Equivalence Classes: Unique Unit Circle Rep.

[hitmeoff]

Homework Statement

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) = (\lambdax',\lambday'). 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).

Homework 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}

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.

 

[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.)

 

[hitmeoff]

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

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

 

[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 = \lambda², or (x/\lambda)² +(y/\lambda)² = 1.
If x' = x/\lambda and y' = y/\lambda then (x,y) = (\lambdax', \lambday')

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

 

[hitmeoff]

anyone else? awkward? anybody?

 

[awkward]

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

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

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.

 

[tinynerdi]

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

 

[HallsofIvy]

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

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.