I wrote up these notes about a month ago and just found them. I could really use a second opinion or two. I notice there's a whole question about how my whole thesis is wrong, but that doesn't automatically mean it's without merit. That said, forgive my layman's take on everything.
Addition is a special case of multiplication. That follows from the whole set theory interpretation where units are indistinct in addition, but have types in multiplication.
By that logic, you could also argue that multiplication is a special case of exponentiation. After all, $(a^b)^c=a^{bc}$, and $a^b \cdot a^c = a^{b+c}$.
So, it's interesting; if you ignore the base, if you're taking a power, you're multiplying, as in the first case; and if you're multiplying, you're adding, as in the second case.
What about if you're adding? It'd be nice if it turned out to be $a^{b+1}$, but I don't see how that's easily compatible. Adding $1$ seems to require $a^b \cdot a=a^b \cdot a^{a^0}=a^{b+1}$.
And on the other side, what operation gives exponentiation? Tetration?
Hmm. Maybe adding requires another level of implicit base below everything. Even if it doesn't "exist" in a useful sense, other than to paint a more complete logical framework, or to serve as an intermediary tool.
Putting this into action, addition is still just exponentiation, just with one deeper level of implicit base. Asking what's below addition is meaningless unless we allow that addition is just a label we've applied to that level of abstraction.
We'll appropriate the symbol $\oplus$ for this hypothetical operation, and let $\gamma$ serve as an exponential base. From powers on down, the four operations we've got now, reframed as exponentiation, look like:
$$ \begin{align} x\textrm{ ^ }y&\Rightarrow&& x^y \\ x\times y&\Rightarrow&& (\alpha^x)^y = \alpha^{xy} \\ x+y&\Rightarrow&& (\beta^{\alpha^x})^{\alpha^y} =\beta^{\alpha^x \cdot \alpha^y} =\beta^{\alpha^{x+y}} \\ x\oplus y&\Rightarrow&& (\gamma^{\beta^{\alpha^x}})^{\beta^{\alpha^y}} =\gamma^{\beta^{\alpha^x} \cdot \beta^{\alpha^y}} =\gamma^{\beta^{(\alpha^x + \alpha^y)}} =\gamma^{\beta^{\alpha^{x \oplus y}}} \end{align} $$
One actual effect of this is we can infer that $x \oplus y$ is arbitrarily close to $\max(x,y)$. This is using $x\oplus y := \log_\alpha (\alpha^x+\alpha^y)$, but there are other possible interpretations of the extrapolation above. Mine was just an illustrative guess.
But if we go ahead with the implication, I'm tempted to assign a real base value for more detail. More specifically, if the base is $e$, it's very close to $\max(x,y)$; if the base is $-e$, it's very close to $-\min(x,y)$.
Upon some reflection though, I think the correct choice of base is $\alpha=2$, because then if $x=y$, you get $x \oplus y = x+1=y+1$, the successor function, which is what one expects for this level of hyperoperation. I acknowledge it's unusual to have a binary successor operation, it being famously unary and all. However, rather than break it, I think we've arguably improved it. By the same token, it might be reasonable to replace all the undefined base variables above with $2$s.
Note we have $x \oplus 0=\log_2 (2^x+1)$. As with every binary operation, all else being equal, the closer the two operands are to each other, the more the value is maximized.
As an aside, it is awfully easy to represent everything computationally as exponentiation (even if it is just wrapping operations up in logs and powers). Observe:
xTimes[a_,b_]:=Log2[(2^a)^b]
xTimes[3,5]
-> 15
xPlus[a_,b_]:=Nest[Log2,(2^2^a)^2^b,2]
xPlus[3,5]
-> 8
xSucc[a_,b_]:=Nest[Log2,(2^2^2^a)^2^2^b,3]
N@xSucc[-100,-50]
-> -50.
N@xSucc[-5,19]
-> 19.
xSucc[-5,-5]
-> -4
xSucc[4,4]
-> 5
FYI - the Nest function $\mathsf{nest}(f,n,k):=f(f(\ldots f(f(n))\ldots \textrm{k times} \ldots))$
So generally, I'm looking for responses like: a) "interesting ideas, worth considering further", or b) "crackpot idea, go read a book". I did find a post alluding to a well-regarded professor that seems to have had a similar thought, although I'm still trying to get details on that.