I want to prove that this relation is equivalence relation on A
about transitivity, there are counter-examples?
for reflexivity is obvious, $(x,x)$ the distance will be the same.
for symmetry $(x,y)\in R , (y,x) \in R$ the distances are the same.
if its Equivalence Relation what are the equivalence classes? and partition set?
I would like to get some suggestions.
Notice that this relation $T$ is defined by the equality relation which's the most natural equivalence relation so $T$ will inherit the same properties and then it's also an equivalence relation.
Remark You can use this method for all relation defined by $$xRy\iff f(x)=f(y)$$
For the class equivalence of $x$: $$[x]=\{y;\, xRy\}=\{y;\, |x|=|y|\}=C(O,|x|) $$ where $C(O,|x|)$ is the circle of center the origin of the plane and radius $|x|$.
Reflexivity:
$$\forall\,(a,b)\in\Bbb R^2\;,\;\;\sqrt{a^2+b^2}=\sqrt{a^2+b^2}\implies (a,b)T(a,b)$$
Symmetry:
$$(a,b)T(c,d)\implies \sqrt{a^2+b^2}=\sqrt{c^2+d^2}\implies\sqrt{c^2+d^2}=\sqrt{a^2+b^2}\implies (c,d)T(a,b) $$
Transitivity:
$$(a,b)T(c,d)\;\wedge\;(c,d)T(x,y)\implies \sqrt{a^2+b^2}=\sqrt{c^2+d^2}=\sqrt{x^2+y^2}\implies$$
$$(a,b)T(x,y)$$
The equivalence classes here : $\;x^2+y^2=R^2 ...$
