Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 1 / c

True. Fix $a\in A$. Reflexivity supplies $a{R}_{1}a$ and $a{R}_{2}a$. In the definition of composition of relations, choose the intermediate element to be $a$ itself. Then $a(R_2\circ R_1)a$. This holds for every $a\in A$, so the composite binary relation is reflexive.

False, even for two equivalence relations. On $A=\{1,2,3\}$, let $R_1$ have equivalence classes $\{1,2\}$ and $\{3\}$, and let $R_2$ have equivalence classes $\{1\}$ and $\{2,3\}$. Both binary relations are transitive. For $S=R_2\circ R_1$, we have $3S2$, using the intermediate element $3$, and $2S1$, using the intermediate element $1$. However, $3S1$ fails: the only $R_1$-successor of $3$ is $3$, and $3R_{2}1$ is false. Thus the composite fails transitivity. The order of composition here means first $R_1$, then $R_2$.

False. Use the same two equivalence relations as in part (ii). They are both symmetric relations. The composite contains $1S3$, since $1R_{1}2$ and $2R_{2}3$. It does not contain $3S1$, as shown there. Hence composition of relations need not preserve symmetry of a relation, even when both factors are also reflexive and transitive.