Here is a question,
A = {1,2,3,4,6} = B, $aRb$ iff $a$ is a multiplier of $b$ .
Now I think the whole cartesian product of AxB should be the relation as every number is somehow a multiplier of another. Please help me out by sharing your review. Thanks
I have Kolman and Busby, Discrete Mathematical Strucxtures in Computer Science. Problem 9 on page 100 is
$A=\lbrace\,1,2,3,4,6\,\rbrace=B$; $a\,R\,b$ if and only if $a$ is a multiple of $b$.
I'm sure that what is intended is $a$ is an integer multiple of $b$. So for example $(6,3)$ will be in the relation, but $(3,6)$ won't be.