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