On $N \times N $ define the relation R, setting $(a,b),(c,d) \in R$ if and only if $a+d=b+c$. Show that $R$ is an equivalence relation. On $N \times N $ define the relation R, setting $(a,b),(c,d) \in R$ if and only if $a+d=b+c$
a. Show that $R$ is an equivalence relation. 
My attempt: By definition 6.2.3 $R$ is an equivalence relation if $R$ is reflexive, symmetric, and transitive. 
R is reflexive if $( \forall x\in S)((x,x) \in R)$
R is symmetric if $(\forall x,y \in S)[(x,y) \in R \rightarrow (y,x) \in R]$
R is transitive if $(\forall x, y, z \in S)[((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R]$
For reflexivity, we have,
$( \forall x \in S)((a,b),(a,b) \in R)$
and $a+b=b+a$
For symmetry, we have
$(\forall x, y, \in S)[(a,b),(c,d) \in R \rightarrow (c,d),(a,b) \in R]$
and $a+d=b+c$ and $c+b=d+a$
For transitivity, we have
$(\forall x,y,z \in S)[(a,b),(c,d) \in R \land (c,d),(e,f) \in R \rightarrow (a,b),(e,f) \in R]$
and $a+d=b+c$ and $c+f=d+e$
$a+d+c+f = b+c+d+e$
Subtracting c and d from both sides, we have
$a+f =b+e$ 
which is
$(a,b),(e,f) \in R$.
We have proven that $R$ is an equivalence relation. 
My question is that how do I put more words into this proof or does this proof look fine? I have arrows drawn (on scratch paper) on the pairs which made it a bit easier to see how the proof process for reflexivity and symmetry is done. 
Another question I have is on this part for transitivity . Why  does 
$a+d=b+c$ and $c+f=d+e$
result in 
$a+d+c+f = b+c+d+e$?
I don't see anything to substitute unless I just focus on gathering the left side of the equations which is $a+d+c+f$ and gather the right hand side of the equations together like  $b+c+d+e$ and that becomes $a+d+c+f = b+c+d+e$. Then subtracting $c$ and $d$ I do have $a+f=b+e$ which becomes $(a,b), (e,f) \in R)$ from the transitivity definition. 
 A: Your proof is pretty much correct. My main problem lies with some notation.

Your relation is on the set $\Bbb N \times \Bbb N$, so the elements of your relation $R$ are elements from $(\Bbb N \times \Bbb N)^2$. This means elements of $R$ have the form
$$\bigl( (a,b),(c,d) \bigr) \text{, where }a,b,c,d \in \Bbb N$$
Notice, this is an ordered pair of ordered pairs. So for your statement of reflexivity, you would not say $(a,b),(a,b) \in R$ but rather $\bigl( (a,b),(a,b) \bigr) \in R$. The first statement specifies two different elements of $R$, and also redundantly says $aRb$ twice; not what you're looking for.

Another issue lies with the proof of symmetry. Notation aside, what you seek to show is
$$\bigl( (a,b),(c,d) \bigr) \in R \implies \bigl( (c,d),(a,b) \bigr) \in R$$
You seem to be stating outright the equalities necessary, rather than showing one implies the other. For instance, this is how I would start it:
$\bigl( (a,b),(c,d) \bigr) \in R$ means that $a+d = b+c$. But by commuting the addition and reversing the equality (since $=$ is a symmetric relation), we get that this also means $c+b=d+a$. But this is the condition such that $\bigl( (c,d),(b,a) \bigr) \in R$, which gives symmetry.


My question is that how do I put more words into this proof or does this proof look fine?

More words aren't strictly necessary, so long as the steps are easy enough to follow. Truthfully, there could have been some cutting down in other respects, but this is totally fine as a proof for a student; over time and with experience, you'll learn where to trim and where to elucidate the details.


Another question I have is on this part for transitivity . Why  does
$$a+d=b+c \text{ and } c+f=d+e$$
result in
$$a+d+c+f = b+c+d+e?$$

You might remember something similar from solving system of equations in your algebra classes, especially if we write the pair of equations like this:
$$\left\{ \begin{matrix}
a+d = b+c \\
c+f = d+e
\end{matrix} \right.$$
You can add these equations together. That is, if $p=q$ and $r=s$, then $p+r=q+s$. Then you'll get a single equation of
$$a+d+c+f = b+c+d+e$$
You'll notice that $c,d$ are on each side as well, so you can subtract it from both sides and cancel them to get
$$a+f=b+e$$
which is what you seek! So yes, you have it quite right.

...though you did post this years ago so I imagine you don't really need this confirmation and help now, but if nothing else this might help others in the future and also gets this off the unanswered list.
