I'm completely lost in discrete mathematics. I have to find out whether $$xRy \iff \exists z\in \mathbb N \;\;[z\mid y \iff z\mid x]$$ where $x,y \in \mathbb N$ is an equivalence.
I know that relation must be reflexive, symmetric and transitive in order to be an equivalence.
If relation is reflexive, then $z\mid x \iff z\mid x$ must be the same, which is true. But I have no idea how to prove symmetry and transitivity of relation. Thanks for your advice
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In plain english. "Two natural numbers are related if they have a factor in common" (which ... as $z$ could be $1$ means all numbers are related but ... I get ahead of myself.)
Reflexive: Is is true that for all $x$ there is an $z$ so that $z|x \iff z|x$? well.... yeah. $poo \iff poo$ is always true so... this is kind of trivial.
Symmetric: Is it true that if $z|x \iff z|y$ then $z|y \iff z|x$? Well, yes... if $A \iff B$ then it's true $B \iff A$.
Transitive: Suppose there is a $z$ so that $z|x \iff z|y$ and that there is a $w$ so the $w|y \iff w|u$. Well.... foey. I'm going to point out if $z = w =1$ we have $1|x$, $1|y$, and $1|u$ so all are always true so transitivity holds.
Let $x,y\in\mathbb Z$ and observe that $z:=\max(x,y)+1$ will not divide $x$ and will not divide $y$.
That means that the statement: $$z\mid x\iff z\mid y$$ is actually a true statement.
Proved is now that for every pair $\langle x,y\rangle\in\mathbb N^2$ we have $xRy$ (or equivalently $R=\mathbb N^2$).
Reflexivity, symmetry and transitivity are evident for this relation.