Assume that the function $f(x,y)$ is not symmetric in $x,y$.

Consider the curve $f(x,y)=0$ in $\mathbb{R}^2$. Is it always true that $f(a,b)=f(b,a)=0$ only if $a=b$?

It does appear to be true when I draw an arbitrary curve $f(x,y)$ in $\mathbb{R}^2$. However, I'm unable to prove this. I also remember seeing an easy counter-example in a similar situation, and want to make sure that I am not making a mistake.

  • $\begingroup$ $f(x,y) = x+y^2-1$. $\endgroup$
    – WimC
    Commented Feb 24 at 20:42
  • $\begingroup$ $f(x,y)=|x|-|y|$ $\endgroup$
    – DARK
    Commented Feb 24 at 20:44

2 Answers 2


Consider the function $f(x,y)=(x+y-1)(y-2x)$. It is not symmetric in $x,y$. $(1,0)$ and $(0,1)$ both are on the plot of $f$, but $0\neq 1$.


Here is a simple counterexample:

Let $f(x,y)=x\cdot y(x-1)$. Then,



but $a\ne b$.


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