Addition and multiplication are the two classic commutative, associative binary operations on the reals. They satisfy a striking property: they are equivalent up to unary operations. By taking a logarithm of the arguments to addition and exponentiating the result, we get multiplication:
$xy = \exp(\log x + \log y)$
In fact, we can create a new commutative, associative operation from every invertible unary function $f$ in the same way:
$(x,y) \mapsto f^{-1} (f(x) + f(y))$
I actually haven't been able to come up with a commutative, associative operation that doesn't follow this form. Are all commutative, associative operations on the reals isomorphic to addition in a similar way?
EDIT – The answers and comments below give several good counterexamples to the problem as stated. Requiring $f$ to be invertible is too strong a condition. What if we allow this slightly more general functional form
$(x,y) \mapsto g (f(x) + f(y))$
where $f$ maps reals to some field and $g$ maps that field to the reals? This seems to capture the counterexamples. For example:
$(x, y) \mapsto 0$ matches with arbitrary $f$ and $g(z) = 0$
$(x, y) \mapsto x + y – \lfloor{x+y}\rfloor$ matches with $f(x) = x$ and $g(z) = z – \lfloor{z}\rfloor$
I think $(x, y) \mapsto max(x,y)$ matches with $f(x) = (x + a)^n$ and $g(z) = z^{1/n} – a$, where $a, n$ are constants we take to infinity. That choice is inspired by the relationship between the max function and the infinity norm. I get that including limits in $f$ and $g$ is pretty suspect, but even with that hack, it's interesting to me that it's possible.
Modifying the question slightly, can every commutative, associative binary operation be cast in the form $(x,y) \mapsto g (f(x) + f(y))$?
Best Answer
You can consider an operation that does not give the structure of a group. For instance $$x\circ y\ \colon = \max(x,y)$$ Here we have $x \circ x = x$. In fact we have a lattice.
But even if the operation would give a commutative group, it may not be isomorphic to $\mathbb{R}$ with addition. For instance $\mathbb{R}/\mathbb{Z}$ is one such group of the same cardinality as $\mathbb{R}$.
You can check that an abelian group with the same cardinatlity as $\mathbb{R}$ is isomorphic to $\mathbb{R}$ with addition if an only if it is a uniquely-divisible group: for every $x$ and $n\ge 1$ integer, there exist a unique $y$ such that $ny = x$. Since then it is a vector space over $\mathbb{Q}$. However, a concrete isomorphism may not be available without some axiom of choice things. For instance, $\mathbb{R}/\mathbb{Q}$ with addition.