Let $S := (R × R) \setminus {(0,0)}$. 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 $x^2 + y^2 = 1$).

First, ~ is obviously reflexive when $\lambda = 1$. Second, $(x, y) = (\lambda x′, \lambda y′)$ implies $(\frac{x}{\lambda},\frac{y}{\lambda}) = (x', y')$, so ~ is symmetric. Third, if $(x, y) = (\lambda_1 x′, \lambda_1 y′)$ and $(x', y') = (\lambda_2 x'', \lambda_2, y'')$, then $(x, y) = (\lambda_1 \lambda_2 x'', \lambda_1 \lambda_2 y'')$, so ~ is transitive. Thus, ~ is an equivalence relation.

To show that each equivalence class contains a unique representative that lies on the unit circle, consider $\lambda = \frac{1}{\sqrt{x^2 + y^2}}$:

$$\begin{aligned} (x_u, y_u) = (\frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}) \end{aligned}$$

Notice

$$\begin{aligned} x_u^2 + y_u^2 = \frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2} = 1 \end{aligned}$$

In other words, $(x_u, y_u)$ is the unique representative that lies on the unit circle for each equivalence class $[(x,y)]$ for all $(x, y) \in S$.

$\Box$

# posted by rot13(Unafba Pune) @ 7:58 PM