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 \(0^0\) really equals. 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 undefined. (Here, \(m\) is a positive real number.)
Therefore, this proves that \(0^{0}\) is undefined. \(_\square\)
See Also
