Why is addition and multiplication associative but not exponentiation and beyond?

Question

The most similar question I found on math.se was concerning commutativity (Why are addition and multiplication commutative, but not exponentiation?). If this ostensibly simple (but not obvious to me) question has already been discussed, please post the link and I apologize for duplicate. I am wondering too how other rules of arithmetic in basic and other algebras (distributivity, transitivity, additive and multiplicative inverses, etc) apply or don't apply to operators and any rhyme to their properties' applicability or hierarchy.

associativity of addition: (α+β)+γ ≡ (α+β+γ) := α+β+γ ≡ α+(β+γ)

associativity of multiplication: (a×b)×c := (ab)⋅c = abc ≡ a(bc)

non-associativity of exponentiation: A^B^C := (A^B)^C := $(A^B)^C$ = ${A^B}^C$ ≢ $A^{B^C}$ = $A^{(B^C)}$ := A^(B^C)

Answer 1

The real question is why $\times$ would associate; properties like that aren't free.

As with the question of commutativity, that $\times$ inherits the property from $+$, but exponentiation doesn't repeat this, can be explained in terms of $\times$ on $\Bbb R^+$ being isomorphic to $+$ on $\Bbb R$. (Distributivity of $\times$ over $+$ extends this to $\times$ on the whole of $\Bbb R$ being commutative & associative.)

Answer 2

Hmmm.... Well, lets think....

in the natural numbers $a+b$ is defined to be $a + \underbrace{1+1+....+1}_b$ and so $a+(b+c) = a+\underbrace{\underbrace{1+1+.....+1}_b + \underbrace{1+1+....+1}_c}_{b+c}=\underbrace{a+\underbrace{1+1+.....+1}_b}_{a+b} + \underbrace{1+1+....+1}_c=(a+b) + c$ which by the construction of natural numbers is consistent and meaningful

(googles Peano Postulates: https://en.wikipedia.org/wiki/Peano_axioms mumbles to self Hmm.... induction...seems to work)

And multiplication defined as $a\times b = \underbrace{b+b+ .....+b}_a$ so

$a\times (b\times c) = \underbrace{b\times c + b\times c+ .... + b\times c}_a=$

$\underbrace{\underbrace{c+c+....c}_b + \underbrace{c+c+....c}_b+..+\underbrace{c+c+....c}_b}_a=$

$\underbrace {c+c+c+c+ .....+ c+c}_{\underbrace{b+b+...+b}_a} =\underbrace{c+c+c+c+...+c+c}_{a\times b}=$

$(a\times b)\times c$.

(mumble,mumble so far so good.)

Now $a^b$ is defined as $\underbrace{a\times a\times ...\times a}_b$ and things just wont work.

We can see that $b^a\times b^c = \underbrace{b\times b....\times b}_a\times \underbrace{b\times b....\times b}_c=\underbrace{b\times b\times ....\times b}_{a+c}=b^{a+c}$,

And that $(b^a)^c = \underbrace{b^a\times b^a\times ..... \times b^a}_c =$

$\underbrace{\underbrace{b\times b\times... \times b}_a\times\underbrace{b\times b\times... \times b}_a\times....\times \underbrace{b\times b\times... \times b}_a}_c=$

$\underbrace{b\times b\times .... \times b}_{\underbrace{a + a+....+a}_c = c\times a}=b^{c\times a}$

but from there it just breaks down.

$(a^b)^c = a^{b\times c}=\underbrace{a\times a\times ....\times a}_{b\times c}$

whereas $a^{(b^c)}=\underbrace{a\times a\times .....\times a}_{b^c}$.

And multiplying $a$ by itself by two different number of times result in different numbers (we haven't proven that it always results in different numbers (although if $a \ne 0$ that will be the case) but we have no reason to think it would result in the same an we can find many examples where they don't) and that $b^c$ need not equal $b\times c$ (ditto about proving it... for all but a finite number of cases and plenty of counter examples) they just don't work.

Actually though.... As addition was defined as simply listing $1$ out first $a$ times, then $b$ times; and multiplication by listing $a$ out $b$ times, they are naturally associative. If we defined an operation by doing the same thing repetitively that would be associative. For example if we defined $a Eb$ as $a^{(a^{(a^{.....a)})})_{b\text{ times}}}$ that would be associative.