probably a good start would be looking up what reflexive, symmetric, and transitive relations are
Let R be an integral domain.
Let Q = { (a, b) | a, b ∈ R and b ≠ 0 }
Deﬁne a relation ∼ on Q by (a, b) ∼ (c, d) if and only if ad = bc.
Prove that ∼ is an equivalence relation (that is, ∼ is reﬂexive, symmetric, and transitive).
So for reﬂexivity, for instance, you must prove that (a, b) ∼ (a, b).
Would this be correct:
1.) For all (a, b) ∈ Q, we have (a, b) ~ (a, b) since ab = ba. So ~ is reflexive.
2.) If (a, b), (c, d) ∈ Q and (a, b) ~ (c, d), then
ad = bc
da = cb
cb = da
(c, d) ~ (a, b)
So, ~ is symmetric
3.) If (a, b), (c, d), (e, f) ∈ Q and (a, b) ~ (c, d), (c, d) ~ (e, f), then
ad = bc and cf = de
adf = bcf and bcf = bde
daf = dbe
d(af - be) = 0
From (c, d) ∈ Q, we know that d is not equal to 0, since R has no zero divisors, we obtain af - be = 0. Then, af = be, so (a, b) ~ (e, f) and ~ is transitive
So ~ is an equivalnce relation on Q
correct
not quite.2.) If (a, b), (c, d) ∈ Q and (a, b) ~ (c, d), then
ad = bc
da = cb
cb = da
(c, d) ~ (a, b)
So, ~ is symmetric
$(a,b)\sim (c,d) \Rightarrow ac=bd$
show $(c,d)\sim (a,b) \Rightarrow ac=db$ is True
no. look at your relation again3.) If (a, b), (c, d), (e, f) ∈ Q and (a, b) ~ (c, d), (c, d) ~ (e, f), then
ad = bc and cf = de