Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Posted on December 4, 2010 by The Mathematician

Clever student:

I know!

$x^{0}$ = $x^{1-1}$ = $x^{1} x^{-1}$ = $\frac{x}{x}$ = $1$ .

Now we just plug in x=0, and we see that zero to the zero is one!

Cleverer student:

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

$0^{x}$ = $0^{1+x-1}$ = $0^{1} \times 0^{x-1}$ = $0 \times 0^{x-1}$ = $0$

which is true since anything times 0 is 0. That means that

$0^{0}$ = $0$ .

Cleverest student :

That doesn’t work either, because if $x=0$ then

$0^{x-1}$ is $0^{-1} = \frac{1}{0}$

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function $x^{x}$ and see what happens as x>0 gets small. We have:

$\lim_{x \to 0^{+}} x^{x}$ = $\lim_{x \to 0^{+}} \exp(\log(x^{x}))$

= $\lim_{x \to 0^{+}} \exp(x \log(x))$

= $\exp( \lim_{x \to 0^{+} } x \log(x) )$

= $\exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )$

= $\exp( \lim_{x \to 0^{+} } -x )$

= $\exp( 0)$

= $1$

So, since $\lim_{x \to 0^{+}} x^{x}$ = 1, that means that $0^{0}$ = 1.

High School Teacher:

Showing that $x^{x}$ approaches 1 as the positive value x gets arbitrarily close to zero does not prove that $0^{0} = 1$ . The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that $0^{0}$ is undefined. $0^{0}$ does not have a value.

Calculus Teacher:

For all $x>0$ , we have

$0^{x} = 0$ .

Hence,

$\lim_{x \to 0^{+}} 0^{x} = 0$

That is, as x gets arbitrarily close to $0$ (but remains positive), $0^{x}$ stays at $0$ .

On the other hand, for real numbers y such that $y \ne 0$ , we have that

$y^{0} = 1$ .

Hence,

$\lim_{y \to 0} y^{0} = 1$

That is, as y gets arbitrarily close to $0$ , $y^{0}$ stays at $1$ .

Therefore, we see that the function $f(x,y) = y^{x}$ has a discontinuity at the point $(x,y) = (0,0)$ . In particular, when we approach (0,0) along the line with x=0 we get

$\lim_{y \to 0} f(0,y) = 1$

but when we approach (0,0) along the line segment with y=0 and x>0 we get

$\lim_{x \to 0^{+}} f(x,0) = 0$ .

Therefore, the value of $\lim_{(x,y) \to (0,0)} y^{x}$ is going to depend on the direction that we take the limit. This means that there is no way to define $0^{0}$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$ .

Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function $f(x,y) = y^x$ for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

$y^x$ := $1 \times y \times y \cdots \times y$

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

$y^{x}$ = $1 \times y$ .

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

$y^{0}$ = $1$

which holds for any y. Hence, when y is zero, we have

$0^0 = 1$ .

Look, we’ve just proved that $0^0 = 1$ ! But this is only for one possible definition of $y^x$ . What if we used another definition? For example, suppose that we decide to define $y^x$ as

$y^x$ := $\lim_{z \to x^{+}} y^{z}$ .

In words, that means that the value of $y^x$ is whatever $y^z$ approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

[Clarification: a reader asked how it is possible that we can use $y^z$ in our definition of $y^x$ , which seems to be recursive. The reason it is okay is because we are working here only with $z>0$ , and everyone agrees about what $y^z$ equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

$0^0$ = $\lim_{x \to 0^{+}} 0^{x}$ = $\lim_{x \to 0^{+}} 0$ = $0$

Hence, we would find that $0^0 = 0$ rather than $0^0 = 1$ . Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what $y^x$ means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is $0^0$ really? Well, for x>0 and y>0 we know what we mean by $y^x$ . But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of $y^x$ is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what $y^x$ means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that $0^0=1$ ? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use $0^0=0$ or if we say that $0^0$ is undefined. For example, consider the binomial theorem, which says that:

$(a+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}$

where $\binom{x}{k}$ means the binomial coefficients.

Now, setting a=0 on both sides and assuming $b \ne 0$ we get

$b^x$

= $(0+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} 0^k b^{x-k}$

= $\binom{x}{0} 0^0 b^{x} + \binom{x}{1} 0^1 b^{x-1} + \binom{x}{2} 0^2 b^{x-2} + \hdots$

= $\binom{x}{0} 0^0 b^{x}$

= $0^0 b^{x}$

where, I’ve used that $0^k = 0$ for k>0, and that $\binom{x}{0} = 1$ . Now, it so happens that the right hand side has the magical factor $0^0$ . Hence, if we do not use $0^0 = 1$ then the binomial theorem (as written) does not hold when a=0 because then $b^x$ does not equal $0^0 b^{x}$ .

If mathematicians were to use $0^0 = 0$ , or to say that $0^0$ is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using $0^0 = 1$ .

There are some further reasons why using $0^0 = 1$ is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.

This entry was posted in -- By the Mathematician, Math, Philosophical. Bookmark the permalink.

1,177 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Mark VA says:

June 17, 2012 at 10:36 pm

Yee and Daniel:

Take a look at this:

http://mathworld.wolfram.com/Zero.html

It seems that at this point there seem to be two valid approaches to this question.
Yee says:

June 18, 2012 at 6:56 am

gerald weinstein and Mark VA,
You are talking about limit.
Limit is not a necessary consideration.
Ladd Webster says:

June 19, 2012 at 5:32 pm

the method of the “cleverer student” works. If x=0, you get (0)((1/0)), and the zeros cancel, so there is no division by zero error. the answer is 1.
Yee says:

June 20, 2012 at 12:15 am

Ladd Webster,
It is not good to cancel zero.
Someguy says:

July 21, 2012 at 8:39 am

So, in other words, 0^0=0 and 0^0=1. But wait! 0=1? This isn’t possible, which means that 0^0 must be undefined!
Yee says:

July 23, 2012 at 8:10 am

Someguy,
Define 0^0=1,
then 0^0 will not equal 0.
Joseph says:

July 31, 2012 at 8:24 pm

I’m Joseph. 11 years old. Your comments are confusing to my math skills. It seems to me that the answer isn’t 0, isn’t 1, isn’t undefined. Settle with an answer so i can use it on my test. In the mean time i’ll use 0 to the 0 equal NOTHING!! And I’ll ask my Teacher which answer she’ll like to hear.
Yee says:

August 1, 2012 at 8:38 am

Joseph,
1 is the best definition.
Undefined is adopted by most people.
No consensus.
I think you will seldom encounter it.
le says:

August 5, 2012 at 4:14 am

this may be the best explanation of ‘0’ that i have ever herd….

“There are some further reasons why using is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.”

As a Mathematician you define the world and the rules as you see fit to the problem at hand… when you apply them to things such as physics you pick the tool that is appropriate…

Cheers,
~/lee
cool kid says:

August 9, 2012 at 4:23 am

well since the limit of the function x^x approaches 0 but never does, doesn’t this mean that it is undefined, like an asymptote?
Yee says:

August 11, 2012 at 9:59 am

cool kid,
Limit is not a necessary consideration.
Defining 0^0=1 is okay.
bailey says:

August 12, 2012 at 5:51 am

if a^b=1*a*a…a*a (there are b a’s)

then 0^0=1 times… wait… there are 0 0’s, so…

0^0=1
rules out all other reasons!
bailey says:

August 12, 2012 at 6:06 am

actually, maybe not all reasons
Mark says:

August 16, 2012 at 8:04 pm

cool kid,

Asymptotes are defined, as they are lines.

Cheers
Mr Tai says:

August 18, 2012 at 2:26 pm

OK E^x=1+x/1!+x^2/2!+x^3/3!+⋯,-∞<x<∞ [E=0 and x=0] 0^0=1
0^0=1 says:

August 18, 2012 at 11:23 pm

The function 0 to x power is quite “unimportant”. So 0^0 = 0^1(0^ (-1)) = 0 times 1/0 = 0/0 = undefined is a wrong statement. X to 0 power is 1, is valid for all x.
Yee says:

August 19, 2012 at 3:28 am

Mr Tai,
E=2.71828…
It is a constant.
You can not make it equal 0.
mythe says:

August 19, 2012 at 10:35 am

A mathematician saying 0/0 = 1 just because it is nice is like saying \pi = 3 because it is easier to remember. I prefer the calculus teachers approach of explaining why 0/0 is indeterminate.
Yee says:

August 19, 2012 at 8:54 pm

pi=3 is not nice.
It is redefinition.
Yee says:

August 20, 2012 at 10:07 pm

The approximation of pi is a trade off between
accuracy and convenience.
Using 3 is for convenience.
3.14 is less convenient but more accurate.
However, 3.14 is not the exact value.
The Power of Zero says:

August 29, 2012 at 6:02 pm

This has been debated forever. First you have to decide if 0 is a place holder, or is 0 a number. Zero: The Biography of a Dangerous Idea is a great book discovering why both are true, why 0 raised to the power of 0=1 and why it does not. It will also give you an insight as to whether Zero is real or not. Can the number 5 be alive without zero? Can we accurately tell time, with our Zero? So in Truth there is no real answer to your question, it is all subjective!!
Yee says:

August 30, 2012 at 5:27 am

The Power of Zero,
Does the existence of 0 have an answer?
Is it subjective?
S K says:

September 5, 2012 at 12:59 pm

I have always been taught it is undefined, and understood it, then a teacher told me it is equal to 1 and could not say why. Now I have seen your arguments why it should be =1, but in the general sense in mathematics, it makes some things untrue…I will give an example.

We know in algebra, 0 is the additive identity, therefor it does not have an inverse, thus we cannot divide by 0. This is accepted by the math community, and have been for a long time.

IF 1=0^0, then using only exponential rules we get
0^0= 0^(x-x) =
=(0^x)/(0^x) =
=(0/0)^x,

but as stated many times here, 0/0 is undefined. But using 1 = 0^0 we can conclude that 0/0 is not undefined but is equal to one from the logic above. Unless I have made a logical fallacy, which is possible that I didn’t find it, please let me know, otherwise I will continue on that 0^0 is undefined conventionally unless otherwise stated.
Yee says:

September 13, 2012 at 12:23 am

S K,
You are abusing index laws.
a^(m-n)=a^m/a^n is invalid when a=0.
If index laws are abused,
then 0^1=0^(2-1)=0^2/0^1=0/0.
0 itself is also undefined.

Why is it “reasonable” to define 0^0=1?
Extend index laws in a reasonable way:
(0^0)^2=0^(0*2)
0^0=0^(-0)=1/0^0

We can list equationS:
x^2=x
x=1/x
1 is the unique solution.
Therefore 1 is the only reasonable definition of 0^0.
Fred Star says:

September 17, 2012 at 3:52 pm

Saying 0^0=1 is inconsistent.
If one plots x^(ln(a)/ln(x)) for x>0 one gets the constant function y=a. After all, ln(a)/ln(x)= log base x of a.
Thus the limit of this as x approaches 0 (from the positive side) would give an indeterminate of the form 0^0 and this limit is a.
Yee says:

September 17, 2012 at 7:09 pm

Fred Star,
You are overemphasizing continuity.
Continuity is not the only consideration.
When the limit does not exist,
the function can also be defined.
Take more viewpoints into consideration,
defining 0^0=1 is more reasonable.
Fred Star says:

September 18, 2012 at 12:31 pm

Yee,
It is certainly true that one could define 0^0=1. Why not define 0^0=pi or 0^0=sqrt (2)? The purpose behind definitions is to provide an agreement on mathematical ideas. Defining 0^0=1 without considering how it relates to other mathematical ideas such as limits, etc. really serves no purpose. Could you please specify where it is fruitful to define 0^0=1. Thank you.
Yee says:

September 18, 2012 at 8:01 pm

Fred Star,
(0^0)^2=0^(0*2)
0^0=0^(-0)=1/0^0

We can list equationS:
x^2=x
x=1/x
1 is the unique solution.
Therefore 1 is the only reasonable definition of 0^0.

Define 0^0=1,
we can simplify a polynomial
a[0]+a[1]*x^1+a[2]*x^2+…+a[n]*x^n
as

n
Σ a[k]*x^k
k=0

These are good reasons to define 0^0=1.
Alex says:

September 19, 2012 at 6:38 pm

“Therefore 1 is the only reasonable definition of 0^0.”

What is the difference between this assertion and the assertion that “it is only reasonable to assume that the product of non-empty sets is nonempty”? While the axiom of choice is most often assumed, there are mathematicians who don’t use it and in fact study axioms that are incompatible with it.

While it is practical to assume that 0^0=1 in most cases it is something that needs to be assumed every time (even if not stated) every time that it is used, as Fred Star’s continuity argument shows. Stating something as a definition makes it universal, and sacrificing the logical consistency of mathematics for convenience hardly seems like a fair trade.
Yee says:

September 20, 2012 at 7:02 pm

Alex,
Continuity needs not be overemphasized.
0^0=1 can be universal without sacrificing logical consistency.
nweze victor says:

September 21, 2012 at 8:50 am

In the real sense of it, it isn’t zero to the power of zero that is undefined, it is zero that is undefined, my reasons are philosophical:
1) If zero where real then every thing should amount to a variable, because all positive numbers are arbitrary in reality.
2) There cannot be zero, where in physical reality nothing is ever diminished to a non-entity.
This problem also surfaces when finding the factorial of zero.
Steve says:

September 26, 2012 at 12:18 pm

I think it is indeterminate.i need more explanation
Yee says:

September 26, 2012 at 10:49 pm

It is reasonable to define 0^0=1.
I have explained much above.
Judah Corsini says:

October 3, 2012 at 10:31 pm

In honor of Aristotle:
A is A
0 = 0
0 = 0^1
0 = 0^(1+0)
0 = (0^1) x (0^0)

0^0 = 0^(1-1) = (0^1) x (0^-1) = 0/0

0 = (0^1) x (0/0)

(0^1) x (0^0) = (0^1) x (0/0)

A x A^A = A x A/A

If A = 1/infinity then A = A
0 = 1/infinity
Judah Corsini says:

October 3, 2012 at 10:35 pm

There cannot be nothing. There can only be something.
Judah Corsini says:

October 3, 2012 at 10:38 pm

It seems that zero itself is undefined…
Yee says:

October 7, 2012 at 6:58 am

Judah Corsini,
You are abusing index laws.
alj cruz says:

October 8, 2012 at 7:05 pm

I tried it to my Casio calculator and it sh0ws syntax err0r, it only means that 0^0 is undefined..or invalid.
Yee says:

October 9, 2012 at 10:15 am

alj cruz,
It depends on the design of the system.
This can not be a reason.
Many systems set 0^0=1.
ChRiS says:

October 12, 2012 at 3:21 pm

Yee,
to work out 0^0 it involves dividing by zero as many other people have proved above
it may be reasonable to define it as 1 but mathematically it has to be undefined
and most systems refer to it as a “math error” not 1
Yee says:

October 14, 2012 at 9:01 pm

ChRiS,
Defining 0^0 with division is only a form, not a proof.
This is not a good form.
This is the result of overemphasis of continuity.
As I know, many systems set 0^0=1.
But the design of systems can not be a reason.
chua says:

October 16, 2012 at 5:00 am

x/x, if you substitute 0 into x, then the answer must be undefined because you shouldn’t forget that zero cannot be the denominator!!! else the value is undefined!!!
Yee says:

October 16, 2012 at 10:19 am

chua,
x^0 does not equal x/x.
Does x equal x^2/x?
Steve says:

October 20, 2012 at 7:03 pm

Zero is the absence of a quantity so the closest number to it is some number “x”. Therefore, x/x will always = 1 because x can not = zero.
chua says:

October 21, 2012 at 10:45 am

But we’re talking about 0^o, not x^0.
As far as i know, any number raise to 0 is equal to 1 except zero raise to zero.
If we sub 0 into x then x/x will be 0/0 which will be undefined!!!
chua says:

October 21, 2012 at 10:48 am

I’m not sure if x^0 is equal to x^x or not. According to the way the clever student solve that on top of this page, x^0 is x/x, so it must be undefined if you sub zero into x.
Yee says:

October 22, 2012 at 6:03 am

Steve and chua,
You have said so much,
but you didn’t explain why x^0 equal x/x?
In my opinion, they are irrelevant.
S K says:

October 22, 2012 at 6:26 pm

Yee you said:

Yee says:
October 16, 2012 at 10:19 am

chua,
x^0 does not equal x/x.
Does x equal x^2/x?

Now in algebra, when solving for x, we always plug in the values to make sure they work. So when we say x = x^2/x, it is implied (though all my teachers made em state) that this is only true for if x is not equal to 0. Now I admit that there are times in which it is perfectly logical to conclude that 0^0=1 (especially in combinatorics), but in general every day algebra, this leads to a lot of potential confusions and problems.

Also, in calculus, this is an UNDETERMINED form, which you apply L’Hopital’s rule to get what is should be, which is rarely ever 1. If you just took 0^0=1 for ALL case, then this would be bad for calculus students for they would put 1 when the answer should be something like 3/4.

A mathematician needs to be aware of ALL cases it can be used in. That is why when dealing with a problem, or a specific area in math, they have their own definitions. I have run across in which the same notation, and same terms mean VERY different things depending on the field of study. Algebra is a very large area, which is what I am assuming that we are talking about, specifically the ring of real numbers. So saying to just define that 0^0 = 1 does not work necessarily unless you define your exponents in such a way that this does not violate any of the axioms to be a ring. And then it can only be applicable in that case because as mathematicians, we cannot ASSUME that everyone will work in that, which is why we define such stuff before hand.

Lastly, I know this carries no weight what so ever to this, but wolfram Alpha says that it is “indeterminate”.
Yee says:

October 22, 2012 at 9:31 pm

S K,
Of course, there are times
in which it is perfectly logical to conclude that 0^0=1.
This definition is irrelevant to limit,
it does not violate L’Hopital’s rule at all.
If you think there are rings which does not include 1,
0^0=1 can be defined only in the rings with 1,
because they are most commonly used.
Thus it will not violate any of the axioms to be a ring.

I know many computer languages use 0^0=1.
How it is defined in computer languages is not a reason at all.
Yee says:

October 22, 2012 at 9:58 pm

Some mathematicians overemphasize continuity and despise other fields.
They don’t define the function value at a point where limit does not exist.
Defining x^0 as x/x is the result of overemphasis of continuity
because it is convenient to define x^0=1 and exclude 0^0=1.
However, it is unreasonable.
Because there are times in which it is perfectly logical.
Continuity should not be overemphasized to such extent.

Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

1,177 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Leave a Reply Cancel reply

Send your questions about math, physics, or anything else you can think of to:

Subscribe via Email

Search the Archives

Categories

Archives

Recent