The question is stated as: Let $A$ be the set of integers, find a relation $R$ which is reflexive and symmetric in $A$ but not transitive in $A$.
By definition we have that.
- $R$ is reflexive in $A$$ \Leftrightarrow (\forall x)(x \in A \Rightarrow xRx)$
- $R$ is symmetric in $A$$ \Leftrightarrow (\forall x)(\forall y)([x\in A \land y \in A \land xRy] \Rightarrow yRx)$
- $R$ is transitive in $A$$ \Leftrightarrow (\forall x)(\forall y)(\forall z)([x\in A \land y \in A \land z \in A \land xRz \land zRy] \Rightarrow xRy)$
What i thinked is to define such a relation using least common multiple, and greatest of two numbers as the following:
- Let $lcm(x,y)$ be the least common multiple of $x$ and $y$
- Let $max(x,y)$ be the greatest number from $\{x,y\}$
- Then let $R = \{(x,y) : x \in A \land y \in A \land lcm(x,y) = max(x,y) \}$
It is transitive because $(\forall x)(x \in A \Rightarrow lcm(x,x) = x = max(x,x))$.
It is symmetric too because if the if $lcm(x,y) = max(x,y)$ holds true, its obvious that $lcm(y,x) = max(y,x)$ will be true too for any integers.
But it is not transitive, i tried to show this with one counter example: $(6,3) \in R \land (3,9) \in R$ but $(6,9) \notin R$.
The way I defined the relation is correct? Its possible to retrieve relations from numerical sets holding choosed properties in a easy way?
One simple example:
$$a R b\iff ab\not\equiv 3\mod 10$$
$aRa$ because $a^2$ cannot end with digit 3.
$aRb \iff bRa$, because $ab=ba$.
But this relation is not transitive. For example, for $a=3$, $b=5$, $c=11$ we have $aRb$, $bRc$ but not $aRc$.
EDIT: Actually the simplest example I have found so far is this one:
$$a R b\iff a+b \ne101 $$
Obviously, it's reflexive, symmetric but not transitive ($a=10, b=20, c=91)$