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

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.

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### 1,029 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

1. ZeroIZsomething says:

I type slow and read even slower..

But if I can opine that true as it may smart people want to be correct ( and wannabe intelligent ones cannot conceive they can ever be wrong ). We not so SMART ones seek comfort in the fact that those intent on arguing will argue even if they were intelligent or not ( correct all the time ).

I also wish to re-state zero is a function. It separates positive and negative numbers, real and imaginary numbers. So if smart people wish to argue 0^0 = 1 or NOT then same said people should arguably disagree that 1^(1/2) =1 … Or NOT
Because -1 x -1 = 1

Again smart people say they are smart…
There are just too too many too smart persons and zero not smart people. ( punning out loud ).

2. Jay Muchmore says:

May I interject an 83 year-old, non-mathematician’s anxiety in dealing with these abstract mathematic manipulations of zero? I was recently faced with accepting: 0! x 0! = 1. (That drove me to this ASK MATEMATICIAN/ASK A PHYSICIST site for answers)
So this would mean that we could perform this math operation (or any of the above which creates 1 from 0) twice and come up with two ones. Then we’d be able to add those: 1 + 1 = 2 providing us with an integer with which to further mathematically carry out to infinity. But, oops, we now can do more brain exercises to deal with the nonexistent infinity.
Seriously, it is difficult to understand how we can take two empty barrels and through mathematics come up with one full one.

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4. dp says:

@Jay Muchmore
“Seriously, it is difficult to understand how we can take two empty barrels and through mathematics come up with one full one.”

there’s the confusion, we aren’t dealing with barrels, we’re dealing with numbers, numbers which exist only in our minds, which obey rules we made up.

5. Sam Iam says:

If you are to properly evaluate polynomials at zero, you must define 0^0 = 1. Can you guess what I do for a living?

6. David Harris says:

x^y is defined to be the number of functions going from a set Y of size y
to a set X of size X.

0^0 is thus the number of functions going from the empty set to the empty set.

How many such functions are there? Only one, the null function.

Hence 0^0 = 1

7. Philip Peterson says:

Jay Muchmore, anything you do to one side of an equation, you must do to the other in order for the statement to stay true.

Yes, 0! + 0! = 1 + 1 = 2. Note that this is different from multiplication (0! * 0! = 1 * 1 = 1). But neither of these says that 0 + 0 = 2.

Just as we can add 1 + 1 + 1 + 1 + … = infinity, we can add 0! + 0! + 0! + 0! + …. = infinity. There is no irregularity or paradox here. The key is that 0! is identical to 1 [but this does not imply 0 = 1].

Similarly, you may be asking why the following is incorrect:

0 + 0 = 0
(0 + 0)^0 = 0^0 + 0^0 = 1 + 1 = 2
Therefore (0 + 0)^0 = 0^0 = 1 = 2

The reason this is incorrect is that exponentiation (the ^ symbol) is not distributive. In other words, (a + b)^c is not equal to (a^c + b^c). Thus the second line is incorrect, because

(0 + 0)^0 ≠ 0^0 + 0^0

8. Musa Ali says:

So it comes down to the notion that there are mathematical functions which have multiple definitions, for integer and real values. For example the common definition of x^n = 1*x*x*x…. (n times) does not apply to real valued n.

Similarly, x! doesn’t ONLY have the definition of x! = 1 * 2 *3 * … * n
Because this can’t work on real valued x. So apparently this one is just a heuristic where as the *real* definition involves calculus. So 0! isn’t simply zero (or “an empty barrel”). it’s 1 for a REASON.

And thus, the conclusion is that certain values of these functions depends on which definition we choose and we choose the one that makes a lot of other things seem elegant.

9. Dr. Jones says:

I enjoyed the article, but even more fun than the article itself are the inevitable wacky comments below that try to philosophise about mathematics with an understanding gained before high school.

Wonderful!

10. Just to throw an algebraist’s approach into the mix…

I find that 0^0 makes more sense sense when you consider x^n as n repeated multiplication of x against itself. The article mentioned this approach as its first “Mathematician” example, but didn’t explain why 1 appeared in the definition: n^0 is an example of an empty product, where you multiply an empty sequence of numbers. We define an empty product to be 1 because 1 is the *multiplicative identity*. This definition is pleasant because it can be extended without effort to arbitrary algebraic operations with identities. For example, the empty sum n*0 = 0, since 0 is the additive identity. It’s interesting that the empty sum is a total non-issue, while the empty product is apparently incomprehensible.

This is also why 0! = 1. When you evaluate 4!, you multiply the sequence 4, 3, 2, 1 to produce 24. 0! multiplies an empty sequence of numbers, and thus defaults to the multiplicative identity.

http://en.wikipedia.org/wiki/Empty_product

11. Warbo says:

I think the difference between the high-school teacher and mathematician is that the teachers are used to working ‘within’ mathematics, ie. trying to explain our existing knowledge, whilst mathematicians are used to defining ‘new’ mathematics, ie. deriving interesting or useful results from appropriate definitions.

In other words, a mathematician will naturally define things in the most convenient way. If the accepted definition was that 0^0 = 0 then many mathematicians would use some definition like “let pow(x, y) = 1 when x = y = 0, and x^y otherwise” and carry on just as they do now. For a high-school teacher, that may be more difficult to justify to her students.

On a related note, it was quite an eye-opening experience when I first encountered abstract algebra, where the various numerical sets (natural, integer, rational, real, complex, etc.) are simply the consequence of particular operations (addition, subtraction, division, exponentiation, etc.), and that they are all important structures with interesting properties.

Prior to this, I thought of ‘numbers’ as a single, absolute, universal thing; the naturals are the ‘easiest’ approximation of the numbers, so we teach them first. Once we’re old enough to handle them, we can switch to integers since they’re a ‘better’ approximation of the numbers, and so on with rationals, reals and complex numbers. Presumably the infinite cardinals would come along at some point, but it all seemed like a quest to understand ‘better’ approximations of ‘the set of all numbers’. In particular I disagreed with Kronecker’s famous quote “God made natural numbers; all else is the work of man”, thinking it to be short-sighted and willfully ignorant (and similarly for the connotations in Alice in Wonderland of mathematics being full of absurdities).

Once I started dabbling in algebra, I started to see how each of these sets is a useful, standalone concept; not an approximation. I also began to see the importance of operators, and in particular their definitions.

In the case of this article, I think the confusion stems from this same treatment of “numbers” and “exponentiation” as single, absolute, things. Finding two different results (0 and 1) might lead someone to look for an inconsistency, or it might lead to the identification and definition of two complementary functions (or many, as found at http://www.haskell.org/haskellwiki/Power_function ). It certainly shouldn’t cause an argument over which is ‘better’ (akin to a ‘better approximation’ of the ‘real’ exponentiation). Likewise, such a viewpoint would reject the mathematician’s definition-by-convenience argument, since it’s ‘made up’.

12. kyrbis says:

helo

0=0- + 0+ and 0=0- – 0+

0=0 * 0+ and 0=0 * 0-

sqrt(0)= 0+ or 0-

sqrt(0+)=0,000….1

(0+)^0=1

curt(0-)=-0,000…1

(0-)^0=-1

0^0 =-1 or 1

13. Raghavendra Tripathi says:

But in our undergraduate classes we are usually told that 0^0 is indeterminate. All the books write the same.
How good is it to defined 0^0 = 1 , as we can see that it is not an absolute definition but depends upon how we define it. One can also define it in some other way.

14. Chris says:

If you don’t want 0^0 to be ambiguous, pick/invent a different mathematical system to use.

15. kyrbis says:

unequivocally

2 4 6 8 ….

negative number only odd

-1 -3 -5 -7 ….

and now 0^0=-1

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17. Stan says:

Let`s take a number, f.e. 62 and see how its value changes through different numeral systems. We will have 62 for decimal system (6*10^1+2*10^0), 56 for nonary (6*9^1+2*9^0), 50 for octal, 44, 38, 32, 26, 20, 14 for binary (6*2^1+2*2^0), 8 for the numeral system based on one. Next in the row is 2 for the numerical system based on zero which means that 6*0^1+2*0^0 =2 or 0^0=1. If we continue the row we will have -4 for system based on -1 (6*-1^1+2*-1^0), -10, -16 and so on.

If we take another number instead of 62, nothing changes. We will just have a different sequence wich will not be broken only if 0^0=1.

The above said of course proofs nothing. It is not necessary for the mentioned sequences to remain unbroken at the point of 0 based num. system, but they are at any other point.

18. Ryder says:

0^0 is 0 because a^0=1 unless a=0

19. Stan says:

Little more “logic”. We know that if we delete 1 by an infinitesimal number we’ll have an infinitely big number as a result. Zero is smaller than the infinitesimal number so 1/0 will be bigger than infinity. Such bigger than infinity number of course does not exist. Actually infinity is just an abstract number bordering something without borders.

So we have: 0 or 0/1 – smaller than an infinitesimal number; 1/0 – bigger than an infinitely big number. We can not define how much bigger or smaller they are, but we can know that they are bigger or smaller. If we multiply the infinitesimal and the infinitely big numbers, the result will be 1 or inf*(1/inf)=1. If “D” is the difference between zero and the infinitesimal number (inf), we’ll have: (inf-D)*[1/(inf-D)]=1.

So in 0*(1/0), we can just eliminate the zeros. The expression is equal to 1*(0/0), so 0/0=1.

Not knowing any exceptions from the rule n^0=n/n, we can conclude that 0^0=0/0=1

20. Anonymous says:

To ask ourselves what 0^0 is, we first must ask ourselves: What is the function of an exponent.

An exponent simply tells us how many of a certain number to multiply by each other. Thusly, 4^4=4*4*4*4; 3^3=3*3*3; and 2^2=2*2. When we keep going, 1^1=1. There is no multiplication to take place, because we are left with a single 1 and nothing else.
0^0 gives us zero zeroes. That means there is nothing at all left. The answer can’t be 1, and it cant even be 0.
The answer is that it is undefined, unsolvable, NaN (not a number), etc., because you are left with absolutely nothing, not even zero.

21. Stan says:

You reject zeroth power not only of zero, but of any other number. I’ll define the things a little bit different. For example: 5^2 tells us that we have to multiply 5 one time with 5. 5^5 tells us that we have to multiply 5 four times with 5. Multiplication needs at least two numbers but it is one math. operation. In the case of 5^1 there is no multiplication. We just have the number 5. What about 5^0? We have to multiply five -1 times with 5 which means to devide 5 one time with 5. And there is no 5 multiplied with itself – that is what the zero means. We have -1 act of multiplication, which is 1 divission. In 5^-1 we have 1/5 or (5/5)/5. Next 5^-2 is 1/(5*5) or [(5/5)/5]/5…

22. Eric Gan says:

Everyone says zero to anything is 0, but that isn’t true for every number. 0^-1 is 1/0, and 0*1/0 is 1. Therefore, 0^0 is 1.
You might be saying (0^2)*(0^-1) equals 0*1/0 which equals 1, but I believe that 0 doesn’t equal 0^2. 0^2 is 0^1/0^-1 which is 0/(1/0) which is a even zeroer zero. It may just be you can’t do it that way, but I don’t know.

23. Anthony Short says:

Shouldn’t 0^0 be 1 because the power of 0 means that there are NO 0′s, so then that means nothing is raised. And when x is raised to nothing, whatever x may be, then the rules say that it will equal to one.

7^3 equals 343 because 7 x 7 x 7 = 343.
7^1 equals 7 because 7 = 7.
7^0 equals 1 because there are no sevens in the equation, so it defaults to 1.
0^0 equals 1 because there are no zeros in the equation, so it defaults to 1.
See where I’m getting at?

BTW, I am 13 years old.

24. Colin says:

Consider the function x^x
2^2 = 4
1.5^1.5 = (3/2)^(3/2) = square root of (3/2) cubed = 1.837117307….
1^1 = 1
0.5^0.5 = square root of (1/2) = 0.707106781….
0.25^0.25 = fourth root of (1/4) = 0.707106781….
0.1^0.1 = tenth root of (1/10) = 0.794328235….
0.01^0.01 = hundredth root of (1/100) = 0.954992586….
0.001^0.001 = thousandth root of (1/1000) = 0.993116048….
0.000001^0.000001 = millionth root of (1/1000000) = 0.999986185….

As x gets closer and closer to zero, x^x gets closer to 1
Therefore it seems reasonable to define 0^0 to be 1

By calculus it can be shown that the minimum value of the curve plotted for x^x for all x > 0 occurs when (ln(x) + 1) = 0, i.e. x = 1/e = 0.367879441….
(1/e)^(1/e) = 0.692200628….

25. marc rogue says:

ok i thought about it. 0^0=1^0 not 1^1 so if u had 1^0 + 1^2=1^2. with this you can make sense of alot of thing, such as it is possible to make something out of nothing if we can find when matter becomes 1^0 we’ll find how the universe came about.

26. Nathan Mitchell says:

If we consider the equation x = 1/0, I believe there are 3 possible answers; 0, +infinity and -infinity. Taking 1/0 = 0 and rearranging gives 0/0 = 1. Stan has previously explained that 0/0 is the same as 0^0, so 0^0 = 0/0 = 1.

I’m currently taking A-level maths for the 2nd time so if anything I have said is incorrect, I apologise in advance.

27. betaneptune says:

I’d like to say that I find the arguments using limits of x^y unconvincing. Why? Consider 1^infinity as an indeterminate form. If you used that to determine 1^1, you could get e: lim (h->0) (1+h)^(1/h), which is of the form 1^infinity, and is equal to e. Surely you wouldn’t say 1^1 equals e because of this! So limit arguments are out.

That leaves you with 1, the empty product. Works great! Using Fortran notation:

3**2 = 1*3*3 = 9
3**1 = 1*3 = 3
3**0 = 1 = 1
3**-1 = 1/3 = 1/3
3**-2 = 1/(3*3) = 1/9

0**2 = 1*0*0 = 0
0**1 = 1*0 = 0
0**0 = 1 = 1
0**-1 = 1/0 = no such number
0**-2 = 1/(0*0) = no such number

Think about it this way: x**-1 = x**0 / x**1. Since we define 0**-1 as 1/0, then 0**0 must be 1 (!). And why do we use 1 as the numerator? Because it’s the empty product, of course!

And, as the article says: there’s the binomial theorem, which has an ugly exception if 0**0 .ne. 1. Such an exception makes computer code using the theorem ugly and cumbersome.

Then there’s the mapping of the empty set to the empty set. One way to do it. (x**y is the number of ways you can map y objects to x objects.)

Yes, it depends on how you define it, but so do other functions and operations in math.

Euler is right.

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