How do we know that tetration is exclusively right-associative?

Question

When we go from multiplication to exponentiation we lose commutativity ($3^2 \neq 2^3$). Perhaps when we go from exponentiation to tetration every operation yields two possible results.

Answer 1

You have a serious but not uncommon confusion about what mathematical notation is. Mathematical objects have interesting properties, and we want to study them. We then invent notation that helps us describe the objects, and reason about the objects. But the notation is not itself the object. It is invented, for particular purposes, by people, to serve those purposes. We do not reason from the behavior of the notation, any more than we try to understand the behavior of an animal by trying to reason about its taxonomic classification.

If you have a theory that Tony Blair is an octopus, you do not then wonder why he has only two arms, and there is no point in philosophizing about how something could be an octopus and yet have only two arms. You instead conclude from the missing arms that your theory is wrong. You can reason from Blair himself and his lack of arms that he is not properly classified as an octopus; you cannot reason in the reverse direction, that because Blair is an octopus, octopuses sometimes have only two arms. The arms are real. Octopuses are real. Our classification of Tony Blair as an octopus is artificial. Mathematical notation is similarly artificial.

There is a convention that the expression $$a^{b^c}$$ means $$a^{\left(b^c\right)}$$ and does not mean $$\left(a^b\right)^c.$$ When we say that exponentiation is "right-associative" we are referring to this convention.

But there is nothing about this convention that is intrinsic to the exponentiation operation itself. It is only a rule about how certain notation will be understood, made up for convenience and utility.

$\def\sup#1#2{\vphantom{#1}^{#2}#1}$ We could just as easily adopt the convention that $$\sup ab$$ henceforth means the same as $a^b$, except that $$\sup a{\sup bc}$$ is defined to mean $$\sup{\left(\sup ab\right)}c$$ and not $$\sup a{\left(\sup bc\right)}.$$ Then instead of $$2^{2^3} = 256$$ we would have $$\sup 2{\sup 23} = 64.$$ Again there is nothing subtle going on here; we are just defining a new notation. It is not even a particularly useful notation, because as it happens $$\sup a{\sup bc} = \sup a{(bc)} = a^{(bc)}$$ so the new notation is not letting us express anything that we could not express before, and this is why we don't bother to do this. Not because of some deep underlying truth, but because it isn't useful to do so.

$\def\bt#1{\underbrace{#1}_{b\text{ times}}}$ Tetration is the same way. We define some tetration operator, say $a\uparrow b$, to mean $\bt{a^{a^{\cdots^a}}}$, where $\bt{a^{a^{\cdots^a}}}$ is understood in the usual way, as $\bt{a^{\left(a^{\left(\cdots^a\right)}\right)}}$. We could define it differently, to mean $\bt{\left(\left(a^a\right)^a\right)^{\cdots a}}$, but then it wouldn't gain us anything new, because we already have a simple notation for $\bt{\left(\left(a^a\right)^a\right)^{\cdots a}}$: $$\bt{\left(\left(a^a\right)^a\right)^{\cdots a}} = a^{a(b-1)}.$$ So we do it the other way, the way that gets us something new, because there is no point in having a second notation for the same thing.

Similarly you could define $a\uparrow b$ to have two values, the values of $\bt{a^{a^{\cdots^a}}}$ and of $\bt{\left(\left(a^a\right)^a\right)^{\cdots a}}$, but what would this get you? You would just find yourself dealing to one or the other, and it would be easier to deal with one at a time. Then you could dispense easily with the second one by pointing out that it was equal to $a^{a(b-1)}$, which we already understand, and so concentrate on the new one, the first one—which is what we do.