Why are addition and multiplication commutative, but not exponentiation?

Question

We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not. My question is why.

As background there are plenty of mathematical schemes that can be used to define these operators. One of these is hyperoperation where
$H_0(a,b) = b+1$ (successor op)
$H_1(a,b) = a+b$ (addition op)
$H_2(a,b) = ab $ (multiplication op)
$H_3(a,b) = a^b$ (exponentiation op)
$H_4(a,b) = a\uparrow \uparrow b$ (tetration op: $a^{(a^{(...a)})}$ nested $b$ times )
etc.

Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$

Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?

Thanks.

Answer 1

Here is a thought, which is not a full answer, but too long for a comment.

Addition $a + b$ means something like: Add $1$ to $a$, $b$ times $= 1\cdot b + a$. Commutativity here means that $1\cdot b + a = 1\cdot a + b$. We can see that it's only through the fortunate use of $1$ that this is commutative; $cb + a \not= ca+b$ in the general case. If we defined successorship to be an increment of two, addition would no longer be commutative!

With multiplication, $a \cdot b $ means add $a$ to itself, $b$ times $= \sum_{i=1}^b a$. It is only through the fact that addition is "hyper-commutative" (i.e. $\sum_{i=1}^b a = \sum_{i=1}^a b$) that multiplication is commutative.

Suppose I define an operation $+_2$ meaning "addition where successorship increases by two" as suggested above. $a+_2 b = 2a + b$. In this case, multiplication loses its commutativity, since $a +_2 a +_2 \dots +_2 a$ $b$ times is not the same as $b +_2 b +_2 \dots$ a times.

So to invert your question: why is it that addition and multiplication are commutative? It seems like it's because of the fortunate choice of successorship.

To stretch it further, we can view multiplication as a weird form of addition. In general, $ab = (a-1)b + b = b +_{a-1} b$. Plugging that in to, say, $b^2$, we would find $b^2=b +_{b-1} b$. So exponentiation is not commutative because its "successor" is not 1. This also explains why exponentiation is commutative for e.g. $2^2$, since $2-1=1$, and so we have a unit successor. It's more complex for $b^b$, but I think the principle holds.

Long story short: it's because "1" is special. Hopefully this is of some interest, even though it's not as "intuitive" as I might like.

---

EDIT: Perhaps the most interesting point in that long story is: if $H_0\not= +1$ then no operations are commutative.

Answer 2

My personal opinion is that the exponential is not naturally regarded as the next step in the progression from addition to multiplication, so there's no reason to expect it to share properties with the other two.

Notice what happens if you demand that all of your quantities have units. Addition is an operation you do to two quantities with the same units: for example, you can add two distances to get another distance. Multiplication is an operation you do to two quantities with units $a$ and $b$ to get a quantity with units $ab$; for example, you can multiply two distances to get an area.

However, you cannot exponentiate two unital quantities. In computing $a^b$, the quantity $b$ needs to be unitless or else there's no sensible way to assign a value to the result. This tells us that $a$ and $b$ are being treated very differently, so there's no reason to expect that switching them is a physically meaningful operation.

As I've discussed on math.SE before, the exponential has many generalizations in mathematics, and in most of them the base and the exponent are very different objects: in fact, in most of them the base is just $e$.

Answer 3

When I first read your question, I expected that it must mean that addition would possess some obscure property that multiplication lacks, after all, both the additive structure and multiplicative structure are abelian groups, so you'd expect something like this to just generalize. But after some thinking, I realized that this wasn't the case, and instead that the problem is that we aren't generalizing properly.

For if we define "applying an operator $f$, $n$ times, ie $f^n$" as the recursive procedure $ f^n(x) = \begin{cases} x & \text{if n = 0} \\ f^1(f^{n - 1}(x)) & \text{otherwise} \end{cases} $

Then this definition actually uses addition, so if we'd want to generalize this procedure properly, we'd need to change our definition of "applying an operator $n$ times" as well. And indeed $a^n$ does equal $(a^2)^{n / 2}$, which induces a better generalization of commutativity.

Answer 4

This might be slightly circular since the definition of exponentiation and logarithm comes after addition and multiplication, but consider: If we perform a certain unary operator on the two arguments, perform addition on them, and then take the inverse, we get $e^{\log a + \log b} = ab$. If we do the same with multiplication, $e^{\log a \log b} = a^{\log b} = b^{\log a}$, which is not exponentiation.