Constructing  $\mathbb Q$ from $\mathbb Z$ Let $X = \{(a,b)|a,b \in \mathbb Z, b \neq 0 \}$.
Define a relation $\sim$ on $X$ by
$(a,b) \sim (c,d)$ iff $ad  = bc $.
a) I'm trying to show that ~ is an equivalence relation.
So, is it reflexive? 
$(a,b) \sim (a,b) $ iff $ab = ab$, with $b$ not equal to zero
so its reflexive
Is it transitive? 
$(a,b) \sim (c,d) $  iff ad = bc and if $(c,d) \sim (e,f)$ then $cf = de$  so, $c = ad/b$ and $d = bc/a$
and we have then $ad/b *f = bc/a *e$
Is it symmetric? 
$(a,b) \sim (a,b)$ iff $ab = ab$,  same as  $ab = ba$, (why?) with $b$ not equal to zero
b) Let $\mathbb Q$ be the set of equivalence classes and denote the equivalence class of $(a,b)$ by $[a,b]$. I'm needing help in showing that the "addition and Multiplication" of $\mathbb Q$ are well defined:
(1) $[a,b] + [c,d] = [ad+bc,bd]$
and 
(2)$ [a,b][c,d] = [ac,bd]$
What I have so far:
(1) $[a,b]  + ([c,d] +[e,f]) = [a,b] + ([cf,de]) = [ade,bcf]$
but this is true for when ?
(2) If $c =b$ and $d = a$
then, 
$[a,b][b,a] = [a^2,b^2]$ iff $a^2 = b^2$, which is not well defined, since $a = -5$ and $b$ could be equal to $5$, with $25 = 25$
 A: Looking at what you have said for (a):


*

*Transitive: I do not like you saying "c = ad/b and d = bc/a and we have then ad/b *f = bc/a *e" because you seem to be using division a little freely. Better to say that if $(a,b) \sim  (c,d)$ and $(c,d) \sim (e,f)$ means $ad = bc$ and $cf = de$ then $bcf = bde$ so $adf = bde$ so (since $d \not = 0$) $af = be$ meaning $(a,b) \sim (e,f)$

*Symmetric: you have confused this with reflexive.  What you want is $(a,b) \sim (c,d)$ iff $ad = bc$ iff $bc = ad$ iff $(c,d) \sim (a,b)$
Now for part (b): 


*

*For addition you need to show that if $(a_1,b_1) \sim  (a_2,b_2)$ then $(a_1 d +b_1 c, b_1 d) \sim  (a_2 d + b_2 c , b_2 d)$

*For multiplication you need to show that if $(a_1,b_1) \sim  (a_2,b_2)$ then $(a_1  c, b_1 d) \sim  (a_2 c , b_2 d)$
It is worth noting that in fact $[a,b][b,a] = [ab,ba] = [1,1]$ 
A: HINT $\ $ To show that addition is well-defined, you need to show that the definition doesn't depend on the choice of equivalence class representatives of the summands. Since addition is commutative it suffices to do this for one summand, i.e. it suffices to show that
$$\rm [A,B]\ =\ [a,b]\ \Rightarrow\ [A,B]\ +\ [c,d]\ =\ [a,b]\ +\ [c,d] $$
$$\rm [A,B]\ =\ [a,b]\ \Rightarrow\ [Ad+Bc,Bd]\ =\ [ad+bc,bd] $$
Unwinding the definition of fraction equality, the implication reduces to simple algebra.  
Applied twice, this immediately implies the sought result for both summands, namely
$$\rm  [A,B]\ =\ [a,b],\ [C,D]\ =\ [c,d]\ \ \Rightarrow\ \ [A,B]\ +\ [C,D]\ =\ [a,b]\ +\ [c,d] $$
which says precisely that addition doesn't depend on the choice of argument representatives.
A: Reflexivity: $(a,b)\sim (a,b)$ since $ab=ab$. 
Symmetry: Assume $(a,b) \sim (c,d)$. This implies $ad=bc$, which implies $cb=da$ by the symmetry of equality of integers. The condition $(c,d) \sim (a,b)$ is equivalent to $cb=da$, which we have already shown. 
Transitivity: Assume $(a,b) \sim (c,d)$ and $(c,d)\sim (e,f)$. Then $ad=bc$ and $cf=ed$. We want to show $af=eb$. Multiplying the two statements together yields $adcf=bced$. By canceling $c$ and $d$ from both sides we see that $af=eb$, which implies $(a,b) \sim (e,f)$.
In your attempt to show rational addition is well defined, you mixed up the definition of addition and multiplication. 
To prove rational addition is well-defined, it is enough to show that if $(a,b)=(a',b')$, then $(a,b)+(c,d)=(a',b')+(c,d)$.
