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.


\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.


\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


= (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.

1,078 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

  1. kyrbis says:

    the shape of the square root


  2. Angel says:

    The argument is a construct for mathematical convenience.
    It is not “true” for all abstracts. But it does work for most models which is why it is so widely used.
    If (in the form a^x)
    which is not “true”

  3. kyrbis says:

    final countdown >>> 0^0=1^sqrt(e)


  4. kyrbis says:

    open design :D

    thanks for Britney ;)


  5. kyrbis says:

    operating scheme:

    0 rotate

    0 plus

    0 minus


  6. kyrbis says:

    negative curvature rescues


  7. kyrbis says:

    strange tangent to displace the variable y


  8. kyrbis says:

    four-dimensional uncertainty with four metrics


  9. kyrbis says:

    Better not ask how this idea


  10. saurabh singh says:

    0^0 is indeterminate quantity : proof
    let 0^0= k ; taking log both side
    0loge 0=loge k
    0(-infinite)= loge k
    hense proof no value of exist

  11. kyrbis vel uneedcallme says:

    oh no! e number is strange


  12. kyrbis vel uneedcallme says:

    calculate the neutral element


  13. kyrbis says:

    Zero to zero was in need of dimensions 28297

    Zero Complex only 13

    harmonize the needs of quaternions referenced to prime numbers

  14. kyrbis says:

    3 neutral elements “0” , “1” , “i”

    The operating matrix


  15. betaneptune says:

    Rao KD,

    Limits are irrelevant. Consider lim(h->0) (1+h)^(1/h). This is an indeterminate form of the type 1^infinity. Clearly 1^infinity = 1, not e. By changing the above to lim(h->0) (1+2/h)^h we get e^2. So limits are not unique, not equal to the correct answer of 1, and therefore are of no help. In fact, we already know that the indeterminate forms produce different answers in different cases, which is why we have, for example, L’Hopital’s rule for the 0/0 cases. (Which is why they’re called “indeterminate”.) So you can’t determine 0^0 via limits any more than you can 0/0.

    As for 3^7 > 7^3 and such: You say that if x,y>2, the exponent rules (gives the larger result). The actual boundary is e, not 2. Example: 2.1 ^ 2.2 = 5.1154… whereas 2.2 ^ 2.1 = 5.2370…

  16. uneedcallme says:

    where you want something dumb to count nothing :(

    +0=(-0)^2 ;)


  17. uneedcallme says:


    cute puzzle


  18. uneedcallme says:

    each features a zero changes e -> sqrt(2)

    eats action by replacing a part number


  19. kyrbis vel uneedcallme says:

    circumvent the uncertainty of the logarithm

    already knows where so many dimensions


  20. kyrbis says:

    Welcome to the Matrix 8)


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