# What is exponentiation?

Last time, we explored how to extend multiplication from an operation on positive integers to an operation on arbitrary real numbers. The three tools we had at our disposal were:

1. Find a nice interpretation for the operation, and see how broadly it applies.
That’s how we moved from multiplying positive integers to multiplying any real number by a positive integer.
2. Write down the laws the operation satisfies, and try to use them to deduce extra cases where the operation must give a certain value.
For example, the law of associativity let us deduce the result of multiplying by a rational number, and the law of distributivity gave us the result of multiplying by zero or a negative number.
3. Use continuity, monotonicity, or other “nice” properties of functions to decide what the operation should give for even more values.
Knowing how to multiply by rational numbers doesn’t tell you how to multiply by irrational ones, but there’s at most one way to do so that makes the multiplication operation continuous.

Today we’ll explore these ideas for exponentiation instead of multiplication, and the story has a twist that finally explains why $(a^2)^{1/2}$ doesn’t always equal $a$.

1. Interpretations of exponentiation

The way most of us were introduced to exponentiation was as repeated multiplication, just as multiplication is repeated addition, so the power $5^3$ means $5\times 5\times 5$, or $125$. This interpretation applies if the base is any real number, as long as the exponent is a positive integer, so already we’ve extended exponentiation to an operation

\begin{aligned}&\mathbb{R}\times \mathbb{N}_{>0} \to \mathbb{R}\\ &(a,n)\mapsto a^n:= a\times\dots\times a\text{ (}n\text{ times),}\end{aligned}

where $\mathbb{R}$ is the set of all real numbers and $\mathbb{N}_{>0}$ is the set of positive integers.

In the rest of this post, we’ll look for other “exponential operations” $B\times E\to \mathbb{R}$, where $B$ is a set of possible bases and $E$ is a set of possible exponents.

There’s also another, less well-known interpretation of exponentiation, involving counting problems. The number $10^6$ is not just one million, it’s the number of six-digit strings $000000$ through $999999$. In other words, it’s the number of ways to make $6$ choices from the $10$-element set $\{0,\dots,9\}$. More generally, if $S$ is an $n$-element set and $T$ is an $m$-element set, then $m^n$ is the number of functions from $S$ to $T$, just like a six-digit string is a function from the set $\{1,\dots,6\}$ to $\{0,\dots,9\}$; the first digit of the string is where the function sends $1$, the second digit is where the function sends $2$, and so on.

The nice thing about this interpretation is that the mathematical definition of a function $S\to T$ (a subset of $S\times T$ that contains one ordered pair for each element of $S$, each with that element as its first entry) applies even when $S$ or $T$ is empty—or both! If $S$ is empty, there’s always exactly one function from $S$ to $T$: the “empty” function with no ordered pairs in it. This means that $m^0$ always equals $1$, including the special case $0^0 = 1$. On the other hand, if $T$ is empty but $S$ isn’t, there are no functions $S\to T$, because such a function would have to contain some ordered pairs in $S\times T$ but there aren’t any to choose from. This means that $0^n =0$ as long as $n> 0$.

In effect, this gives us an exponential operation

\begin{aligned} &\mathbb{N} \times \mathbb{N} \to \mathbb{N}\\ &(m,n) \mapsto m^n := \text{ \# of functions }\{1,\dots,n\}\to\{1,\dots,m\},\end{aligned}

where $\mathbb{N}$ is the set of nonnegative integers.

These two interpretations have given us two different exponential operations: the first lets us have arbitrary real numbers as bases; the second lets us have $0$ as an exponent. If we want, we can combine these into a single exponential operation

$\mathbb{R} \times \mathbb{N} \to \mathbb{R}$

just by declaring that $a^0$ should always equal $1$, which makes sense with $1$ as the “empty product.” What other cases of exponentiations can we define?

2. Laws of exponents

From the two definitions of exponentiation we already have, we can abstract out the following three pairs of laws:

1. $a^m\cdot a^n = a^{m+n}$ and $a^0 = 1$.
2. $(a^m)^n = a^{m\cdot n}$ and $a^1 = a$.
3. $a^n\cdot b^n = (a\cdot b)^n$ and $1^n = 1$.

We will define an exponential operation to be a function $B\times E\to \mathbb{R}$ obeying the above laws, where $B\subseteq \mathbb{R}$ is a set of allowed bases and $E\subseteq\mathbb{R}$ is a set of allowed exponents. Our goal is to find the exponential operations that are defined on as broad a range of input values as possible. But before we get started trying to find such extended exponential operations, here are a few observations we can make already:

• For the third law to make sense, $B$ can’t be just any subset of $R$; we need $1\in B$ and $a\cdot b\in B$ whenever $a,b\in B$. (This makes $B$ what’s called a “submonoid of $\mathbb{R}$ under multiplication.”)
• Similarly, for the first and second pairs of laws to make sense, we also need to have $0,1\in E$ and $m+n,m\cdot n\in E$ whenever $m$ and $n$ are in $E$. In particular, $E$ must contain at least the nonnegative integers. (In general, these facts make $E$ a “subrig” of $\mathbb{R}$.)
• The second law also tells us that every output of the exponential operator must also be an allowed base, in order for combinations like $(a^m)^n$ to make sense. This means that we’re not just looking for functions $B\times E\to \mathbb{R}$, we’re looking for functions $B\times E\to B$.

Now what values of $a^n$ can we deduce just from the three pairs of exponential laws?

• We know $E$ must contain the set of nonnegative integers, and in fact, for those values of $n$ we can deduce directly from the laws that $a^n$ must always have the value given by multiplying $a$ together $n$ times. Indeed, either $n=0$, in which case we have the law $a^0 = 1$, or we can write $n = 1+\dots+1$ with $n$ $1$‘s, in which case

$a^n = a^{1+\dots+1} = a^1\times\dots\times a^1 = a\times\dots\times a$ with $n$ $a$‘s.

So if $E=\mathbb{N}$, we can make $B$ all of $\mathbb{R}$. This gives us the exponential operation $\mathbb{R}\times\mathbb{N}\to\mathbb{R}$ we had before. Are there any other values of $a^n$ we can deduce from the exponential laws?

We can deduce that $a^{-1}$ must always be a multiplicative inverse for $a$, since $a^{-1}\cdot a = a^{-1}\cdot a^1 = a^{-1+1}=a^0=1$. But this means that if all real numbers are allowed bases, we can’t allow $-1$ to be a possible exponent, or we would be required to produce a multiplicative inverse for $0$.

We can also deduce that $a^{1/2}$ must always be a square root of $a$, since by the second law we must have $(a^{1/2})^2 = a^{(1/2)\cdot 2} = a^1 = a$. But again, this means that if we want to allow all real numbers as bases, we cannot allow $1/2$ to be a possible exponent, or else we would be required to produce square roots of negative numbers.

However, every real number has a unique cube root, and a unique fifth root, and a unique seventh root, etc., and we can use those to define $a^{1/3}$, $a^{1/5}$, and $a^{1/7}$, etc., for any real number $a$. In fact, we can define $a^{m/n}$ for any real number $a$, and any nonnegative rational number $m/n$ with $n$ odd, by setting $a^{m/n}$ equal to the $n$th root of $a^m$. Denoting the set of nonnegative rational numbers with odd denominator by $\mathbb{N}_{(2)}$, this gives us an exponential operation with $B=\mathbb{R}$ and $E=\mathbb{N}_{(2)}$:

$\mathbb{R}\times\mathbb{N}_{(2)}\to \mathbb{R}$

There aren’t any more exponents we can add to $E$ while still keeping $B=\mathbb{R}$: having $0$ as a possible base rules out negative exponents, and having negative numbers as possible bases rules out rational exponents with even denominators. (I don’t think you can deduce any specific cases of irrational powers directly from the exponential laws, but I’d love to be proven wrong in the comments!)

However, by ruling out those troublesome bases, we can write down some exponential operations with more possible exponents:

• If we rule out $0$ as a base, we can define $a^{-n} = 1/a^n$ and obtain an exponential operation

$\mathbb{R}_{\neq 0} \times \mathbb{Z}_{(2)} \to \mathbb{R}_{\neq 0},$

where $\mathbb{Z}_{(2)}$ is the set of all rational numbers (positive, negative, or zero) that can be written with odd denominators.

• If we rule out negative numbers as bases, then we can define $a^{1/2}$ as a square root of $a$. Note that since we’re ruling out negative numbers as bases, we’re ruling out negative numbers as powers too, so $a^{1/2}$ must be the nonnegative square root of $a$. Using the same root-of-power procedure as before, this lets us deduce the value of any nonnegative rational power of $a$: just let $a^{m/n}$ be the $n$th root of $a^m$. Thus we get an exponential operation

$\mathbb{R}_{\geq 0} \times \mathbb{Q}_{\geq 0} \to \mathbb{R}_{\geq 0}.$

• If we do both and restrict just to positive bases, we can use arbitrary rational powers, so we get an exponential operation

$\mathbb{R}_{>0} \times \mathbb{Q} \to \mathbb{R}_{> 0}.$

So we’ve ended up with several exponential operations, and the larger we want to make our set of bases, the smaller we have to make our set of exponents, and vice versa.

3. Extending by continuity

Our last tool is to try to pin down more outputs of our exponential operations by asking that they be continuous. We’ll investigate each function of the type $x\mapsto a^x$ to see if we can extend it to new exponents.

• For a positive base $a$, we’ve defined $a^x$ for all rational numbers $x$, and indeed this gives a continuous function $\mathbb{Q}\to \mathbb{R}_{>0}$ that we can extend to a continuous function $\mathbb{R}\to \mathbb{R}_{>0}$. The result is an exponential operation $\mathbb{R}_{>0}\times\mathbb{R}\to\mathbb{R}_{>0}$!
• For the case $a=0$, we’ve defined $0^x$ for all nonnegative rational numbers $x$. The result is $1$ if $x=0$ and $0$ if $x>0$—unfortunately that’s already discontinuous at $0$. But everywhere else, we can extend by continuity to get $0^x=0$ for all positive real numbers $x$. This gives us an exponential operation $\mathbb{R}_{\geq 0}\times\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$.
• If the base $a$ is negative, its possible exponents are rational numbers with odd denominator. Can we fill in any values for the missing real exponents between them? No: this function is already discontinuous everywhere. Because $a^{m/n}$ is positive if $m$ is even and negative if $m$ is odd, the function $x\mapsto a^x$ is constantly flipping between comparably sized positive and negative values, and no definition of $a^x$ for irrational $x$ can make that continuous.

All in all, we’ve ended up with four exponential operations, giving us four contexts in which $a^n$ is always defined and the the resulting exponential operation satisfies the exponential laws:

1. When $a$ is a positive real number and $n$ is any real number. Then $a^n$ is positive, like $a$.
2. When $a$ and $n$ are both nonnegative real numbers. Then $a^n$ is too. (This includes the special case $0^0 = 1$.)
3. When $a$ is a nonzero real number and $n$ is a rational number with odd denominator. Then $a^n$ is nonzero, like $a$.
4. When $a$ is an arbitrary real number and $n$ is a nonnegative rational number with odd denomeinator.

These definitions of $a^n$ all agree when they overlap, and indeed, the fourth context doesn’t add anything new once you have the first three. So why are we bothering with contexts and sets of allowed bases and exponents at all—why not just declare that $a^n$ is defined for these values of $a$ and $n$ and be done with it?

Because that resulting function doesn’t obey the exponential laws.

The reason is because $\sqrt{a^2} = |a|$ instead of $a$. When we write down something like $((-3)^2)^{1/2}$, on the one hand the exponential laws say it should equal $(-3)^{2\cdot 1/2} = (-3)^1 = -3$, but on the other hand the result we’re forced to compute is $((-3)^2)^{1/2} = (9)^{1/2} = \sqrt{9}= 3$.

One way of thinking about this is that $((-3)^2)^{1/2}$ is mixing the different contexts: to form $(-3)^2$, that negative base means we must be in context 3 or 4, so we’re guaranteed never to violate the exponential laws as long as we stick to nonzero bases and rational odd-denominator exponents, or to arbitrary bases and nonnegative rational odd-denominator exponents. In particular, that rules out using another exponent of $1/2$, which is only allowed in contexts 1 and 2.

Another way of thinking about it is that the exponential operations don’t form a sheaf: you can’t patch them together to make a bigger one, even if they agree where they overlap. Instead, to use the exponential laws carefully you must think about what values of base and exponent you are generally allowing, not just whether what you’re writing down is well-defined.

And this is only the story for real exponentiation—there’s also complex exponentiation, cardinal exponentiation, ordinal exponentiaton… Quite a tricky subject for just being repeated multiplication, eh?