Sets: How to show that relation R is not transitive? I have a question which states:

$R = \{ (t,t), (t,v), (t,z), (u,t), (u,u), (u,v), (u,x), (u,z), (v,v), (w,w), (w,z), (x,t), (x,v), (x,w), (x,x), (x,y), (x,z), (y,w), (y,y), (y,z), (z,z) \}$
"Give a counter example to show $R$ is not transitive ... "

I understand that $R$ is said to be transitive if $(a,b) \in R$  and $(b,c) \in R$, implies $(a,c) \in R$. If these conditions aren't met then the relation is not transitive. 
How could apply this in the example? Thank you.

Since $R$ is small, you can actually consider all possibilities for $a,b$, and $c$. However, this is easier if you have a systematic arrangement of the ordered pairs. One way is to represent them by a matrix:
$$\begin{array}{c|ccc}
&t&u&v&w&x&y&z\\ \hline
t&1&&1&&&&1\\
u&1&1&1&&1&&1\\
v&&&1\\
w&&&&1&&&1\\
x&1&&1&1&1&1&1\\
y&&&&1&&1&1\\
z&&&&&&&1
\end{array}$$
Here there is a $1$ in row $t$, column $v$, for instance, because the pair $\langle t,v\rangle$ is in $R$. Each of the other $1$s represents one of the other ordered pairs in $R$. Now you want to find $a,b,c\in\{t,u,v,w,x,y,z\}$ such that $\langle a,b\rangle$ and $\langle b,c\rangle$ are in $R$, but $\langle a,c\rangle$ is not in $R$. 
Notice that $x$ is related by $R$ to almost everything – everything except $u$, in fact – so it’s easy to find pairs $\langle x,\star\rangle\in R$. The only pair with first element $x$ that is not in $R$ is $\langle x,u\rangle$; can you find a $\star$ such that $\langle x,\star\rangle$ and $\langle\star,u\rangle$ are in $R$?
Added: That attempt fails, because the $u$ column has only $u$ in it, so we might try for $\langle u,\star\rangle$ and $\langle\star,w\rangle$, or $\langle u,\star\rangle$ and $\langle\star,y\rangle$; this time it turns out that both yield counterexamples.
