# Geometrical Description

• May 26th 2009, 04:16 PM
Sampras
Geometrical Description
Let $S = \{(x,y)| x,y \in \mathbb{R} \}$. If $(a,b)$ and $(c,d)$ belong to $S$, define $(a,b)R(c,d)$ if $a^2+b^2 = c^2+d^2$. Prove that $R$ is an equivalence relation on $S$. Give a geometrical description of the equivalence classes of $S$.

Proof. So $(a,b)R(a,b)$ (reflexivity). Also $(a,b)R(b,a)$ (symmetry). Also if $(a,b)R(c,d)$ and $(c,d)R(e,f)$, then $(a,b)R(e,f)$ (transitivity). $\square$

Consider $a \in S$ where $a = (a_1, a_2)$. Then $[a] = \{x \in S| x \sim a \}$. In this case, $a$ is an ordered pair. So it would be a circle?
• May 26th 2009, 04:24 PM
amitface
For the first part, you essentially just stated the definition of an equivalence relation. Sure, it may be obvious, but you should show the (very elementary, I know) algebra to prove it.

For the second part, yes, it's a circle, but you did not explain why.
• May 26th 2009, 04:38 PM
Sampras
Quote:

Originally Posted by amitface
For the first part, you essentially just stated the definition of an equivalence relation. Sure, it may be obvious, but you should show the (very elementary, I know) algebra to prove it.

For the second part, yes, it's a circle, but you did not explain why.

Because you are essentially fixing $r^2$ and varying the points $(x,y)$ that satisfy that equation. Hence you get a circle.
• May 27th 2009, 05:25 AM
alunw
This is not just any equivalence relation
The equivalence classes are fairly obviously the family of circles centred on the origin, and you could view them as being the cosets of the kernel in the very natural homomorphism of (C,*)->(R,*) that sends z to |z|.

What about if R was (a,b)R(c,d) if (aa+bb+ab) == (cc+dd+cd)?
Or indeed (aa+bb-ab) == (cc+dd-cd)
Both those equivalences have nice geometric interpretations as well.