hello i have a upcoming quiz and I was solving practice problems that the instructor gave us. But Im not sure how to approach this problem
the problem is: Let $A = \{1,2,3,4\}$, and $\mathcal{R}$ be a binary relation on $A\times A$ given by: $(a,b)\mathcal{R}(c,d) \iff a|c$ and $b|c$. Show that $\mathcal{R}$ is a order and draw its diagram.
I really don't know how to solve for transitivity and antisymmetricity.... detailed explanation will be very much appreaciated.. Thank you
Well it's not too hard, I mean, it's easy to think that: If $(a,b) \textbf{R} (c,d) and (c,d) \textbf{R} (e,f) \implies a \lvert c , b \lvert d , c \lvert e \wedge d \lvert f$ so if $c$ divides $e$ and $a$ divides $c$ then $a$ divides $e$, because the relation "divides" is transitive, the same with $b$. Therefore $(a,b) \textbf{R} (e,f)$ so it is transitive. For antisymmetry is routine for a positive set, I mean. Clearly if $(a,b) \textbf{R} (c,d) \wedge (c,d) \textbf{R} (a,b) \implies a \lvert c, b \lvert d \wedge c \lvert a, d \lvert b$ therefore $ a = c \wedge b = d$. So it is transitive, anti symmetric and reflexive. Note that the anti symmetry works in this case because it is a positive set. If it is not a positive set then you have: If $ a \lvert b \wedge b\lvert a \implies \lvert a \rvert = \lvert b \rvert$. Hope this helped!