Is the below relation transitive? The theorem says if (a,b) and (b,c) belong to a relation, then (a,c) also does.
I am confused because (3,2) and (2,3) is like (a,b) with (b,a) and (3,3) like (a,a), and something is wrong here, or the relation or the theorem
$$R\:=\left\{\left(3,2\right),\:\left(2,3\right),\:\left(3,3\right)\right\}\:$$
Table for checking: $$ \begin{array}{|c|c|c|c|}\hline (a,b) &(b,c) &(a,c) &(a,c)\in\text{set?} \\ \hline (3,2) &(2,3) &(3,3) &\text{yes} \\ \hline (2,3) &(3,2) &(2,2) &\text{no} \\ \hline (2,3) &(3,3) &(2,3) &\text{yes} \\ \hline (3,3) &(3,2) &(3,2) &\text{yes} \\ \hline (3,3) &(3,3) &(3,3) &\text{yes} \\ \hline \end{array} $$ These are all the possibilities, because the second position of the first ordered pair must equal the first position of the second ordered pair. Since the second line fails to be in the set, the relation is not transitive. I think if you added $(2,2)$ to the relation, it would be transitive.