Why are addition and multiplication commutative, but not exponentiation?

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
$H_0(a,b) = b+1$ (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 )

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?



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 $e$.

Source : Link , Question Author : rob levin , Answer Author : Community

Leave a Comment