*Oh, the ambiguity**! *

**Introduction**

Most of us learned in high
school that 0^{0} is somehow undefined or ambiguous. In college or
university, your calculus professor will confirm this, citing the ambiguity
resulting from different limits. We have Lim (*x* à 0): *x*^{0} = 1. But we also have Lim
(*y* à 0): 0* ^{y}* = 0. There is an obvious
discontinuity in the function

Your algebra professor, on the
other hand, may tell you that you can assume that 0^{0} = 1 on the
natural numbers--for convenience mostly. They may justify it with analogies to
various conventions, e.g. usually the convention of so-called empty
products--the product of *no* numbers?
Many simply define it to be 1. It’s apparently not something you can actually
prove. As for 0^{0} being undefined on the real numbers, and
exponentiation being entirely consistent in both domains, that is a *mere coincidence*. We are talking about *entirely different* functions here, they
will say. If that strikes you as being just a bit too, well, “hand wavy” for
your liking, read on!

Here, I will develop the
exponentiation function on the natural numbers with 0^{0} undefined,
given only the operations of addition and multiplication on *N. *I*
*use what I believe to be a novel approach that looks at all possible
functions that satisfy the usual requirements for an exponentiation function on
*N*. In so doing, we can justify
leaving 0^{0} undefined, as it is on the set of real numbers *R*. I will also look at some implications
for the usual laws of exponents on *N*
for undefined 0^{0}.

**Exponentiation Defined as
Repeated Multiplication on N**

When you were first introduced to
exponents in elementary or high school, you probably started with the exponents
greater than or equal to two. After all, you need at least 2 numbers to
multiply. For all *a* ε *N*, we have:

a^{2}=a.a

a^{3}=a.a.a=a^{2}.a

a^{4}=a.a.a.a=a^{3}.a

a^{5}=a.a.a.a.a=a^{4}.a

and so on.

This infinite sequence of
equations can be recursively summarized in just two equations for all *a, b* ε *N* as follows:

1. *a*^{2} = *a. a*

2. *a ^{b}*

These two equations, by themselves, do not, however, tell us
anything about exponents 0 or 1. It turns out that there are infinitely many
such exponent-like functions on *N*
that satisfy these equations. Proof (21 lines)

Fortunately, these infinitely
many functions differ *only* in the value
assigned to 0^{0}. Proof (194 lines)

This suggests that, in our
definition of exponentiation on N, we should simply leave 0^{0}
undefined. To this end, we can construct (i.e. prove the existence of) a unique
partial function
for exponentiation on *N*. Proof
(618 lines)

Thus we can define
exponentiation on *N* as follows:

1.
*a ^{b}* ε

2.
0^{1} = 0

3.
*a*^{0} = 1 (for a ≠ 0)

4.
*a ^{b}*

**The Laws of Exponents on N
for undefined 0^{0}**

Using the above definition, we
can derive the 3 Laws of Exponents on *N*:

1.
The Product of Powers Rule: *a ^{b}*.

2.
The Power of a Power Rule: *(**a ^{b})^{c}*
=

3. The Power of a Product Rule: (*a.b*)* ^{c}*
=

Interestingly, these restrictions would *not* apply if 0^{0}
was defined to be either 0 or 1. So, adding these laws to the requirements for
exponentiation would narrow down the infinitely many possibilities to only two.
But we would *still* be left with *some* ambiguity—is 0^{0}
equal to 0 or 1? *Oh, the ambiguity!*

**My Recommendation**

If, as in most well used,
centuries-old textbook applications where an informal proof will usually
suffice, it is convenient and *probably*
safe to assume 0^{0}=1 on the natural numbers, if not on the reals. If,
however, you are pushing the limits of number theory and only a formal proof
will do, you should take the extra care and avoid assuming any particular value
for 0^{0} in *both* the real
numbers *and* the natural numbers.
There are easy work-arounds in many cases. Applications of the binomial theorem to
obtain the usual expansions of (*x*+*y*)* ^{n}*, for example, will result
in terms including 0