# Homework Help: Proving equivalence classes bijective to the set of points on the unit circle

1. Nov 1, 2009

### ams2990

1. The problem statement, all variables and given/known data
De fine a relation on R as follows. Two real numbers $$x, y$$ are
equivalent if $$x - y$$ $$\epsilon Z$$ . Show that the set of equivalence classes of this relation is bijective to the set of points on the unit circle.

2. Relevant equations
N/A? I don't think there are any special equations that are needed for this problem.

3. The attempt at a solution
A part of the problem that I've omitted asked us to prove that the relation is an equivalence one -- I've done that. I've also defined the set of points on the unit circle, which is $$\{ a,b \epsilon R | \sqrt{x^{2}+y^{2}} \}$$ Damned if I know where to go from here, though.

2. Nov 1, 2009

### tiny-tim

Welcome to PF!

Hi ams2990! Welcome to PF!
How would you label each class?

(in other words, what are the classes?)

3. Nov 1, 2009

### ams2990

Re: Welcome to PF!

Not really sure what you're asking...the equivalence class of $$a \in A$$ is $$[ a ]$$, which is the set of all elements equivalent to $$a$$.

4. Nov 1, 2009

### tiny-tim

ok, and what is that?

5. Nov 1, 2009

### jambaugh

Bijective means there exists a bijection. You need to construct the bijection (one-to one and onto mapping.)

(Consider ways to iterate over all the points on the unit circle).

To index your equivalence classes it is sufficient to refer to representative elements. Can you think of a good set of representative elements one element for each equivalence class?

6. Nov 1, 2009

### ams2990

Uh, really vague question, but I'm assuming you're asking what the contents of the class are. For any $$a$$ and $$b \in Z$$, the contents will be $$a + b$$, for all $$b$$. Stated another way, the class will be $$a$$ + every integer value. So if $$a = 1.5$$, $$[a]$$ will be $$\{...,-1.5,-.5,.5,1.5,2.5,3.5,4.5,...\}$$

7. Nov 1, 2009

### tiny-tim

ok, so how would you label that class (what name would you give it, to distinguish it from any different class)?

8. Nov 1, 2009

### ams2990

I was thinking about a good way to traverse the points. The obvious solution of choice are sine and cosine. I'm not sure how to create a function from what would seem like $$R \times R$$ when the relation is only defined on $$R$$. Can I "pretend" it's $$R \times R$$ because I have $$a,b \in R$$?
[EDIT]That was extremely poorly worded. Let me try to rephrase. The function as I see it would be from $$R \times R$$ to $$R$$. Is there a distinction between a function based on $$R \times R$$ and one based on two elements from $$R$$? Are these the same thing?

Also, how do I enforce "bounds checking" on my function? (i.e., not traverse the circle more than once) Do I even want this? [/EDIT]

As to the indices, obviously the name of the class would be the element for which the equivalence is true...so for $$[a]$$, it would be $$a$$, right?

$$[a]$$?

Last edited: Nov 1, 2009
9. Nov 1, 2009

### tiny-tim

ah, now I see you write R x R, I can see what's confused you.

You're making this especially complicated by defining the unit circle with two parameters (a and b or x and y), as if it were part of the (2D) plane.

The unit circle is 1D … find a way of defining it with one parameter.

10. Nov 1, 2009

### ams2990

Sorry if I'm being incredibly dense, but if you only had one parameter, then it would be $$R \Rightarrow R \times R$$, right? Like a parametric function?

11. Nov 1, 2009

### tiny-tim

Forget R x R …

you're only interested in the unit circle for itself, not for anything it's related to.

The unit circle is a 1D line, just like the unit interval is a 1D line (you wouldn't insist on using two parameters for the unit interval, would you, even though you can embed in in 2D?) …

we normally label it with a parameter going from 0 to 2π.

12. Nov 1, 2009

### jambaugh

Yea, remember its only points themselves not the shape. Imagine you can cut the circle and bend it however you like ... or even [red herring] chop it into pieces if necessary. [\red herring]