The document defines an equivalence relation R on a set of ordered pairs of positive integers. It then shows that R is reflexive by showing that any ordered pair is related to itself under multiplication. It is symmetric since if (a,b) is related to (c,d), then (c,d) is related to (a,b). It is transitive, since if (a,b) is related to (c,d) and (c,d) is related to (e,f), then (a,b) is related to (e,f).

1. Given a binary relation R on a set A, the set of all equivalence classes
Solution
I. Reflexive: Let (a, b) be an ordered pair of positive integers. To show R is reflexive we must
show ((a, b),(a, b)) R. Multiplication of integers is commutative, so ab = ba. Thus ((a, b),(a, b))
R. II. Symmetric: Let (a, b) and (c, d) be ordered pairs of positive integers such that (a, b)R(c, d)
(recall this notation is equivalent to ((a, b),(c, d)) R). Then ad = bc. This equation is equivalent
to cb = da, so (c, d)R(a, b). This shows R is symmetric. III. Transitive: Let (a, b), (c, d), and (e, f)
be ordered pairs of positive integers such that (a, b)R(c, d) and (c, d)R(e, f). Then ad = bc and cf
= de. Thus, adf = bcf and bcf = bde, which implies adf = bde. Since d 6= 0, we can cancel it from
both sides of this equation to get af = be. This shows (a, b)R(e, f), and so R is transitive.