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 $xRz$ ?
You can obtain a counterexample for $x=3, y=4, z=6$.
So the relationship is not tranistive.
Bare in mind that: $$xRy: x − 2 ≤ y ≤ x + 2 \\ yRz: y− 2 ≤ z ≤ y + 2\\ xRz: x − 2 ≤ z ≤ x + 2$$
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.
