# Why Zero Raised to the Zero Power IS One

Updated! February 5, 2017

The value of zero raised to the zero power, $(0^0)$, has been discussed since the time of Euler in the 18th century (1700s). There are three reasonable choices: 1,0, or “indeterminate”. Despite consensus amongst mathematicians that the correct answer is one, computing platforms seem to have reached a variety of conclusions: Google, R, Octave, Ruby, and Microsoft Calculator choose 1; Hexelon Max and TI-36 calculator choose 0; and Maxima and Excel throw an error (indeterminate). In this article, I’ll explain why, for discrete mathematics, the correct answer cannot be anything other than 0^0=1, for reasons that go beyond consistency with the Binomial Theorem (Knuth’s argument).

#### Context of the Debate: Continuous Mathematics

The three choices for the value of $0^0$ appear because $x^y$, as a function of two continuous variables, is discontinuous at (0,0) and takes three different values depending on the direction of approach to the discontinuity:

1. Fixing y=0, we have $x^0=1$ for all $x \neq 0$. (Proof: $x^0 = x^{(1-1)} = x^1 x^{-1} = x/x = 1$, each statement holding for all $x \neq 0$). Indeed, $x^y \rightarrow 1$ as $x \rightarrow 0$, approaching from left or right, with y=0. (This was Euler’s reason.)
2. Fixing x=0, we have $0^y=0$ for $y >0$. (When y < 0 we have division by zero which is undefined in the reals and $+\infty$ in the extended reals). Taking limits, $x^y \rightarrow 0$ as $y \searrow 0$, approaching from above only, with x=0.
3. Fixing x=0, we have an undefined value when y < 0 due to division by zero.

Notice that the discontinuity is not a simple (point) discontinuity, but rather a pole discontinuity due to the approach from below. (Exercise: what happens as the origin is approached from 45 degrees?)

#### Principles for a Decision in Mathematics: Extension and Consistency

In mathematics, when there is more than one choice, a decision is typically made by extending an existing precedent to maintain consistency with the evidence that is already accumulated and accepted.

An elementary example is the way ordinary multiplication is extended from two positive numbers to a positive and a negative number, then to two negative numbers, i.e. $(-1)(-1) = 1$.

“Minus times minus is plus.
The reason for this we need not discuss!”
— W.H. Auden

Empirically, multiplication of two positive numbers has a well-defined, tangible meaning as repeated addition. This meaning holds when one of the numbers is negative. But when both are negative, the empirical meaning fails.

For the mathematician, declaring something to be undefined (throwing an error) means a loss of efficiency because every instance now has to be checked for the undefined case, and this must be treated separately. If a definition could be found that remains consistent with all other empirically obtained rules, and if that definition means that calculation can proceed indifferent to the decision, then that is a big win.

The consistency in this particular case is the distributivity of multiplication over addition, a law which, for positive numbers, can be accepted on entirely empirical grounds. (See the footnote for the full argument.1.)

#### Turning to Discrete Mathematics – Consistency with the Binomial Theorem

In discrete mathematics, there is no notion of “approaching” — one is either at $0^0$ or away from it, in which case $1^0 = 1$ or $0^1 = 0$.

The case of $0^0$ can be decided on consistency grounds with respect to the binomial theorem, i.e. loss of computational efficiency to have to treat this case separately. This is the argument of Knuth (of The Art of Computer Programming, and TeX fame), based on maintaining consistency with the binomial theorem $(a+b)^x$ when x=0, due to its fundamental place in both discrete and continuous mathematics:

“Some textbooks leave the quantity $0^0$ undefined, because the functions $0^x$ and $x^0$ have different limiting values when $x$ decreases to 0. But this is a mistake. We must define $x^0=1$ for all $x$ , if the binomial theorem is to be valid when $x=0, y=0$ , and/or $x=-y$. The theorem is too important to be arbitrarily restricted! By contrast, the function $0^x$ is quite unimportant.”
– from Concrete Mathematics, p.162, R. Graham, D. Knuth, O. Patashnik, Addison-Wesley, 1988

#### Different Conventions Among Mathematical Computing Platforms

Given the universality of the $0^0=1$ convention amongst mathematicians, it is surprising to find that various computing platforms have implemented different values:

• Value one: Google Calculator, R, Octave, Ruby, and Microsoft’s Calculator all give $0^0 = 1$.
• Value zero: Hexalon Max (calculator) and a physical TI-36 hand calculator give $0^0 = 0$.
• Value indeterminate: Maxima and Microsoft Excel (2000) give $0^0$ is indeterminate, i.e. an error is thrown.

#### An Alternative Decision Criteria – tangible computation with verifiable count that requires the answer

While Knuth’s argument of convenient extension works, the finite summation of integer powers provides us with a real, tangible result (a finite sum), whose value (an empirically determinable fact) depends unavoidably on the chosen value of $0^0$. So here we have a consistency argument that does not rely on efficiency.

The crucial step in this argument occurs in the derivation of (*1b) from (*1a) in Finite Summation of Integer Powers, Part 2.

Extracting the relevant part of that derivation, we have:

$\sum_{k=1}^{N-1} (N-K)(K+1)^{p-1}\ \ \ \ \ (*1a)$

After expanding the binomial power using the binomial formula and further manipulation, we arrive at:

$= \sum_{j=0}^{p-1} \binom{p-1}{j} \sum_{k=0}^{N-1} \left[NK^j -K^{j+1}\right]$
(Pull the $K=0$ term out of both summations. Note: $0^0 = 1\ \ \ \ \ \mbox{(***)}$)
$= N + \sum_{j=0}^{p-1} \binom{p-1}{j} \sum_{k=1}^{N-1} \left[NK^j -K^{j+1}\right]$
$= N + N(N-1) - S_p(N-1) + \sum_{j=1}^{p-1} S_j(N-1)\left[N\binom{p-1}{j} - \binom{p-1}{j-1}\right]$
$= N^2 - S_p(N-1) + \sum_{j=1}^{p-1} S_j(N-1)\left[N\binom{p-1}{j} - \binom{p-1}{j-1}\right]\ \ \ \ \ \mbox{(*1b)}$

The key step happens in (***) above: we peel off the $K=0$ term of the inner summation to get: $N0^j - 0^{j+1}$. Peeling this out of the outer summation requires considering the expression for all $j$. Now, 0 raised to any positive power is 0, so we can dispel the case of $j>0$. But what is $0^0$? A decision must be made: it is either $0$ or $1$. Indeterminacy is not an option, since the situation is real and is required to continue the simplification.

#### The Argument for $0^0 = 1$

What are the consequences of choosing the other definition, i.e. $0^0 = 0$? In this case, the final formula for $S_p(N)$ is off by a linear constant $N$, while the choice $0^0 = 1$ leads to the exact formula and a computed value that matches a brute force summation.

For $S_5(10) = 1^5+2^5+3^5+4^5+5^5+6^5+7^5+8^5+9^5+10^5$, the difference is between 220,825 (the correct, verifiable answer), and 220,815 (verifiably NOT correct). The correct definition is clear: 0^0 = 1 is for empirical reasons that have to do with counting and summing. While it is the binomial theorem that provides the detail, the argument is one of verifiable necessity and not one of consistency.

For discrete mathematics, the empirical evidence shows that 0^0=1 is required:2

References
The Math Forum

(If you’re a software developer of a mathematical package, I’d be interested in how you arrived at your decision. You can send me an email using the Comments link below.)

#### Footnotes

1. Considering that any quantity times zero is zero, and that one times any quantity is the quantity, we have no hesitation in granting $-1 \times 0 = 0$. But then observe that we way write $0 = (-1 + 1)$, which means, combining the two expressions, we have $-1 (1 + (-1)) = 0$. If we accept the law of distribution of multiplication over addition for positive whole numbers, purely on empirical grounds, and if we wish negative numbers to behave in the same manner as our empirically accepted positive whole numbers, then we want the distributive law to hold as well. And therefore we have $0 = (-1)(1) + (-1)(-1) = -1 + (-1)(-1).$ Which means that $(-1)(-1)$ must be the oppositive (additive inverse) of $(-1)$, and hence $(-1)(-1) = +1.$
2. The implications for continuous mathematics are a consideration for another discussion. The statement that a discontinuity exists at the origin $(x,y) = (0,0)$ isn’t quite enough.

### 8 comments to Why Zero Raised to the Zero Power IS One

• John

I am performing a polynomial regression analysis with arbitrary powers and found that 0^0 must be 1 for it work correctly. I am having to use the IF function in Excel to get around its #NUM! error for 0^0, i.e., =IF(B1=0,1,A1^B1) instead of simply =A1^B1.

P.S. My TI-36X Pro calculator gives Domain Error for 0^0. NARS2000 APL gives 1 for 0^0.

Hi Sid, your argument, as I understand it, has two flaws. To see this clearly, we must first rewrite it to make the logic explicit. This boils it down to 3 lines: #1,4,5 — the others don’t add anything.

Sid’s argument:
“We will assume that x^0 = 1 for all numbers x and show that this leads to a contradiction.
We know that 1 > -1.
But taking 0th powers of both sides and apply the assumption does not preserve the inequality, i.e. 1^0 = 1 = (-1)^0.
So the assumption is shown not to hold for any x, so in particular 0^1 cannot be 1.”

The two (fatal) flaws are:

1) the first line is in fact a provably true statement when x is non-zero. (Proof: x^(1-1) = x^1 x^-1 = x/x = 1, each of which holds for all x != 0)

2) taking powers does not preserve inequality (Consider 1 > -1 but 1^2 is not > than (-1)^2.) So even a failure here would not generate the desired contradiction.

Hope that helps. (But good effort in writing up an argument.)

• Sid Dasari

I am still a high school student (Algebra II) so please correct me If I’m incorrect in the steps of any of the math I do*

Wouldn’t any number^0 be 0″? As it wouldn’t make sense if not?
For example
Let’s look at the statement
0+1>0-1
Now let’s say one decides to put this in parentheses
(0+1)>(0-1)
Now what if we make both sides (parenthesis) have to be to the power of 0
(0+1)^0>(0-1)^0
Which in turn simplifies to
(1)^0>(-1)^0
Making
1>1?
This can’t be true no matter what
1=1
But if x^0=0 then this statement would be valid, right?

• Anonymous

X^0=1
If x =0, it would be undefined due to people arguing about whether or not it should be 0 or 1.
So calculators can’t show 2 results and it can’t find out the answer, thus the errors
We should stop this fighting though, 0^0=1 makes not sense.
0^0=/= 0*0 which may cause confusion
X^1=X
X^0=1
But
0^0 is 0*unknown
So it’s should be 0

• It is a basic principle in mathematics that consequences of widely accepted theorems should be accepted as well. The binomial theorem, as well as the empty product rule, power series notation, combinatorial arguments, and cardinal exponentiation, all imply the same value 0^0 = 1. Hence, if we want mathematics to be consistent, then 0^0 must be 1.

It is sometimes claimed that defining 0^0=1 once and for all could lead to contradictions. This claim is wrong. It is based on a general distrust of 0, combined with faulty reasoning, such as: (a) arguments that require the impossible (requiring a discontinuous function to be continuous) or (b) arguments that use erroneous claims such as 0^x = 0 for all x (to see that this claim is wrong, try x=-1).