Can a self-inverse function $y=f(x)$ always be expressed as an equation that is symmetric in $x$ and $y$? For example, the simple self-inverse function $y = 6 - x$ can be written as $x+y=6$, which is symmetric in $x$ and $y$.  Less apparent (to me at least), $y = (x+1)/(1-x)$ can be expanded and written as $x + y = xy - 1$, which is also symmetric in $x$ and $y$, and must therefore be self-inverse. Can all self-inverse functions be expressed this way?
 A: The answer is yes. First, notice that $f$ must satisfy $f:I\subset\mathbb{R}\to I\subset\mathbb{R}$, since it is its own inverse. The best way of justifying it is by looking at the $graph$ of the function, i.e., the set
$$G = \{(x,f(x))\in\mathbb{R}^2:x\in I\}=\{(f(y),y)\in\mathbb{R}^2:y\in I\}$$
This equality holds, again, because $f$ is its own inverse.
Therefore, the graph of $f$ is symmetric, that is to say, it satisfies $G=\{(y,x)\in\mathbb{R}^2:(x,y)\in G)\}$ (it is immediate to see it from the above equality).
Hence, if $G$ can be expressed as the set of points that follow an equation, $G=\{(x,y)\in\mathbb{R}^2:g(x,y)=0\}$, this equation $g(x,y)=0$ must be symmetric in the variables $(x,y)$ (again, it is immediate to see it since $G$ is symmetric)
If you have any problems seeing what I said, don't hesitate to write a comment.
A: No, because most functions, even most self-inverse ones, cannot be expressed by any equation at all.  We can choose any subset $A$ of $\Bbb R^{\gt 0}$ and define $$f_A(x)=\begin {cases} x&|x| \in A\\-x&|x| \not \in A \end {cases}$$
There are uncountably many subsets, each of which defines a self-inverse function.  Most of them cannot be expressed as there are only countably many ways to express such a function.
