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 $(x,y) \mapsto g (f(x) + f(y))$?

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    $\begingroup$ I think you have a counterexample hiding right in front of you: what exactly happens with negative numbers or zero under your claimed isomorphism between addition and multiplication? $\endgroup$ Jan 12 at 22:21
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    $\begingroup$ max is commutative and associative, but not solvable. $\endgroup$
    – user694818
    Jan 12 at 22:22
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    $\begingroup$ what is the domain of $f$ meant to be? in your example, the domain of $f$ is only $\mathbb{R}_{>0}$, not all of $\mathbb{R}$. anyway, denoting your binary operation by $\bullet$, you want an injective function $f$ on some domain such that $f(x\bullet y)=f(x)+f(y)$ for any $x,y$. what about the operation $\bullet:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ given by $x\bullet y=0$? there is no injective function $f:\mathbb{R}\to\mathbb{R}$ satisfying what you want here, since we would need $f(0)=f(x)+f(y)$ for any $x,y\in\mathbb{R}$. in particular, $f(x)=0$ for any $x\in\mathbb{R}$ (why?) $\endgroup$ Jan 12 at 22:38
  • $\begingroup$ In the power set $\mathcal{P}(\Bbb{N})$ we have the binary commutative, associative operation $\Delta$ $$S \Delta R=(S\setminus R)\cup (R\setminus S).$$ That is, the symmetric difference of the subsets. This operation turns $\mathcal{P}(\Bbb{N})$ into a group. Furthermore, it has the property $S\Delta S=\emptyset$ for all $S$. We have a bijection between $\mathcal{P}(\Bbb{N})$ and $\Bbb{R}$.Similar transportation of structure gives you a binary commutative group operation on $\Bbb{R}$ with the extra property that $x\Delta x=0$ for all $x\in\Bbb{R}$. Therefore it cannot come from addition. $\endgroup$ Jan 12 at 22:47
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    $\begingroup$ Addition modulo one, that is, $a*b=\{\,a+b\,\}$, the fractional part of $a+b$. $\endgroup$ Jan 12 at 22:50

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.


Edit: The new version of your question is, I think, much more general than you intended; the answer is yes, due to the following cheap trick, but it doesn't really have anything to do with addition anymore.

Let $m : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be any commutative operation (we will not need associativity, and operations of the form $g(f(x) + f(y))$ will not be associative in general). We want to find $g, f$ such that $m(x, y) = g(f(x) + f(y))$. You say that $f$ can take values in "some field." We'll take this field to be the function field $K = \mathbb{Q}(f_x, x \in \mathbb{R})$ over $\mathbb{Q}$ on uncountably many variables, one for every real number, and $f : \mathbb{R} \to K$ will be the function $f(x) = f_x$. This guarantees that $f(x) + f(y) = f(z) + f(w)$ iff either $x = z$ and $y = w$ or $x = w$ and $y = z$; in other words, the value of $f(x) + f(y)$ uniquely determines the unordered pair $\{ x, y \}$.

Now we can take $g : K \to \mathbb{R}$ to be any function such that $g(f_x + f_y) = m(x, y)$; since $f_x + f_y$ uniquely determines $\{ x, y \}$ and $m$ is commutative this is well-defined. Of course we have not used any features of $\mathbb{R}$ in this argument, which can be replaced by an arbitrary set.

(This responds to the original question only.)

Your statement about addition and multiplication is not quite right; you only get an isomorphism between $(\mathbb{R}, +)$ and $(\mathbb{R}_{+}, \times)$ (where the subscript denotes taking positive reals). However, it is true that addition and multiplication are locally isomorphic in a suitable sense.

Are all commutative, associative operations on the reals isomorphic to addition in a similar way?

As stated, since you haven't asked for compatibility with any other structure on the reals, you are just asking for commutative monoids with the same cardinality as the reals. There is an enormous diversity of such things, even if we restrict further to abelian groups (which means requiring inverses); as a random example, any countable product $\prod_{i=1}^{\infty} A_i$ of at-most-countable but nontrivial abelian groups has the same cardinality as the reals, for example $\prod_{i=1}^{\infty} \mathbb{Z}$, but such a group is isomorphic to the reals as an abstract group iff each $A_i$ is an at-most-countable-dimensional $\mathbb{Q}$-vector space.

However, we can get a positive result by requiring continuity.

Proposition: Every continuous abelian group structure on $\mathbb{R}$ (with its usual topology) is continuously isomorphic to addition.

This came up recently on MathOverflow. It follows from Gleason-Montgomery-Zippin's solution to Hilbert's fifth problem, which is almost certainly overkill; probably it follows from an easier and known structure theorem for locally compact (Hausdorff) abelian groups but I couldn't quite finish the argument. Maybe it can also be done "by hand."

(Edit: This is overkill, as expected. There is a totally elementary argument that a topological group (not necessarily abelian!) homeomorphic to $\mathbb{R}$ must be isomorphic to $(\mathbb{R}, +)$; see this math.SE question.)

If you further require smoothness then the proof is easier and can probably be done "by hand" more easily; abstractly, $\mathbb{R}$ is the unique $1$-dimensional Lie algebra and hence the unique $1$-dimensional simply connected Lie group, so:

Corollary: Any smooth abelian group structure on $\mathbb{R}$ is smoothly isomorphic to addition.

A related result you may be interested in is the classification of $1$-dimensional formal group laws; over a $\mathbb{Q}$-algebra, and in particular over $\mathbb{R}$, they are all isomorphic to the additive formal group law.

A question I don't know the answer to is whether it suffices to require measurability; that is, I don't know whether a measurable abelian group structure on $\mathbb{R}$ (with the Borel $\sigma$-algebra, say) is measurably isomorphic to addition.

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    $\begingroup$ It can be shown "by hand" that every continuous group law $*$ on $\mathbf R$ (for its usual topology) is essentially addition: $x*y = f^{-1}(f(x) + f(y))$ for some homeomorphism $f \colon \mathbf R \to \mathbf R$. The details are a bit tedious; one useful result is "Hion's lemma". Concerning your question about measurability of a group law, see mathoverflow.net/questions/120738/…. $\endgroup$
    – KCd
    Jan 13 at 1:10
  • $\begingroup$ Thanks for the detailed response! Would you mind explaining the notation 𝐾=ℚ(𝑓𝑥,𝑥∈ℝ) ? $\endgroup$
    – mrfish
    Jan 13 at 20:06
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    $\begingroup$ @mrfish: it's a function field (en.wikipedia.org/wiki/Algebraic_function_field) of rational functions in uncountably many variables, one for each real number. This might be slightly overkill; I just need to be working somewhere big enough that $f(x) + f(y)$ uniquely determines $\{ x, y \}$. I might be able to get away with something like the field of meromorphic functions instead but then it'd be slightly less obvious that $g$ exists. $\endgroup$ Jan 13 at 20:21

"Isomorphic" refers to entire mathematical structure, not to an operation in that system. One could ask whether, given any set $S$ and an operation $*$, whether $(S,\star)$ is isomorphic to $(\mathbb R, +)$, but the answer would be a rather trivial; for instance, $(\mathbb C,+)$ is not isomorphic to $(\mathbb R, +)$. If you want to restrict it to $(\mathbb R, \star)$ being isomorphic to $(\mathbb R, +)$, that doesn't really change anything essential; $\mathbb C$ has the same cardinality as $\mathbb R$, so there exists bijections between them. Take any such bijection $\phi: \mathbb R \rightarrow \mathbb C$, and we can define $a\star b$ for real numbers $a, b$ as $a\star b = \phi^{-1}(\phi(a)+ \phi(b))$ , and $(\mathbb R, \star)$ will not be isomorphic to $(\mathbb R, +)$.


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