# Thread: Having Trouble with Relation Proof

1. ## Having Trouble with Relation Proof

So, I'm working on a relation proof, which is as follows.

Let < be the usual relation on Z. Prove that < o < is not equal to <.

So, I know that I want to prove that one is not a subset of another. And, I feel like multiplication can somehow be involved in this, but I'm not sure.

Please help guys. I'm retaking this class, and aside from set theory, this doesn't make much sense to me.

2. ## Re: Having Trouble with Relation Proof

By definition, $(x, z)\in S\circ R$ if there exists a $y$ such that $(x,y)\in R$ and $(y,z)\in S$. This is easy to visualize if $R=S$ and the relation $R$ is depicted as a directed graph (possibly with loops). That is, if $R$ is a binary relation on a set $A$, then elements of $A$ are drawn as vertices (or nodes, or points), and there is an arrow from vertex $x$ to vertex $y$ iff $(x,y)\in R$. One can take walks in such a graph by following arrows. Then $(x,z)\in R\circ R=R^2$ if one can make two steps following arrows of $R$: one from $x$ to some $y$ and the other from $y$ to $z$.

Applied to the case when $R$ is $<$, this means that $(x,z)\in{<}\circ{<}$ if there exists some $y$ such that $x<y<z$. How does this compare with simply $x<z$ when the underlying set is $\Bbb Z$? Which statement implies the other?