1. Let - be a relation over Zx (Z\ {0}) defined by (a,b)-(c,d) iff ad = bc. Prove that this is an equivalence relation and describe the resulting quotient set

Need help with this Foundations of Mathematics homework problem. I received feedback from my professor for this problem and he said Quotient set description needs considerably more detail. Hint: Can you relate the quotient set to a well-known number system? Below is my work for this problem. The image uploaded is the homework problem. Make sure handwriting is neat and readable. 

1. K = Ζ x (Ζ ∖ {0}) is the provided set which a relation ∼ is determined by (a,b) ∼ (c,d) iff ad = bc.

We will prove that ∼ is an equivalence relation. If you want to prove that ∼ is an equivalence relation, we are required to prove that ∼ is reflexive , symmetric, and transitive.

(i) Reflexive: Let (a,b)∈K, we are required to prove that (a,b) ∼ (a,b).

Now since the commutative under multiplication is Ζ, we will be able to write

                                    ab = ba             (where a,b ∈ Ζ∖{0})

                             ⇒ (a,b) ∼ (b,a)

For this reason, this proves that ∼ is reflexive.

(ii) Symmetric: Let (a,b) ∼ (c,d), where (a,b) and (c,d) are elements of K, we are required to prove that (c,d) ∼ (a,b).

Now, (a,b) ∼ (c,d)

        ⇒ ad = bc

With Z being the commutative under multiplication, we can therefore write it as

                  da =  cb                      (since ad = da and bc = cb) or              cb = da

⇒              (c,d) ∼ (a,b)

For this reason, this proves that ∼ is symmetric.

(iii) Transitive: Let (a,b), (c,d) and (e,f) be in K such that

(a,b) ∼ (c,d) and (c,d) ∼ (e,f)

Then we are required to prove that

(a,b) ∼ (e,f)

Now we have

                  (a,b) ∼ (c,d) ⇒ ad = bc

          and (c,d) ∼ (e,f) ⇒ cf = de

Here, ef = de gives us           c = de/f      (As f ≠ 0)

         Using this in

                  ad = bc

         ⇒     ad = b(de/f)

                  ⇒     a = be/f                    (cancel d on both sides and note that d ≠ 0)

                  ⇒     af = be

         ⇒     (a,b) ∼ (f,e)

Earlier in the problem, ∼ is reflexive, which allows us to now write

               (a,b) ∼ (f,e) ∼ (e,f)

that is,  (a,b) ∼ (e,f)

⇒ ∼ is transitive.

For this reason, it follows that ∼ is an equivalence relation.

Now we are required to explain the subsequent quotient set. Relation ∼ that is on set K is the quotient set, which is described as the set of all equivalence classes of relation ∼.

Ex: x = (a,b)∈K

We define

                    [x] = [(a,b)] = {y∈K ∣ y∼K}

                                            = {y = (c,d)∈K ∣ (c,d) ∼ (a,b)}

                                            = {y = (c,d)∈K ∣ (a,b) ∼ (c,d)}

                                            = {y = (c,d)∈K ∣ ad = bc}

Then the quotient set is defined as

K∕∼ = {[x] ∣ x∈K}