    Next: Name for f(x)^(f(x)) = Up: Special Numbers and Functions Previous: Euler's formula: e^(i pi)

# What is 0^0

According to some Calculus textbooks, 0^0 is an ``indeterminate form''. When evaluating a limit of the form 0^0, then you need to know that limits of that form are called ``indeterminate forms'', and that you need to use a special technique such as L'Hopital's rule to evaluate them. Otherwise, 0^0 = 1 seems to be the most useful choice for 0^0. This convention allows us to extend definitions in different areas of mathematics that otherwise would require treating 0 as a special case. Notice that 0^0 is a discontinuity of the function x^y. More importantly, keep in mind that the value of a function and its limit need not be the same thing, and functions need not be continous, if that serves a purpose (see Dirac's delta).

This means that depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent.

Some people feel that giving a value to a function with an essential discontinuity at a point, such as x^y at (0,0), is an inelegant patch and should not be done. Others point out correctly that in mathematics, usefulness and consistency are very important, and that under these parameters 0^0 = 1 is the natural choice.

The following is a list of reasons why 0^0 should be 1.

Rotando & Korn show that if f and g are real functions that vanish at the origin and are analytic at 0 (infinitely differentiable is not sufficient), then f(x)^(g(x)) approaches 1 as x approaches 0 from the right.

From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik):

Some textbooks leave the quantity 0^0 undefined, because the functions x^0 and 0^x 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.
Published by Addison-Wesley, 2nd printing Dec, 1988.

As a rule of thumb, one can say that 0^0 = 1, but 0.0^(0.0) is undefined, meaning that when approaching from a different direction there is no clearly predetermined value to assign to 0.0^(0.0); but Kahan has argued that 0.0^(0.0) should be 1, because if f(x), g(x) --> 0 as x approaches some limit, and f(x) and g(x) are analytic functions, then f(x)^g(x) --> 1.

The discussion on 0^0 is very old, Euler argues for 0^0 = 1 since a^0 = 1 for a != 0. The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitschrift für Mathematik und Physik. Consensus has recently been built around setting the value of 0^0 = 1.

On a discussion of the use of the function 0^(0^x) by an Italian mathematician named Guglielmo Libri.

[T]he paper  did produce several ripples in mathematical waters when it originally appeared, because it stirred up a controversy about whether 0^0 is defined. Most mathematicians agreed that 0^0 = 1, but Cauchy [5, page 70] had listed 0^0 together with other expressions like 0/0 and oo - oo in a table of undefined forms. Libri's justification for the equation 0^0 = 1 was far from convincing, and a commentator who signed his name simply ``S'' rose to the attack . August Möbius  defended Libri, by presenting his former professor's reason for believing that 0^0 = 1 (basically a proof that lim_(x --> 0+) x^x = 1). Möbius also went further and presented a supposed proof that lim_(x --> 0+) f(x)^(g(x)) whenever lim_(x --> 0+) f(x) = lim_(x --> 0+) g(x) = 0. Of course ``S'' then asked  whether Möbius knew about functions such as f(x) = e^(-1/x) and g(x) = x. (And paper  was quietly omitted from the historical record when the collected words of Möbius were ultimately published.) The debate stopped there, apparently with the conclusion that 0^0 should be undefined.

But no, no, ten thousand times no! Anybody who wants the binomial theorem (x + y)^n = sum_(k = 0)^n (n k) x^k y^(n - k) to hold for at least one nonnegative integer n must believe that 0^0 = 1, for we can plug in x = 0 and y = 1 to get 1 on the left and 0^0 on the right.

The number of mappings from the empty set to the empty set is 0^0. It has to be 1.

On the other hand, Cauchy had good reason to consider 0^0 as an undefined limiting form, in the sense that the limiting value of f(x)^(g(x)) is not known a priori when f(x) and g(x) approach 0 independently. In this much stronger sense, the value of 0^0 is less defined than, say, the value of 0+0. Both Cauchy and Libri were right, but Libri and his defenders did not understand why truth was on their side.

 Anonymous and S... Bemerkungen zu den Aufsatze überschrieben, `Beweis der Gleichung 0^0 = 1, nach J. F. Pfaff', im zweiten Hefte dieses Bandes, S. 134, Journal für die reine und angewandte Mathematik, 12 (1834), 292--294. uvres Complètes. Augustin-Louis Cauchy. Cours d'Analyse de l'Ecole Royale Polytechnique (1821). Series 2, volume 3.

 Guillaume Libri. Mémoire sur les fonctions discontinues, Journal für die reine und angewandte Mathematik, 10 (1833), 303--316.

 A. F. Möbius. Beweis der Gleichung 0^0 = 1, nach J. F. Pfaff. Journal für die reine und angewandte Mathematik,

12 (1834), 134--136.

 S... Sur la valeur de 0^0. Journal für die reine und angewandte Mathematik 11, (1834), 272--273.

##### References

Knuth. Two notes on notation. (AMM 99 no. 5 (May 1992), 403--422).

H. E. Vaughan. The expression '0^0'. Mathematics Teacher 63 (1970), pp.111-112.

Kahan, W. Branch Cuts for Complex Elementary Functions or Much Ado about Nothing's Sign Bit, The State of the Art in Numerical Analysis, editors A. Iserles and M. J. D. Powell, Clarendon Press, Oxford, pp. 165--212. \

article Louis M. Rotando and Henry Korn.The Indeterminate Form 0^0. Mathematics Magazine,Vol. 50, No. 1 (January 1977), pp. 41-42.

L. J. Paige,. A note on indeterminate forms. American Mathematical Monthly, 61 (1954), 189-190; reprinted in the Mathematical Association of America's 1969 volume, Selected Papers on Calculus, pp. 210-211.

Baxley & Hayashi. A note on indeterminate forms. American Mathematical Monthly, 85 (1978), pp. 484-486.

Crimes and Misdemeanors in the Computer Algebra Trade. Notices of the American Mathematical Society, September 1991, volume 38, number 7, pp.778-785

## Why is 0.9999... = 1?

In modern mathematics, the string of symbols 0.9999... is understood to be a shorthand for ``the infinite sum 9/10 + 9/100 + 9/1000 + ...''. This in turn is shorthand for ``the limit of the sequence of real numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, ...''. Using the well-known epsilon-delta definition of the limit (you can find it in any of the given references on analysis), one can easily show that this limit is 1. The statement that 0.9999... = 1 is simply an abbreviation of this fact.

0.9999... = sum_(n = 1)^(oo) (9)/(10^n) = lim_(m --> oo) sum_(n = 1)^m (9)/(10^n)

Choose varepsilon > 0. Suppose delta = 1/- log_(10) varepsilon , thus varepsilon = 10^(-1/delta). For every m > 1/delta we have that

sum_(n = 1)^m (9)/(10^n) - 1 = (1)/(10^m) < (1)/(10^(1/delta)) = varepsilon

So by the varepsilon - delta definition of the limit we have

lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1

Not formal enough? In that case you need to go back to the construction of the number system. After you have constructed the reals (Cauchy sequences are well suited for this case, see [Shapiro75]), you can indeed verify that the preceding proof correctly shows 0.9999... = 1.

An informal argument could be given by noticing that the following sequence of ``natural'' operations has as a consequence 0.9999... = 1. Therefore it's ``natural'' to assume 0.9999... = 1.

```
x = 0.9999....
10 x = 10 * 0.9999...
10 x = 9.9999...
10 x - x = 9.99999... - 0.9999...
```

Thus 0.9999... = 1.

An even easier argument multiplies both sides of 0.3333... = 1/3 by 3. The result is 0.9999... = 3/3 = 1.

Another informal argument is to notice that all periodic numbers such as 0.46464646... are equal to the period divided over the same number of 9s. Thus 0.46464646... = 46/99. Applying the same argument to 0.9999... = 9/9 = 1.

Although the three informal arguments might convince you that 0.9999... = 1, they are not complete proofs. Basically, you need to prove that each step on the way is allowed and is correct. They are also ``clumsy'' ways to prove the equality since they go around the bush: proving 0.9999... = 1 directly is much easier.

You can even have that while you are proving it the ``clumsy'' way, you get proof of the result in another way. For instance, in the first argument the first step is showing that 0.9999... is real indeed. You can do this by giving the formal proof stated in the beginning of this FAQ question. But then you have 0.9999... = 1 as corollary. So the rest of the argument is irrelevant: you already proved what you wanted to prove.

##### References

R.V. Churchill and J.W. Brown. Complex Variables and Applications. 5^(th) ed., McGraw-Hill, 1990.

E. Hewitt and K. Stromberg. Real and Abstract Analysis. Springer-Verlag, Berlin, 1965.

W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, 1976.

L. Shapiro. Introduction to Abstract Algebra. McGraw-Hill, 1975.    Next: Name for f(x)^(f(x)) = Up: Special Numbers and Functions Previous: Euler's formula: e^(i pi)

Alex Lopez-Ortiz
Fri Feb 20 21:45:30 EST 1998