The question: determine if p is reflective, symmetric, transitive and/or anti-symmetric, if p is an equivalence relation, describe the equivalence classes
A = Z , and $apb$ if and only if $5 | (2 a + 3 b )$
This is what i have done:
reflexive: if a,b are an element of integers $2*5 + 3*1 = 13$ and this is not a multiple of 5 therefore $apb$ therefore is not reflective.
symmetric: if $apb$, $5|(2a+3b)$ therefore $5|(3b+2a)$ therefore $bpa$ therefore symmetric
transitive: if $apb$ then $bpc$ $5|(2a+3b)$ Then $5|(3b + c)$
therefore $5|[(2a+3b)+(3b+c)] = 2a+c$ therefore $apc$
and since the relation is not reflexive there is no equivalence relation.
I think ive done transitive wrong, the numbers with throw me off. Any help would be greatly appreciated.
HINT: I think you misunderstood the relation. The relation statement is this: \begin{equation} aPb\Longleftrightarrow5|(2a+3b) \end{equation} $a$ and $b$ are just placeholders. You can think of $P$ like a relation between two objects, lke so: \begin{equation} \heartsuit P\diamondsuit\Longleftrightarrow5|(2\heartsuit+3\diamondsuit) \end{equation}
Read it carefully and pay attention to the roles of $a$ and $b$ (or, equivalently, $\heartsuit$ and $\diamondsuit$) in the definition. To check properties, you must properly write down the relation.
1) reflectivity: $aPa\Longrightarrow\cdots$
2) symmetry: $aPb\Longrightarrow\cdots$
3) transitivity: $aPb,\ bPc\Longrightarrow\cdots$