Show that $(x, y) \sim (a,b)$ if there exists $\lambda >0$ such that $(a,b) = (\lambda x, \lambda y) $, defines an equivalence relation

Question

I am trying to prove the equivalence relation of $(x, y) $ ~ $ (a,b)\ \ \lambda >0$ such that $(a,b) = (\lambda x, \lambda y) $

And describe its equivalence classes. So far I proved that

1. reflexive: $(x,y)$ ~ $(x, y) \longrightarrow (x, y) = (\lambda x, \lambda y)$ when $\lambda = 1$

2. symmetry: $(x,y)$ ~ $(a, b) \longrightarrow (x, y) = (\lambda a, \lambda b)$ and $(a,b)$ ~ $(x, y) \longrightarrow (a, b) = (\lambda x, \lambda y)$ are the same when $\lambda = 1$

3. transitive: $(x,y)$ ~ $(a, b) $ and $(a,b)$ ~ $(c, d) \longrightarrow (x, y)$ ~ $(c, d)$ when $\lambda = 1$

I am not so sure about the equivalence classes and how they would look geometrically.

Answer 1

Your proof for $2$ and $3$ is wrong.

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For symmetry:

• Yes, from $(x,y)\sim (a,b)$, you can conclude that $(x,y)=(\lambda a, \lambda b)$ for some $\lambda$.

however, what you then did was assume that $(a,b)\sim (x,y)$, when in fact, that's what you want to prove.

So, I suggest you try this again. And just to recap:

What you know:

You know that $(x,y)\sim (a,b)$, i.e. you know that there exists some $\lambda>0$ such that $(a,b) =(\lambda x, \lambda y)$

What you need to prove:

You need to prove that $(a,b)\sim (x,y)$, that means you need to prove that there exists some $\lambda > 0$ such that $(x,y)=(\lambda a, \lambda b)$.

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For transitivity:

You do not know that $(x,y)\sim (c,d)$ when $\lambda=1$. For example, if $(x,y)=(1,1)$, and $(a,b)=(2,2)$, and $(c,d)=(4,4)$, then $(1,1)\sim (4,4)$ even though $\lambda$ in that case is not equal to $1$.

Again, try this proof again, and again, here is a recap:

So, I suggest you try this again. And just to recap:

What you know:

You know that $(x,y)\sim (a,b)$ and you know that $(a,b)\sim (c,d)$. In other words, you know that there exists some $\lambda_1>0$ such that $(a,b) =(\lambda_1 x, \lambda_1 y)$. And you know that there exists some $\lambda_2>0$ such that $(c,d)=(\lambda_2 a, \lambda_2 b)$.

What you need to prove:

You need to prove that $(x,y)\sim (c,d)$, that means you need to prove that there exists some $\lambda > 0$ such that $(c, d)=(\lambda x, \lambda y)$.