Let R be a relation on \[Z \times Z\] defined by (a,b)R(c,d) if and only if ad−bc is divisible by 5. Then R is
Solutions · 1
Official solution
(a,b)R(a,b) as ab−ab=0
Therefore reflexive
Let \[(a, b) R(c, d) \Rightarrow a d-b c\] is divisible by 5
\[\Rightarrow \mathrm{bc}-\mathrm{ad}\] is divisible by \[5 \Rightarrow(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})\]
Therefore symmetric
Relation not transitive as \[(3,1) \mathrm{R}(10,5)\] and \[(10,5) \mathrm{R}(1,1)\] but (3,1) is not related to (1,1)