1. ## Geometrical Description

Let $\displaystyle S = \{(x,y)| x,y \in \mathbb{R} \}$. If $\displaystyle (a,b)$ and $\displaystyle (c,d)$ belong to $\displaystyle S$, define $\displaystyle (a,b)R(c,d)$ if $\displaystyle a^2+b^2 = c^2+d^2$. Prove that $\displaystyle R$ is an equivalence relation on $\displaystyle S$. Give a geometrical description of the equivalence classes of $\displaystyle S$.

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

Consider $\displaystyle a \in S$ where $\displaystyle a = (a_1, a_2)$. Then $\displaystyle [a] = \{x \in S| x \sim a \}$. In this case, $\displaystyle a$ is an ordered pair. So it would be a circle?

2. 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.

3. 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 $\displaystyle r^2$ and varying the points $\displaystyle (x,y)$ that satisfy that equation. Hence you get a circle.

4. ## 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.