Problems:
1.) R = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)}
2.) R = {(x, y) ¦ y - x is an odd integer}
3.) R = {(x, y) ¦ y - x is a multiple of 3}
Attempt:
By definition, a relation is equivalent if it is transitive, reflexive, and symmetrical. If either one of these sets fail any of these tests, then the relation is not an equivalence relation.
1.)
Checking if it's reflexive.
(0,0) is present
(0,0) is present
(0,0) is present
(1,1) is present
(1,1) is present
(2,2) is present
Thus this relation is reflexive.
Checking if symmetrical.
(0,0) is present
(1,0) is not present
Thus this is not symmetrical and it is not an equivalence relation.
2.)
Checking if reflexive
This relation fails because of two axioms:
1st axiom: Adding two odd integers will result in an even integer.
2nd axiom: Adding two even integers will also result in an even integer.
Not to mention that you are always going to get zero, an even number, when you subtract the same x values every time.
The only time it is ever odd is when you have an odd and an even number.
Thus this relations fails the reflexive test and it is not an equivalence relation.
3.)
Checking if reflexive:
This relation is reflexive because of this rule.
An integer a is a multiple of an integer b means that ab=integer. so, as 0 divided by any integer (except zero itself) yields an integer.
Checking if symmetrical:
No this is not symmetrical simply because not every number is a multiple of three.
Here is one case: 9 - 5 = 4 and 5 - 9 = -4. Neither of these are multiples of 3.
Thus this relations fails the symmetrical test and it is not an equivalence relation.
Is all of my work correct?
You're fine up until the analysis of the last relation.
Recall that $R$ is symmetric iff $(\forall x,y)[(x,y) \in R \Rightarrow (y,x) \in R]$. Now, suppose $(x,y) \in R$. Then $y-x$ is a multiple of $3$. What can you say about $x-y$?