If $R$ is a transitive relation, then $R\circ R\subseteq R$ Here's the question I'm struggling with:
Let R be a transitive relation on a set A.  Prove the R composed with R is a subset of R.
I'm kind of lost on how to prove this.
I've started with saying:
"If R is transitive, then R is the subset of A such that (a,b) is in R and (b,c) is in R, and, due to transitivity, (a,c) is in R when (a,b) and (b,c) have the same b for all a, b, c in A."
Now, for R composed with R, I understand that there is an (a,b) in R and an (b,c) in R, and R composed with R is the set of all (a,c) in A such that (a,b) and (b,c) have the same b.
So to continue from here, this is what I've written, continuing from the previous quotation:
"The composition of R and R on the set A is the subset of all (a,c) in A such that (a,b) in R and (b,c) in R have the same b."
And now I'm lost - I'm not sure how to show that R composed with R is a subset of R from here.
I was going to write out my attempt but I realized I was simply repeating myself and didn't make much sense, so I'd like to ask for some direction before I try again.
Thanks for any help!
 A: You are very close. For brevity, denote $R$ composed with $R$ by $R\circ R$.  
By the definition of composition, the pair $(a,c)$ is in $R\circ R$ if and only if there exists a $b$ such that $(a,b)$ and $(b,c)$ are both in $R$. 
But if there is such a $b$, then by transitivity $(a,c)$ is in $R$. Thus if  $(a,c)$ is in $R\circ R$, then $(a,c)$  is in $R$. It follows that $R\circ R\subset R$.  
A: The first thing you need to do is to get the definitions absolutely clear.
A (binary) relation $R$ on a set $A$ is not a subset of $A$, it is a subset of $A\times A$.  In other words, $R$ does not consist of elements of $A$, it consists of pairs of elements of $A$.
A relation $R$ on a set $A$ is transitive means: for all $a,b,c\in A$, if $(a,b)\in R$ and $(b,c)\in R$ then $(a,c)\in R$.
The composition of $R$ with $R$ is the relation (which I will write as $R^2$, don't know if your instructor uses this notation) defined as follows: for all $a,b\in A$ we have $(a,b)\in R^2$ if and only if there exists $c\in A$ such that $(a,c)\in R$ and $(c,b)\in R$.
Please learn these thoroughly: having a rough idea of what they mean is just not good enough.
So to the problem: if $R$ is transitive then $R^2\subseteq R$.  We will prove this in the same way we would prove any subset statement: show that any element of the LHS is also an element of the RHS.
Assume that $R$ is transitive, and let $(a,b)\in R^2$.  By definition, there exists $c\in A$ such that $(a,c)\in R$ and $(c,b)\in R$.  Since $R$ is transitive, $(a,b)\in R$.  We have proved that if $(a,b)\in R^2$, then $(a,b)\in R$.  That is, $R^2\subseteq R$.
