 0^0 is an empty product, and empty products are necessarily equal to 1. As Galileo points out above, we need 0^0=1 for series to have compact formulas.

Everyone agrees that $$x^0=1$$ for $$x\neq0$$, but there's no reason to think that it should be different at 0 -- $$0^x$$ is only 0 for x > 0, since it's not defined for negative x.

There's no problem accepting 0^0 as 1, and there are many good reasons to think it shouldn't be undefined or 0.
That's not mathematical at all. I've always had a simple way of looking at it:

$$x^0 = \frac{x}{x}$$.

Everyone also agrees that $$0^x=0$$ for all $$x>0$$, so there should be no reason to think it's different at 0, so $$0^0=0$$ right? We have a problem with the limit of $$x^y$$ as $$(x,y)\rightarrow (0,0)$$, it doesn't exist (different values depending on how x and y are approaching zero) so there is no obvious or natural choice of a value for $$0^0$$, so it's usually left as undefined (barring notational convenience).

The reasons to call the symbol 0^0 1 is usually for notational convenience, like empty products, binomial theorem, power series, etc.

Most of the mathematicians I know seem to take the position that 0^0 is technically undefined, but there's nothing wrong with letting it equal 1 for notational convenience.

It's very rare for it to be convenient for 0^0 to be defined as any other value. The standard reason it needs to be something other than 1 is to extend 0^x to x=0, but in practice that doesn't seem to happen very often.

Of course, if you have a limit of the form f(x)^g(x) with f(x),g(x) -> 0 you have to be careful; it won't always converge to one.

What about $$x^x$$ as $$x \rightarrow 0$$? I have always found $$x^x$$ fascinating for some obscure reason.

Icebreaker said:
What about $$x^x$$ as $$x \rightarrow 0$$? I have always found $$x^x$$ fascinating for some obscure reason.

x^x -> 1 as x -> 0, of course.

In fact, I believe there's a result that says that if f(x),g(x) -> 0 as x approaches some limit, then f(x)^g(x) -> 1 as long as f and g are analytic.

COUNTEREXAMPLE:

The functions $\frac{1}{3x+5}$ and $\frac{1}{x^{2}+4}$ are analytical in every point from their domain...

However,

$$\lim_{x\rightarrow +\infty}\left(\frac{1}{x^{2}+4}\right)^{\frac{1}{3x+5}} =-\infty$$

Daniel.

My calculator symbolically evaluates that limit to 1...

There's a more obvious counterexample:

$$\lim_{x \rightarrow 0} 0^x = 0$$.

The whole thing about 0^0 depends precisely what ^ means. As an operation on real numbers, 0^0 is undefined. The reason is that (0, 0) is not in the domain of ^.

Okay, so you want the "philosophical" reason. A crucial property about real operations is that they're continuous within their domain.

However, ^ cannot be continuous at (0, 0) -- the classic examples demonstrating this fact are 0^x --> 0 and x^0 --> 1 as x --> 0.

However, there are other meanings to ^. For example, there's a definition of ^ that means repeated multiplication. The empty product is, by definition, 1, so anything to the zero-th power is equal to 1. In polynomials and power series, this definition is what ^ "really" means -- that's why one uses 0^0 = 1.

Icebreaker said:
What about $$x^x$$ as $$x \rightarrow 0$$? I have always found $$x^x$$ fascinating for some obscure reason.
It is a very nice function though, when I first was thinking about it I tried to think about it in 4D in complex space and was pleasantly surprised when I started graphing it on mathematica recently I had quite a good idea oh how it looked.

Zurtex said:
That's not mathematical at all. I've always had a simple way of looking at it:

$$x^0 = \frac{x}{x}$$.

Empty products aren't mathematical?

A less trivial example is:

$$\lim_{x \rightarrow 0} \left( e^{-1/x^2} \right)^{x^2} = \frac{1}{e}$$

Both the base and the exponent approach 0 from the positive side, but as we can see, the limit is not 1.

(Note the base is not an analytic function of x, despite its infinite differentiability at 0!)

CRGreathouse said:
Empty products aren't mathematical?
No, I meant your approch to the problem.

Zurtex said:
No, I meant your approch to the problem.

My approach was just stating that it's an empty product, and that empty products are always 1. The rest of my post was descriptive/informative (and perhaps poorly worded).

take log both side...
u get
0 = (log 1)/log(x)
log 1 = 0
0/log(x) as x > 0
= 0/inf
interminate^_^

That's what I get for quoting a theorem from memory.

I was way too general it seems. After looking it up, it seems that the theorem actually requires that:

1) f and g are non-zero
2) f and g are analytic at zero
3) f(x) -> 0 and g(x) -> 0 as x -> 0 from the right
4) f(x) is positive from all positive x <= some value close to zero

Then you get that f(x)^g(x) -> 1 as x -> 0 from the right.

Of course, that result is much less impressive. But I'm pretty sure this one is true.

Edited to remove the counterexample that wasn't really a counterexample.

Last edited:
I'm pretty sure the limit dex quoted goes to 1, not -&infin;.

Hurkyl said:
I'm pretty sure the limit dex quoted goes to 1, not -∞.

Yes, it seems that it does. Still, your trivial example of 0^x was a good counterexample.

its the hundreth time the same question has been asked

Yes,you're both right.It goes to 1.

Daniel.

We cannot divide by 0.

Take the classic example:

a = b
a^2 = b^2
a^2 + a^2 = a^2 + ab
2(a^2) = a^2 + ab
2(a^2) - 2ab = 2(a^2) + ab - 2ab
2(a^2) - 2ab = 2(a^2) - ab
2((a^2) - ab) = 1((a^2) - ab)

then divide by (a^2) - ab) to give 1 =2

The fallacy comes right at the end when dividing by 0. So, we cannot divide by zero, and 0^0 is 0/0 therefore it is Mathematically undefined.

Regards,

M

BenGoodchild said:
The fallacy comes right at the end when dividing by 0. So, we cannot divide by zero, and 0^0 is 0/0 therefore it is Mathematically undefined.

0^0 isn't 0/0.

Why not?

x^0 = x/x

Yes,but $$x^{0}=\frac{x}{x} \Leftrightarrow x\neq 0$$.So master coda was right.

Daniel.

dextercioby said:
Yes,but $$x^{0}=\frac{x}{x} \Leftrightarrow x\neq 0$$.So master coda was right.

I understand that this is true, and is infact what i said.

BenGoodchild said:
So, we cannot divide by zero, and 0^0 is 0/0 therefore it is Mathematically undefined.
I say we cannot divide by zero,
and as x^0 is x/x
which becomes, when x = 0, 0/0 cannot be omputed if we cannot divide by 0 it either doesn't exist or as I put it it is mathematically undefined.

Regards,

Ben

BenGoodchild said:
I understand that this is true, and is infact what i said.

I say we cannot divide by zero,
and as x^0 is x/x
which becomes, when x = 0, 0/0 cannot be omputed if we cannot divide by 0 it either doesn't exist or as I put it it is mathematically undefined.

Regards,

Ben

x/x is not the definition of x^0. So the fact that x/x is not defined for x=0 doesn't automatically mean that x^0 is not defined for x=0.

I thought x^0 was x^(m-n) where m=n, so x^0 = x^m/x^n. If x=0, then
0^0 must be 0/0 = undefined.

Correct me if I am wrong.

Leopold Infeld said:
I thought x^0 was x^(m-n) where m=n, so x^0 = x^m/x^n. If x=0, then
0^0 must be 0/0 = undefined.

Correct me if I am wrong.

No. x^(m-n) = x^m/x^n doesn't hold for x=0, at least if you want something sensible like 0^1=0 to be true.

It is merely a convention that 0^0 is taken as 1,and not universal.

Think carefully about what it even means to raise numbers to powers.

if x is a whole number and n is a whole number then x^n is straight foward, we can even do this for negative integer x , and negative integer n too, and this extends nicly to x^0=1 when x=/=0 by your above argument. We can even extend to rational x and thence to real x.

We can even take roots occasionally ie we can square root +ve numbers (let's not worry abuot complex ones yet), and so we can raise real numbers to rational powers by, say, for 4/5 rasing to the power 4 then taknig the 5th root. All well and good. But what does it even mean to raise a number to the power sqrt(2)?

To be honest we do not need to consider that.

The most common time we use ^0 is in taylor series, when we want a nice formula, and then it makes sense to define x^0=1 for all x even 0. This makes the function f(x)=x^0 a nice continuous function on the whole of R.

Defining 0^0 to be 1 is useful and consistent with extending x^0 to a continuous function on the whole of R.

I thought x^0 was x^(m-n) where m=n,

No... there are some subtleties here -- we're talking about these expressions as strings of symbols in the language of mathematics.

In particular, in this context, x^(m-n) is never x^m / x^n. They're different strings of symbols.

What we have is that, under the right conditions, x^(m-n) = x^m / x^n is a true statement.

Those conditions require x to be nonzero (and positive, if we're talking about the usual exponentiation operation on the reals, and being strict).

This means that trying to substitute x = 0 into this equation to get 0^(m-n) = 0^m / 0^n is meaningless. It makes as much sense as if I were to write "+ 1 & *(".

Now, it might be of some interest to provide a rigorous, although tedious method of defining "exponentiation".
1) First, by induction, it is unproblematic to define
$$x\equiv{x}^{1}, x^{n}=x^{n-1}*x, n\geq{2}, n\in\mathbb{N}, x\in\mathbb{R}$$

2) Let us now define a function Exp(x) on $$\mathbb{R}$$:
$$Exp(x)=1+\sum_{n=1}^{\infty}\frac{x^{n}}{n!}$$
Exp(x) can be shown to be larger than zero, and strictly increasing.
We also have Exp(0)=1
Since it is strictly increasing, we may define its inverse, Log(x), defined on the positive half-axis (zero not included).
3) We may now, for any $$b\in\mathbb{R}, a\in(0,\infty}$$ define:
$$a^{b}=Exp(b*Log(a))$$
Thus, we have for any a>0, $$a^{0}=Exp(0*Log(a))=Exp(0)=1$$

4) We might extend our idea of exponentiation to give, for example, meaning to the function $$0^{x}$$ but the value of the expression $$0^{0}$$ will, at best, always be one from a "provisional" definition dictated by convenience, rather than that we may deduce its value from first principles.

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