If $f$ and $g$ are commutative operations $$\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R},$$ then for any constants $a,b \in \mathbb{R}$, the system of equations $$f(x,y) = f(a,b), \qquad g(x,y) = g(a,b)$$ must include $(x,y) = (a,b)$ and $(x,y) = (b,a)$ among its solutions.

I conjecture that if $f$ is given by $f(x,y) = xy$, and if $g$ can be expressed as $g(x,y) = G(x) + G(y)$ for some $G : \mathbb{R} \rightarrow \mathbb{R}$ that's continuous and injective, then the above system has only these solutions.

The injectivity requirement in particular is designed in particular to block $G(x) = x^2,$ which would break the theorem if it were allowed, while allowing the choices $G(x) = x^3$ and $G(x) = \mathrm{tan}^{-1}(x),$ which seem to be consistent with the theorem (based on graphical considerations).

Here's a graphical example illustrating the case $a = 2, b = 3, G(x) = x^3$:

enter image description here

Question. Is this true? If not, what would be a counterexample? If so, how might we prove it?

  • $\begingroup$ The term "commutative function" is rather unusual : Do you mean $f(x,y)=f(y,x)$ for all $x,y$ ? $\endgroup$ – Jean Marie May 9 at 11:57
  • $\begingroup$ @JeanMarie, yep. I've just changed it. $\endgroup$ – goblin May 9 at 11:59

It's not true.


$$G(x)= \begin{cases} x, & \text{for } x \le 4 \\ \frac12x+2, & \text{for } x > 4. \\ \end{cases} $$

This is a monotonically increasing, continous (check $x=4$ on both 'branches') function, thus injective.

Especially note that

$$G(2)=2, G(3)=3, G(4)=4, G(6)=5$$

and thus $g(2,6)=7=g(3,4)$, and of course $2\times6=3\times4$.


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