# What is 0 to the power of 0?

This is part of a series on common misconceptions.

Is this true or false?

\[ 0^0 = 1 \]

**Why some people say it's true:** A base to the power of \(0\) is \(1\).

**Why some people say it's false:** An exponent with the base of \(0\) is \(0\).

The statement \(0^0 = 1\) is

ambiguous, and has been long debated in mathematics.This is mostly a matter of definition. Mathematicians love to define things. (After all, how else can we talk about mathematics if we don't know the definitions?)

Many sources consider \(0^0\) to be an "indeterminate form," or say that \(0^0\) is "undefined." On the other hand, other sources/branches of mathematics define \(0^0 = 1.\) Note that, certainly, \(0^0 \ne 0.\)

Some of the arguments for why \(0^0\) is indeterminate or undefined are as follows:

Argument 1:We know that \(a^0 = 1\) \((\)for all \(a \ne 0),\) but \(0^a = 0\) \((\)for all \(a>0).\) This contradiction means \(0^0\) should be left undefined.

Argument 2:With respect to limits, if \(\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0\), then \(\lim_{x \to a} f(x)^{g(x)}\) doesn't necessarily tend to any particular value. For example, \(\lim_{x \to 0} e^{-\frac{1}{|x|}} = \lim_{x \to 0} |x| = 0,\) but \[\lim_{x \to 0} \left( e^{-\frac{1}{|x|}} \right)^{|x|} = e^{-\frac{1}{|x|} \cdot |x|} = e^{-1}.\] Most of the arguments for why defining \(0^0=1\) is useful surround the fact that in some formulas, \(0^0=1\) makes the formula true for special cases involving 0.

Example 1: The binomial theorem says that \( (x+1)^n \equiv \sum_{k=0}^n \binom{n}{k} x^k\). In order for this to hold for \(x = 0\), we need \(0^0 = 1\).

Example 2: The power rule in differentiation states that \(\frac{d}{dx} x^n = n x^{n-1}\). In order for this to hold for \(x = 0\) and \( n = 1\), we need \(0^0 = 1\).

Example 3: \(0^0\) represents the empty product (the number of sets of 0 elements that can be chosen from a set of 0 elements), which by definition is 1. This is also the same reason why anything else raised to the power of 0 is 1.

Rebuttal: \(0^0\) has to be \(0\), since \(0^3 = 0\), \(0^2 = 0\), \(0^1 = 0\).

Reply: We cannot generalize from that pattern alone. Certainly, \(0^{-1}\) does not equal 0, since we cannot divide by 0.

Rebuttal: \(0^0\) has to be \(1,\) since formulas like the binomial theorem would not work when \(x=0.\)

Reply: Mathematics is a subject built upon definitions--there is no "universal truth" of what \(0^0\)reallyequals. If it is convenient for the binomial theorem to assume \(0^0=1,\) that is fine. On the other hand, if a mathematician who works with limits chooses to leave \(0^0\) as undefined, that is fine too!

Rebuttal: Why do some problems on Brilliant say that \(0^0\) is undefined?

Reply: As explained in this wiki, some sources argue that \(0^0\) is undefined. In particular, many algebra courses (prior to the university level) choose to define \(0^0\) in this way. Thus, some problem authors--especially in basic algebra problems--may use this definition of \(0^0.\)

Firstly, how did we come to know that \(a^{0} = 1?\)

Well, using the third law of indices, \(a^{0} = a^{m-m} = \frac{a^{m}}{a^{m}} = 1.\)

But think what if \(a\) had been equal to \(0?\)

Then, \(0^{0} = 0^{m-m} = \frac{0^{m}}{0^{m}} = \frac{0}{0}\), which is nothing but

indeterminate. (Here, \(m\) is a positive real number.)Therefore, this suggests that \(0^{0}\) is indeterminate. \(_\square\)

If your proof was valid \(0^{n}\) would be undefined for ALL positive \(n\) which it clearly is NOT (also you're using the second law of indices not the first).

Following your logic, \(0^{n} = 0^{(m+n)-m} = \frac{0^{m+n}}{0^{m}} = \frac{0^{m}\times 0^{n}}{0^{m}} = \frac{0\times 0}{0} = \frac{0}{0}\), which is nothing but

undefined. (Here, \(m\) and \(n\) are positive real numbers.)The laws of indices assume positive bases once you have a non-positive base the rules are no longer valid. If you use those laws on a negative base you can "prove" that \(1=-1\) because \(-1 = (-1)^{1} = (-1)^{2 \times \frac{1}{2}} = \big((-1)^{2}\big)^{\frac{1}{2}} = 1^{\frac{1}{2}} = 1\). Also in your particular case behind the scenes you're dividing by 0 which is going to force your proof into an indeterminate state regardless of the actual value of the initial supposition.

The reason why \(0^{0} = 1\) is definitional:

- \(a^{b}\) is equal to the product of a set of numbers with cardinality \(b\) where each element is equal to \(a.\)
- The product of an empty set of numbers is defined as being equal to 1.
- If the set doesn't contain any elements changing the value of each of those elements doesn't change the set at all and therefore also doesn't change the product of that set: \[0^{0} = 1^{0} = 2^{0} = 3^{0} = n^{0} = i^{0} = ∞^{0} = \left(\frac{0}{0}\right)^{0} = \left(\frac{∞}{0}\right)^{0}.\] These are all equivalent because they are all equal to the product of Ø.

**See Also**

**Cite as:**What is 0 to the power of 0?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/what-is-00/