Does R^2 has the same property as R?

Question

If R is a relation on set A, define $R^2$ by $aR^2b$ if and only if there exists c with aRc and cRb. If R is reflexive/symmetric/transitive does $R^2$ have the same property ?

I'm not sure how to do this question? Any help is appreciated!

Answer 1

I'll be simple-minded and try to prove it.

Reflexive: Since $aRa$, $aR^2a$ immediately.

Symmetric: If $aRb$ then $bRa$, we want if $aR^2b$ then $bR^2a$. If $aR^2b$ there is a $c$ with $aRc$ and $cRb$. Since $R$ is symmetric, just turn this around to get $bRc$ and $cRa$ so $bR^2a$.

Transitive: We want if $aR^2b$ and $bR^2c$ then $aR^2c$. From $aR^2b$ and $bR^2c$, there are $d$ and $e$ such that $aRd$, $dRb$, $bRe$, and $eRc$.

Since $R$ is transitive, if $aRd$ and $dRb$ then $aRb$; if $bRe$, and $eRc$ then $bRc$. Therefore, since $aRb$ and $bRc$, $aR^2c$.

A variation on this: if $R^2$ is an equivalence relation, is $R$ also? My guess is not, but I could be wrong.

Answer 2

If $R$ is reflexive then $R^2$ is also reflexive trivially. Suppose $aR^2b$ and $R$ is symmetric:

$aR^2b\rightarrow$ there exists a $c$ such that $aRc$ and $cRb$. But $R$ is symmetric so $bRc$ and $cRa$ therefore $bR^2a$.

Now if $aR^2b$ and $bR^2c$ and $R$ is transitive:

$aR^2b\rightarrow$ there is an $e$ , $aRe$ and $eRb$ so by transitivity of $R$ we have $aRb$. Similarly $bRc$ and therefor $aRc$.Which by definition of $R^2$ implys $aR^2c$.