# The equivalence classes and their representatives

A relation on $$\mathbb{R}^3 \backslash \{(0,0,0)\}$$ is defined by $$(a, b, c)\sim(d, e, f)$$ if and only if there is $$\lambda \in \mathbb{R}$$ with $$\lambda > 0$$, and $$a=\lambda d$$, $$b=\lambda e$$ and $$c=\lambda f$$. Prove that $$\sim$$ is an equivalence relation and find a transversal for the relation that is geometrically pleasant.

I can show that $$\sim$$ is an equivalence relation. For the reflexivity, simmetry and transitivity to hold simultaneously, $$\lambda$$ should be 1, except if there's a mistake in my reasoning.

Now if $$\lambda$$ is indeed 1 that also implies that if $$(a, b, c)\sim(d, e, f)$$ then $$a=d$$, $$b=e$$ and $$c=f$$, which means that every $$(a, b, c)$$ is related only to itself.

Now the transversal is apparently a sphere, but I don't see how that could be if every point represents its own equivalence class?

"I can show that ∼ is an equivalence relation. For the reflexivity, simmetry and transitivity to hold simultaneously, λ should be 1, except if there's a mistake in my reasoning."

No, $$\lambda$$ is arbitrary. For reflexivity, $$\lambda=1$$. But to prove symmetry, let $$(a,b,c)=\lambda(x,y,z)$$ for some $$\lambda\ne 0$$. Then $$(x,y,z)=\frac{1}{\lambda}(a,b,c)$$. Thus $$(a,b,c)\sim (x,y,z)$$ implies $$(x,y,z)\sim(a,b,c)$$. Similar for transitivity.

The quotient set is the projective plane over the reals. A transversal is given by $$\{(1,b,c)\mid b,c\in\Bbb R\}\cup \{(0,1,c)\mid c\in\Bbb R\}\cup \{(0,0,1)\}.$$ The points with first component 0 are the points at infinity.

• The OP states: "For $\lambda>0$". This answer seems to be for $\lambda \ne 0$. – paw88789 Oct 20 '20 at 12:13
• It works also for $\lambda >0$. But then the transversal has ''twice'' as much elements. – Wuestenfux Oct 20 '20 at 15:34
• Okay, thank you, but I'd need some further clarification. How so that for reflexivity $\lambda$ takes a certain value, but that value doesn't have to stay the same for the symmetry and transitivity? Also I thought that the definition of symmetry - if $x \sim y$ then $y \sim x$ - in this case means if $(a, b, c)= \lambda (d, e, f)$ then $(d, e, f)= \lambda (a, b, c)$. This question is similar to the above, but why can there be a certain value ($\lambda$) before $(d, e, f)$ in the first part of the if clause and a different one before $(a, b, c)$, specifically $\frac{1}{\lambda}$ in 2nd part? – Treex Oct 20 '20 at 19:50
• The definition is as follows: $(a,b,c)\sim (x,y,z)$ if there exists $\lambda\ne 0$ (or $\lambda >0$) such that $(a,b,c) = \lambda(x,y,z)$. For reflexivity, you can choose $\lambda$ appropriately. – Wuestenfux Oct 21 '20 at 6:41

Reflexive: $$(a,b,c)=1\cdot(a,b,c)$$, so $$(a,b,c)\sim (a,b,c)$$.

Symmetric: If $$(a,b,c)\sim (d,e,f)$$, then $$\exists \lambda>0$$ with $$(a,b,c)=\lambda\cdot(d,e,f)$$. Then$$(d,e,f)=\frac{1}{\lambda}\cdot (a,b,c)$$. So $$(d,e,f)\sim (a,b,c).$$.

Transitive: To you. (But it amounts to noting that the product of two positive values of $$\lambda$$ is again positive.)

Equivalence classes are rays (half lines) with endpoint at the origin. So the unit sphere (centered at the origin) indeed forms a transversal.

Edit: To be accurate: The origin referred to above as the endpoint of each ray is not included in each ray. (I.e., they are open rays.)