Consider two relations on the set of real numbers
$R_1=\{(a,b) \in R^2\,|\,a>b\}$ $R_2=\{(a,b) \in R^2 \,|\, a\geq b\}$
What is $R_2 \circ R_1 ?$
I know that $(a,b) \in R_2\circ R_1$ iff $(a,c) \in R_1$ and $(c,b) \in R_2$.
This implies that $(a,b) \in R_2 \circ R_1$ iff $a\gt c$ and $c \geq b$ for all a,b,c that have their domain as real numbers. The answer to this in Rosen is given to be $R_1$ but how did they conclude this?
On
$R_2\circ R_1=\{(a,b)\in \Bbb R^2:\exists c\in \Bbb R~(a>c\wedge c\geq b)\}$
So for $a, b$ to be related there must exist some $c$ less than $a$ and at least $b$. That is $c\in [b;a)$
Noting that $b$ itself will be in that interval if $a>b$, we can surely say $$R_2\circ R_1=\{(a,b)\in \Bbb R^2:a>b)\}$$
If $(a,b)\in R_2 o R_1$ then for some $c$, $a>c$ and $c\ge b$ then the transitive property implies that $a>b$ that is $(a,b)\in R_1$
On the other hand if $a>b$, you can find a $c$ such that $a>c$ and $c\ge b$ which means $(a,b)\in R_2 o R_1$
Thus $R_2 o R_1 = R_1$