Let R be the relation on ℤ+→ℤ+ defined by (a,b)R(c,d) if and only if a-2d=c-2b. List all the elements of the equivalence class [(3,3)]. I'm confused on how to find all the elements. I know how to find some but not all, wouldn't they be infinite? This is affecting me with the other questions as well. Thanks in advance!
 A: Instead of writing $(a,b)R(c,d)$, we will write that $(c,d)$ is equivalent to $(a,b)$. 
We have $(c,d)$ is equivalent to $(3,3)$ precisely if $3-2d=c-6$. (I am just reading this from the definition, replacing $a$ by $3$ and $b$ by $3$.)
For clarity rewrite the equation $3-2d=c-6$ as $c+2d=9$. Now recall that we are working in $\mathbb{Z}^+$, which in your course probably means the positive integers. 
So we want to find all pairs $(c,d)$ of positive integers such that $c+2d=9$.  These ordered pairs will be the full equivalence class of $(3,3)$, the set of all members of $(3,3)$'s family. 
The possibilities are $c=1$, $d=4$; $c=3$, $d=3$; $c=5$, $d=2$; and $c=7$, $d=1$. 
If we allow $0$ in $\mathbb{Z}^+$, we also have $c=9$, $d=0$.
On the first interpretation of $\mathbb{Z}^+$, the equivalence class of $(3,3)$ is the set
$$\{ (1,4), (3,3), (5,2), (7,1)\}.$$
Remark: If the relation were defined on ordered pairs $(x,y)$ where $x$ and $y$ are in $\mathbb{Z}$, then indeed the equivalence class of $(3,3)$ would be infinite. For it would for example contain the ordered pair $(99,-45)$, and infinitely many others. 
