# If relation $R^2$ is symmetric, R does not need to be symmetric

I was checking a question that said the following

If $$R^2$$ is symmetric, does that mean that R is symmetric?

This being R a relation on the set $$A$$.

What I thought the proof would be is:

$$aR^2c \land cR^2a$$ This because $$R^2$$ is symmetric.

From the definition of $$R^2$$:

$$aRb \land bRc \iff aR^2c$$ $$bRa \land cRb \iff cR^2a$$

Here we can see that R is symmetric $$aRb \land bRa$$ $$bRc \land cRb$$

However... this is not true... it does not mean that R is symmetric (I can give an example).

What did I miss in my proof?

• $R^2$ being symmetric does not mean that $aR^2c\land cR^2a$ for arbitrary $a,c\in A$. It means that if $aR^2c$, then $cR^2a$. Equivalently, it means that if $a,c\in A$, then either $aR^2c\land cR^2a$, or $a\not R^2c\land c\not R^2a$. – Brian M. Scott Apr 11 '20 at 14:48
$$R^2$$ being symmetric does not mean that $$aR^2c\land cR^2a$$ for arbitrary $$a,c\in A$$. It means that if $$aR^2c$$, then $$cR^2a$$. Equivalently, it means that if $$a,c\in A$$, then either $$aR^2c\land cR^2a$$, or $$a\not R^2c\land c\not R^2a$$.