1
$\begingroup$

Let $R$ be a binary relation on $\mathbb{N}$ defined by $xRy$ if and only if $x − 2 ≤ y ≤ x + 2$

How do you find if it is a transitive relation when there is only $xRy$? Isn't transitivity the relation between 2 conditions, for example $xRy$, $yRz$ therefore $xRz$ ?

$\endgroup$
  • 1
    $\begingroup$ You have the right definition of transitivity. Now what do you think the answer to your question is? Just take three numbers in $x,y,z \in \mathbb{N}$ and suppose $xRy$ and $yRz$. Does $xRz$ or is there a counter-example? $\endgroup$ – Bill O'Haran Apr 29 at 15:29
  • $\begingroup$ Also, for your next posts here please use MathJax. $\endgroup$ – Bill O'Haran Apr 29 at 15:34
1
$\begingroup$

You can obtain a counterexample for $x=3, y=4, z=6$.

So the relationship is not tranistive.

Bare in mind that: $$xRy: x − 2 ≤ y ≤ x + 2 \\ yRz: y− 2 ≤ z ≤ y + 2\\ xRz: x − 2 ≤ z ≤ x + 2$$

$\endgroup$
1
$\begingroup$

Transitivity of $R$ means that if we have, for some $x, y, z \in \mathbb{N}$, $xRy$ and $yRz$, then we automatically have $xRz$. You are right to say that transitivity is a ‘relation between two conditions’, but it isn’t a relation between two relations.

In this case, if $R$ is transitive then whenever $x-2 \leq y \leq x+2$ and $y-2 \leq z \leq y+2$, it should also be true that $x-2 \leq z \leq x+2$. Intuitively, this means that whenever $y$ is a distance of at most $2$ from $x$, and $z$ is a distance of at most $2$ from $y$, then $z$ must be a distance of at most $2$ from $x$. Let’s think. Does this seem reasonable? It does not, because clearly $x$ and $y$ can be separated by $2$, and then $z$ can be more than $2$ away from $x$, if it is greater than $y$.

A concrete counterexample would be: $a=1,b=3,c=5$. $a$ and $b$ are at most $2$ apart, and so are $b$ and $c$, but $c$ is more than $2$ away from $a$. So we have $aRb$ and $bRc$ but not $aRc$, so $R$ is not transitive. A single counterexample is all that is required to show that $R$ is not transitive, but if it was transitive, then we would need to prove this.

To prove that a relation is transitive, we need to start with the assumptions (that $xRy$ and $yRz$ for some arbitrary $x,y,z$ in the set) and deduce rigorously, perhaps using other known theorems, that $xRz$ must be true.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.