= Solution
The answer is \b[no], even when both original <binary relations> are <equivalence relations>. On $X=\{a,b,c\}$, let $R_1$ have <equivalence classes> $\{a,b\},\{c\}$ and $R_2$ have <equivalence classes> $\{a\},\{b,c\}$. Each is a <transitive relation> and, by the earlier proof, a <Euclidean relation>.
Use exactly the paper's convention for <composition of relations>: $xRy$ means $xR_1z$ and $zR_2y$ for some $z$. The possible successors are
$$
R(a)=R(b)=\{a,b,c\},\qquad R(c)=\{b,c\}.
$$
Indeed, from $a$ or $b$ the first step can reach $a$ or $b$, whose $R_2$ successors together exhaust $X$; from $c$ the first step must stay at $c$, whose $R_2$ successors are only $b,c$. Consequently $cRb$ and $bRa$, but not $cRa$. Therefore
$$
\boxed{R_1\text{ and }R_2\text{ transitive and Euclidean does not force }R\text{ transitive}.}
$$
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