Show this relation on $\mathbb{R}^2\setminus \left \{ (0,0) \right \}$ is an equivalence relation.

Question

Could someone please tell me whether my solution is okay?

Let $\sim$ be a relation on $\mathbb{R}^2\setminus \left \{ (0,0) \right \}$ where $(x_{1},y_{1})\sim (x_{2},y_{2})$ if and only if there exists $\lambda \neq 0$ such that $(x_{1},y_{1})=(\lambda x_{2}, \lambda y_{2})$. Prove $\sim$ is an equivalence relation.

Reflexive: Since $(x_{1}, y_{1})=(x_{1},y_{1})$ for $\lambda =1$, $(x_{1},y_{1})\sim (x_{1},y_{1})$.

Symmetric: Let $(x_{1},y_{2})\sim (x_{2},y_{2})$. Then $(x_{1},y_{1})=(\lambda x_{2},\lambda y_{2})=\lambda (x_{2},y_{2})$. Since $\lambda \neq 0$, $(x_{2},y_{2})=\frac{1}{\lambda}(x_{1},y_{1})$. Then $(x_{2},y_{2})\sim (x_{1},y_{1})$.

Transitive: Let $(x_{1},y_{2})\sim (x_{2},y_{2})$ and let $(x_{2},y_{2})\sim (x_{3},y_{3})$. Then $(x_{1},y_{1})=\lambda _{1} (x_{2},y_{2})$ and $(x_{2},y_{2})=\lambda _{2} (x_{3},y_{3})$. Then $(x_{2},y_{2})=\frac{1}{\lambda _{1}}(x_{1},y_{1})$. Then $\frac{1}{\lambda_{1}}(x_{1},y_{1})=\lambda _{2}(x_{3},y_{3})$. Then $(x_{1},y_{1})=\lambda _{1} \lambda _{2}(x_{3},y_{3})$. Then $(x_{1},y_{1})\sim (x_{3},y_{3})$.

Answer 1

Your solution is fine except in transitive part you have some statements which could have been dropped.

All you need is $$(x_{1},y_{1})=\lambda _{1} (x_{2},y_{2})$$ and $$(x_{2},y_{2})=\lambda _{2} (x_{3},y_{3})$$

Thus $$(x_{1},y_{1})=\lambda _{1}\lambda _{2} (x_{3},y_{3})= \lambda _{3} (x_{3},y_{3})$$

Answer 2

Your solution is okay. You might want to emphasize that each of your $\lambda$'s is a nonzero real number.