Equivalence relation $(a,b) R (c,d) \Leftrightarrow a + d = b + c$ Suppose $A$ is the set composed of all ordered pairs of positive integers. Let $R$ be the relation defined on $A$ where $(a,b) R (c,d)$ means that $a + d = b + c$.
(a) Prove that $R$ is an equivalence relation.
Here is what I have so far. Is this correct?
Reflexive:
$a \sim a$
$\implies$ $a+b=a+b$; $(a,b) R (c,d)$
Symmetric:
if $a \sim b$ then $b \sim a$
$\implies$ if $a+d=b+c$ then $c+b=d+a$
Transitive:
if $a \sim b$ and $b \sim c$ then $a \sim c$
$\implies$ if $a+d=b+c$ and $c+f=d+e$ then $a+d=d+e$
(b) Find $[(1,1)]$.
I'm not sure how to approach this.
 A: Your proof regarding the fact that $R$ is in fact an equivalence relation is indeed correct.
Now think what the equivalence class of $(1,1)$ could be, since $(1,1) R (1,1)$ you know that everyone in the equivalence class has the property that $(a,b)$ holds $a+1=b+1$, therefore $a=b$.
This means that the equivalence class of $(1,1)$ is $(a,a)$ for $a\in\mathbb{N}$.
A: HINT: The equivalence class of $(1,1)$ is made up of all pairs $(x,y)\sim(1,1)$. Write explicitely what the latter means and get a relation that need to be satisfied by $x$ and $y$.
A: HINT $\ $ This is geometrically  obvious once you note that the equivalence class containing $\rm\ (a,b)\ $ is simply the slope $1$ line through it (with positive integer coordinates). In effect you are discovering the negative integers as those lines having no positive $x$-intercept (= normal-form), i.e. whose intercept is "negative". This "group of differences" construction is the additive analog of the field of fractions construction.
A: Let
    (a,b)R(a,b)
a+b=b+a  => R is reflexive (a,b are integers)
(a,b)R(c,d) if a+d=b+c and 
(c,d)R(a,b) if c+b=d+a;
both are equal(a,b,c,d are integers) hence R is symmetric
(a,b)R(c,d) if  a+d=b+c --->(1)
(c,d)R(e,f) if  c+f=d+e --->(2)
add (1)&(2)
            a+c+d+f=b+d+c+e, ==> a+f=b+e
ie. (a,b)R(e,f)
R is transitive.
Hence R is an equivalence relation
A: Let (a,b)=x, (b,c)=y and (c,d)=z;
Now to prove that R on A is equivalence. We'll have prove that it is also reflexive(xRx), symmetric[(xRy)=(yRx)] and transitive[(xRy) and (yRz) implies (xRz)].
And the condition is: (a,b)R(c,d) implies a+d=b+c
1. xRx=(a,b)R(a,b) implies a+b=b+a which is true; therefore it is reflexive.

2. xRy=(a,b)R(b,c) implies a+c=b+b….1 & yRx=(b,c)R(a,b) implies b+b=c+a…2 here 1 and 2 are equal therefore R is symmetric.

3. xRy=a+c=b+b…1 & yRz=(b,c)R(c,d) implies b+d=2c…..2 implies that xRz=(a,b)R(c,d) implies a+d=b+c……3
Now Adding 1 and 2 we get:
a+c+b+d=2b+2c implies a+d=b+c which is to be proved for transitivity.

