Prove by contradiction that for a transitive relation $R$ on $A$, $R^2$ is also transitive

Question

Prove by contradiction that for a transitive relation $R$ on $A$, $R^2$ is also transitive.

In order for a relation to be transitive it must satisfy $$aRc \wedge bRc \rightarrow aRc$$ for all $a,b,c \in A$.

Answer 1

Proving this by contradiction seems weird, but here goes: Assume to the contrary that $R$ is transitive, but $R^2$ is not transitive.

Since $R^2$ is not transitive, there exists $x,y,z\in A$, such that $(x,y)\in R^2$, $(y,z)\in R^2$, but $(x,z)\not\in R^2$.

By definition of $R^2$ there exists $u_1,u_2\in A$, such that $(x,u_1)\in R$, $(u_1,y)\in R$, $(y,u_2)\in R$, and $(u_2,z)\in R$. Since $R$ is transitive, we obtain also that $(x,y)\in R$ and $(y,z)\in R$.

By definition of $R^2$, this implies $(x,z)\in R^2$, which contradicts $(x,z)\not\in R^2$.