Review for Group Theory Final Exam:
Define a relation on $\Bbb{R}^2 \setminus (0, 0)$ by letting $(x_1, y_1) \sim (x_2, y_2)$ if there exists a nonzero real number $\lambda$ such that $(x_1, y_1) = (\lambda x_2, \lambda y_2)$. Prove that $\sim$ defines an equivalence
relation on $\Bbb{R}^2 \setminus (0, 0)$. What are the corresponding equivalence classes?
I know we need to test for:
(i) Reflexive: $(x_1, y_1) \sim (x_1, y_1)$
$(x_1, y_1) = (\lambda x_1, \lambda y_1)$ holds true if $\lambda = 1$.
(ii) Symmetric: if $(x_1, y_1) \sim (x_2, y_2)$ then $(x_2, y_2) \sim (x_1, y_1)$.
This ends up being: if $(x_1, y_1) = (\lambda x_2, \lambda y_2)$ then is $(x_2, y_2) = (\lambda x_2, \lambda y_2)$?
This is true if λ = 1.
(iii) Transitive: if $(x_1, y_1) \sim (x_2, y_2)$ and $(x_2, y_2) \sim (x_3, y_3)$, then $(x_1, y_1) \sim (x_3, y_3)$?
Once again, this is true if $\lambda = 1$.
Is this the correct way of doing this? And I am stuck on the corresponding equivalence classes. How do I define those?
Best Answer
(i) looks good.
(ii) doesn't look so good. You didn't prove the general case, you only showed it for $\lambda = 1$. You want it to be true for any nonzero real value of $\lambda$. You should have if $(x_1,y_1) \sim (x_2,y_2)$, then by definition $(x_1,y_1)=(\lambda x_2,\lambda y_2)$, so then let $\gamma = \frac{1}{\lambda}$, and then we have $(x_2,y_2) = (\gamma x_1,\gamma y_1)$, so $(x_2,y_2) \sim (x_1,y_1)$.
(iii). Again, you didn't prove the general case. You should have: if $(x_1,y_1) \sim (x_2,y_2)$, then $(x_1,y_1) = (\lambda x_2, y_2)$. If $(x_2,y_2) \sim (x_3,y_3)$, then we have, where $\alpha$ is some nonzero real, $(x_2,y_2) = (\alpha x_3, \alpha y_3)$. So then $(x_1,y_1) = (\lambda \alpha x_3, \lambda \alpha y_3)$. The reals are closed under multiplication so $\lambda\alpha$ is a real, so it works.
An equivalence class is just a set of elements that are equivalent under the equivalence relation. For example, from (iii), all $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ would be in an equivalence class, because they are all equivalent. If there are any other $(x_n,y_n)$ that are equivalent to them, then they would also be in that class.