Finding the unique representative that lies on the unit circle in an equivalence class of a given equivalence relation.

Question

I am doing an exercise from a number theory textbook for practice and not sure how to approach this problem.

Let $S:=(\Bbb R \times \Bbb R)\setminus{(0,0)}$. For $(x,y),(x',y') \in S$ let $(x,y)~\sim (x',y')$ if and only if there exists a real number $\lambda \gt 0$ such that $(x,y)=(\lambda x', \lambda y')$. Show that $\sim$ is an equivalence relation; moreover, show that each equivalence class contains a unique representative that lies on the unit circle.

So I was able to prove that $\sim$ is an equivalence relation which was pretty straightforward. I am having trouble on how to show that each equivalence class has a unique representative of a point on the unit circle.

I understand that $(x,y)=(\lambda x', \lambda y')$ implies that $(x,y)$ and $(x',y')$ are on the same line that passes through the origin and hence for each unique $\lambda$ there is a unique line passing through the origin containing both of these points. My thinking was to plug in $(\lambda x', \lambda y')$ into $x^2 + y^2=1$ which would yield $(x')^2 + (y')^2=\frac{1}{\lambda^2}$ hence for our equivalence classes taking the new constant $\frac{1}{\lambda^2}$ would be our unique representative. This is because for unique $\lambda \gt 0$ we have that each of the $\frac{1}{\lambda^2}$ are unique.

I'm not sure if this is at least on the right track but it's all the progress I've made so far.

Answer 1

If $(x,y)\in\Bbb R^2$, then $(x,y)\sim\left(\frac x{\sqrt{x^2+y^2}},\frac y{\sqrt{x^2+y^2}}\right)$ and $\left(\frac x{\sqrt{x^2+y^2}},\frac y{\sqrt{x^2+y^2}}\right)$ belongs to the unit circle. Besides, if $(x,y)\sim(x',y')$ and $(x',y')$ belongs to the unit circle, then $(x')^2+(y')^2=1$ and, for some $\lambda\in(0,\infty)$, $x'=\lambda x$ and $y'=\lambda y$. But then$$1=(x')^2+(y')^2=\lambda^2(x^2+y^2),$$and therefore$$\lambda=\frac1{\sqrt{x^2+y^2}}.$$So, $\left(\frac x{\sqrt{x^2+y^2}},\frac y{\sqrt{x^2+y^2}}\right)$ is the only element of the equivalence class of $(x,y)$ which belongs to the unit circle.