Let S be the relation on the set of ordered pairs of positive integers such that ((a,b),(c,d)) ∈ S if and only if ac = bd. Show that S is not an equivalence relation.
So far I got:
Let (a,b) ∈ A, Since ab=ba, ((a,b),(b,a)) ∈ S, therefore, S is reflexive.
Now let ((a,b),(c,d)) ∈ S, ac = bd -> bd = ac -> db = ca. This implies ((d,b),(c,a)) ∈ S. Therefore, S is not symmetric.
Any issues with this? Do I need to show any more work? Thanks!
Your proof for non-symmetry isn't valid since there's multiple conclusions to be had.
Suppose $((a,b),(c,d)) \in S$. Then $ac=bd$. Equivalently, $ca=db$ since multiplication commutes. Therefore $((c,d),(a,b)) \in S$, giving symmetry. That other pairs are implied to be in $S$ isn't relevant.
More generally, $R$ is a symmetric relation if $(a,b) \in R \implies (b,a) \in R$.
So, we know the relation $S$ is reflexive and symmetric... If it's truly not an equivalence, it must not be transitive.
Except it's not reflexive.
If it is, then $((a,b),(a,b)) \in S$. But then $a^2 = b^2$. Does this always hold?