Points defined by relations (an exercise from "System of Coordinates")? An exercise from "System of Coordinates" (by Gelfand, Glagoleva and Kirilov) asks me to "[t]ry to decide by yourself which sets of points are defined by these relations" and relations given are:
a. $|x| = |y|$;
b. $\frac{x}{|x|} = \frac{y}{|y|}$;
c. $|x| + x = |y| + y$;
d. $[x] = [y]$ ($[\gamma]$ is defined as giving whole part of $\gamma$);
e. $x - [x] = y - [y]$;
f. $x - [x] > y - [y]$.  
What I'm having most trouble with is probably the notation as I don't think I understand the question(s) in the exercise properly. As an example a straight line, $x = y$ has been given, together with $x^2 - y^2 = 0$ which is also graphed as a straight line in example.
Case (a) could be reduced to $f(x) = |x|$ to be graphed, I think, but after that I'm at loss.
For example for case (b) I can graph $f(x) = \frac{x}{|x|}$ but what on earth am I to do with $\frac{x}{|x|} = \frac{y}{|y|}$? How can I graph something with two unknowns to a two-dimensional plane? It goes without a mention that I've got nowhere with cases c-f because I'm clearly not even getting the question.
The exercise is from page 18 of the book.
 A: This is about points $(x,y)$ on the plane which satisfy the given equations (or inequalities). For examplex, $x^2=y^2$ is the union of  two lines, $y=x$ and $y=-x$. This example was easy because $0=(x^2-y^2)=(x-y)(x+y)$, but in general one needs another approach. 
When the equation is of the form $f(x)=f(y)$ for some function $f$, we always have $x=y$ as one possibility. Any other points will have to come from $f$ not being injective, that is, taking the same value more than once. For example,   $f(x)=x^2$ takes the same value at $\pm x$. So does  $f(x)=|x|$, which answers your question (a): the set is the union of $y=x$ and $y=-x$. 
Let's try  (b). The function $x/|x|$ takes on value $1$ for all positive $x$, and value $-1$ for all  negative $x$. Therefore, the equation is satisfies if both $x,y$ are positive (this means taking the whole first quadrant) or both of them are negative (the whole third quadrant). This union of two quadrants is the set you want.
I'll do one more, (d). The function $[x]$ is constant on every interval of the form $[n,n)$ for integer $n$. So we get the union of squares of unit sidelength placed along the diagonal $y=x$, like this: 

