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.
On
Yes, you are right. You need to find three specific elements (we'll call them $a$, $b$, and $c$) such that $(a,b)$ and $(b,c)$ belong to $R$, but $(a,c)$ does not. Here's a hint: suppose $u$ is our $a$, and $x$ is our $b$. $(u,x$) belongs to $R$. Now see if there is some $c$ such that $(x,c)$ belongs to $R$, but $(u,c)$ doesn't.
On
Whatever the set you are provided. If relation on the set is transitive then it must satisfy the condition i.e. if $(a,b)\in R$ and $(b,c) \in R$ then $(a,c)$ does $\forall a,b,c \in S$
You can use definition to prove or disprove a statement given for a Relation. Nothing changes in your reasoning but the elements.
Well, it may take enough time to check whether your set is transitive or not. But you have to do it. :)
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.