# Equivalence Relations

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?

(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.

The equivalence relation is defined by $$(x_1, y_1) \sim (x_2, y_2) \quad \text{if} \quad (x_1, y_1) = (\lambda x_2, \lambda y_2) \text{ for} \textbf{ some } \lambda \in \Bbb{R}^\times.$$

To fix your symmetry argument, note that if $$(x_1, y_1) = (\lambda x_2, \lambda y_2)$$ for some $\lambda \in \Bbb{R}^\times$, then $$(x_2, y_2) = (\lambda^{-1} x_1, \lambda^{-1} y_1)$$ and $\lambda^{-1} \in \Bbb{R}^\times$.

To fix your transitivity argument, note that if $$(x_1, y_1) = (\lambda x_2, \lambda y_2) \quad \text{and} \quad (x_2, y_2) = (\mu x_3, \mu y_3)$$ for some $\lambda, \mu \in \Bbb{R}^\times$, then $$(x_1, y_1) = (\lambda \mu x_3, \lambda \mu y_3)$$ and $\lambda \mu \in \Bbb{R}^\times$.

For a geometric interpretation of the relation $R$, notice that $P_1(x_1,y_1) R P_2(x_2,y_2)$ means that the points $P_1,P_2$ are in the same line through the origin in $\mathbb R^2$, since the equation of a line thru the origin in $\mathbb R^2$ is $(0,0)+t(x,y); t$ in $(-\infty, \infty)$. This means that the equivalence classes are lines going thru the origin. If you want to read more on this, you may want to read about the projective plane.

For symmetric, use 1/lambda to show the reverse. Because if lambda is in R - {0}, then so is 1/lambda.

Ex) $(x, y) = 2(x_2, y_2)$ then $(x_2, y_2) = 1/2(x,y)$.

For transitive, you would know there are 2 lambdas. One for each relation. So take the product of the 2 for the lambda needed to show transitivity.

An equivalence class is all elements related to each other by the equivalence relation using some base (x,y). For example (1,2) is related to all (L, 2L), where L is in R.