# Homework Help: Equivalence Classes

1. Aug 14, 2010

### hitmeoff

1. The problem statement, all variables and given/known data
Let S := ($$\Re$$ x $$\Re$$ \ {(0,0)}. For (x,y), (x',y') $$\in$$ S, let us say (x,y) ~ (x',y') if there exists a real number $$\lambda$$ > 0 such that (x,y) = ($$\lambda$$x',$$\lambda$$y'). Show that ~ is an equivalence relation; moreover, show that each equivalence class contains a unique representative that lies on the unit circle (i.e., the set of points (x,y) such that x2 + y2 = 1).

2. Relevant equations
I know in order to show that something is an equivalence relation if the following 3 properties hold

reflexive: a ~ a for all a $$\in$$ S
symmetric: a ~ b implies b ~ a for all a,b $$\in$$ S
transitive: a ~ b and b ~ c implies a ~ c for all a, b, c $$\in$$ S

and for an equivalence relation ~ the equivalence class as the set {x $$\in$$ : x ~ a}

3. The attempt at a solution
What I don't get is, if there is no operation defined how do we show equivalence? Or is the operation scalar multiplication?

Im not sure where to go from here, especially showing that the solution to the unit circle is in the equivalent class.

2. Aug 14, 2010

### awkward

This is easy, so I think you are just confused by the fact that the members of your space are ordered pairs instead of numbers. So instead of showing that a ~ a, say, you have to show that for arbitrary (x,y), we have (x,y) ~ (x,y).

Hint: can you think of a $$\lambda$$ such that $$(x, y) = (\lambda x, \lambda y)$$?

(I told you it was easy.)

3. Aug 14, 2010

### hitmeoff

well for the reflexive property if $$\lambda$$ = 1 which is > 0, then (x, y) = (1$$\cdot$$x, 1$$\cdot$$y) so that holds.

But now for symmetric:
If (x, y) = ($$\lambda$$x', $$\lambda$$y') then ($$\lambda$$x', $$\lambda$$y') = (x,y), so (x,y) ~ ($$\lambda$$x', $$\lambda$$y') implies ($$\lambda$$x', $$\lambda$$y') ~ (x, y) but is this "showing it"?

Last edited: Aug 14, 2010
4. Aug 14, 2010

### hitmeoff

and for the part: "moreover, show that each equivalence class contains a unique representative that lies on the unit circle (i.e., the set of points (x,y) such that x2 + y2 = 1)."

Could we say:

For any (x,y) in R x R there exist a $$\lambda$$ s.t x2 + y2 = $$\lambda$$2, or (x/$$\lambda$$)2 +(y/$$\lambda$$)2 = 1.
If x' = x/$$\lambda$$ and y' = y/$$\lambda$$ then (x,y) = ($$\lambda$$x', $$\lambda$$y')

So for any equivalence class [(x,y)] there is a unique representative (x',y') s.t. x'2 + y'2 = 1

Last edited: Aug 14, 2010
5. Aug 14, 2010

### hitmeoff

anyone else? awkward? anybody?

6. Aug 14, 2010

### awkward

Not quite. If
$$x' = \lambda x$$,
then what is x in terms of x'?

For the part about the representative on the circle, you are OK.

7. Aug 15, 2010

### tinynerdi

I have the same problem. How would do you symmetric and transitive? Do you suppose to use integral?

8. Aug 16, 2010

### HallsofIvy

There is no reason to use calculus at all!

Here, we say that (x, y) is equivalent to (u, v) if and only if $(x, y)= \lambda(u, v)$ for some non-zero number $\lambda$.
If $x= \lambda u$ and $y= \lambda v$, then $u= (1/\lambda)x$ and $v= (1/\lambda) y$. Since $\lambda> 0$, $1/\lambda> 0$.

Similarly, if (x, y) is equivalent to (u, v) and (u, v) is equivalent to (a, b), then $x= \lambda_1u$ and $y= \lambda_1v$ for some $\lambda_1> 0$ and $u= \lambda_2a$ and $v= \lambda_2b$ for some $\lambda_2> 0$.

It should be obvious then that $x= \lambda_1(\lambda_2a)$$= (\lambda_1\lambda_2)a$ and $y= \lambda_1(\lambda_2 b)= (\lambda_1\lambda_2)b$ so that $x= \lambda a$ and $y= \lambda b$ with $\lambda= \lambda_1\lambda_2> 0$.