Recent content by seismichills

If ##(x,z) \in R_{1}## then ##(x,w) \in R_{1};R_{2}## and ##(z,w) \in R_{2}## for some ##w \in S##. But ##R_{1};R_{2}=I## so ##x=w\,.## Now, suppose ##(y,z) \in R_{1}## then ##(y,w)\in R_{1};R_{2}## since ##(z,w) \in R_{2}\,.## Hence ##y=w=x\,.## Same argument for ##R_{2}\,.## Now to prove...

Hey, thanks for your comments, everyone. I've figured out how to do it so I'll post the solution here

<Moderator's note: Moved from a technical forum and thus no template. Also re-edited: Please use ## instead of $$.> If ##R_{1}## and ##R_{2}## are relations on a set S with ##R_{1};R_{2}=I=R_{2};R_{1}##. Then ##R_{1}## and ##R_{2}## are bijective maps ##R_{1};R_{2}## is a composition of two...