Updated! February 5, 2017
The value of zero raised to the zero power, 
Context of the Debate: Continuous Mathematics
The three choices for the value of 
- Fixing y=0, we have
for all
. (Proof:
, each statement holding for all
as
, approaching from left or right, with y=0. (This was Euler’s reason.)
- Fixing x=0, we have
for
. (When y < 0 we have division by zero which is undefined in the reals and
in the extended reals). Taking limits,
as
, approaching from above only, with x=0.
- 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. 
“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 
The case of 
“Some textbooks leave the quantity
and
have different limiting values when
decreases to 0. But this is a mistake. We must define
, and/or
. The theorem is too important to be arbitrarily restricted! By contrast, the function
– 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 
- Value one: Google Calculator, R, Octave, Ruby, and Microsoft’s Calculator all give
.
- Value zero: Hexalon Max (calculator) and a physical TI-36 hand calculator give
.
- Value indeterminate: Maxima and Microsoft Excel (2000) give
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
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:
After expanding the binomial power using the binomial formula and further manipulation, we arrive at:
(Pull the 
(which, after additional manipulation, yields)
![= \sum_{j=0}^{p-1} \binom{p-1}{j} \sum_{k=0}^{N-1} \left[NK^j -K^{j+1}\right]](https://s0.wp.com/latex.php?latex=%3D+%5Csum_%7Bj%3D0%7D%5E%7Bp-1%7D+%5Cbinom%7Bp-1%7D%7Bj%7D+%5Csum_%7Bk%3D0%7D%5E%7BN-1%7D+%5Cleft%5BNK%5Ej+-K%5E%7Bj%2B1%7D%5Cright%5D&bg=ffffff&fg=000000&s=0 "= \sum_{j=0}^{p-1} \binom{p-1}{j} \sum_{k=0}^{N-1} \left[NK^j -K^{j+1}\right]")
![= N + \sum_{j=0}^{p-1} \binom{p-1}{j} \sum_{k=1}^{N-1} \left[NK^j -K^{j+1}\right]](https://s0.wp.com/latex.php?latex=%3D+N+%2B+%5Csum_%7Bj%3D0%7D%5E%7Bp-1%7D+%5Cbinom%7Bp-1%7D%7Bj%7D+%5Csum_%7Bk%3D1%7D%5E%7BN-1%7D+%5Cleft%5BNK%5Ej+-K%5E%7Bj%2B1%7D%5Cright%5D&bg=ffffff&fg=000000&s=0 "= 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]](https://s0.wp.com/latex.php?latex=%3D+N+%2B+N%28N-1%29+-+S_p%28N-1%29+%2B+%5Csum_%7Bj%3D1%7D%5E%7Bp-1%7D+S_j%28N-1%29%5Cleft%5BN%5Cbinom%7Bp-1%7D%7Bj%7D+-+%5Cbinom%7Bp-1%7D%7Bj-1%7D%5Cright%5D&bg=ffffff&fg=000000&s=0 "= 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)}](https://s0.wp.com/latex.php?latex=%3D+N%5E2+-+S_p%28N-1%29+%2B+%5Csum_%7Bj%3D1%7D%5E%7Bp-1%7D+S_j%28N-1%29%5Cleft%5BN%5Cbinom%7Bp-1%7D%7Bj%7D+-+%5Cbinom%7Bp-1%7D%7Bj-1%7D%5Cright%5D%5C+%5C+%5C+%5C+%5C+%5Cmbox%7B%28%2A1b%29%7D&bg=ffffff&fg=000000&s=0 "= 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 
The Argument for
What are the consequences of choosing the other definition, i.e. 
For 
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.)
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Footnotes
- Considering that any quantity times zero is zero, and that one times any quantity is the quantity, we have no hesitation in granting
. But then observe that we way write
, which means, combining the two expressions, we have
. 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
Which means that
must be the oppositive (additive inverse) of
, and hence
↩
- The implications for continuous mathematics are a consideration for another discussion. The statement that a discontinuity exists at the origin
isn’t quite enough. ↩
0^0 must be indeteriminate. The binomial equation is not that important. The future of computer science is, and setting it equal to 1 simply upsets it.
I’m still trying to understand the whole thing, as described above. One thing I don’t understand is why you towards the top wrote Sigma E’ with b above, k=a below, f(k) to the right, and then a little bit later in your writing, moved the small numbers to the right of the E’. Sorry if this makes me seem obtuse, but are you just writing it another way with the top and bottom numbers just to the right of the sigma symbol corresponding to those that were above and below it before, or does this new style of formula mean something else?
@Jeffrey: Not sure which line you’re referring to. Sounds like you might be confused by the typographical arrangement? The sigma notation in mathematics just means sum up the argument (expression to the right of the sigma) a certain number of times, each time with the next value of the index variable, in this case k running from a to b. Whether the indices are directly below the sigma, or just to the right makes no difference. Hope this helps.
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).
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
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?
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.)
Assad–
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.
0/0 is undefined.
By Law of Exponent, 0/0 = 0^0
So, if 0/0 is undefined, 0^0 is also undefined.
Any number except zero, when you raised to zero is one.
That’s what my simple mind thinks.
If you are willing to stick to simple ways of explaining this, let n be a very small number that I will have approach 0 at a “consistent rate”—something that is not true when we look at various functions approaching 0 such as e^-n as n approaches a large number.
Now, N^n and N^-n = 1/N^n will both have N^n approaching 0 at the same rate and then are always “the same number”. Thus, 0/0 is not undetermined because it continues it ratio of n/n always the same as n approaches 0. That to me is consistency rather than relying on the very nebulous 0/0 when we don’t know the rate at which we reached 0 for the numerator compared to the numerator for the denominator.
Calculators and computers have no way of distinguishing one zero from the next because they have no idea of the origin for deriving that particular 0. Remember, 0 is an infinitely small dot and so what matters is how fast you are sliding into that particular infinitely small space because what happens just outside 0 really does make a difference on the comparative value of that specific zero. Ignore that difference at your own peril and confusion!
Follow on conclusion: 0^0 = 1 as long as you recognize your 0 in each position as being derived by the same limiting rate.
Harold, yes, your argument using limiting ratios is sound when dealing with continuous variables. This is essentially L’Hopital’s argument from first year calculus (L’Hopital’s Rule). However, in discrete mathematics and combinatorics, we are dealing with integer functions, so we don’t have the benefit of limiting values. In such cases how should we define 0^0? That’s where the definition by convention appears. The convention is chosen because it allows a general law to be expressed without having to call out 0 as an edge case.
Thanks for the comment!
Assad.