Does anyone have a convincing story about why exponentiation isn't commutative?

Like, what happens here:
a + b: repeat "add 1" b times
a × b: repeat "add a" b times
a^b: repeat "times a" b times

Are there other sequences of operations built by repeating the previous one that are all commutative?

· · Web · 4 · 2 · 1

@pkra I could have started with "in Peano arithmetic", but I don't know if the answer depends on that

@christianp there I was, so proud that I didn't immediately write "how is multiplying real-valued matrices like 'adding a, b times'?"

i think you are not strict enough about the grammar

lets define two operations. add with two arguments and times with two arguments. Now let the only numbers we have be 0 and 1 (really not that important here but for completeness sake)

a + b is add (times 1 a) (times 1 b)
a * b is times (times 1 a) (times 1 b)
observe that up to this point we are able to express the statements above in our grammar
a^b however is not expressible in our limited grammar, except if we define a new operation (or add ambiguity to the times operation - which is what happened in your post);

@christianp well it is kind of expressible but not in a linear manner. The expression grows with b, so i have to write an new statement for each value of b - which was not the case with * and +. Sorry for being unclear about that.

@christianp No but it sounds interessting from what i can tell by reading the wikipedia article. How do you think does the Grzegorczyk Hierachie relate to your question?

@christianp I got nothing. But you have hit on a key part of what drives me _up a wall_ in number theory. These 'number' things... they are not well behaved. And as soon as you have three operations in this confounded sequence (it takes three, not two, so this isn't a 'field' thing per se, it's something more and weirder), it goes all to pickles.

something to do with Category theory, coproducts, products, Hom sets and that C^{op} is not isomorphic to C ?

Obviously the coproduct A+B is isomorphic to B+A and the product AxB is to BxA

but A^B is the functions from B -> A and that is not the same as the functions A->B
because in general C^{op} is not isomorphic to C

I dunno .. out of my depth!
Love to hear what you discover.

@kat I don't know enough category theory to know if you've just repeated my question back at me! 😆 A common feeling when commutative diagrams start appearing.

I feel like I'm doing the same thing in each step: starting from either 0 or 1, repeat the previous one b times. Why does that mean each step produces a different picture?

Sign in to participate in the conversation

The social network of the future: No ads, no corporate surveillance, ethical design, and decentralization! Own your data with Mastodon!