Question

Let P be a plane. If p, q are points in P, then $\mathit{p}\mathbf{~}\mathit{q}$means p and q are the same distance from the origin. Prove that $\mathbf{~}$is an equivalence relation on P.

Short answer

It is proved that given relation $~$is equivalence.

Step-by-step solution

Definitions of Equivalence relation

A relation R defined on a set is called an equivalence relation if it is reflexive, symmetric and transitive.

Prove that ~is an equivalence relation on P 

Objective is to prove that $~$is an equivalence relation. That is to show that the relation$~$ is reflexive, symmetric and transitive.

For reflexive,

Let $p,p\in P$then clearly, $p~p$is reflexive as$p$ is equidistance from origin

Therefore, the relation$~$ is reflexive.

For symmetric,

Let$p,q\in P$ and that$p~q$, that is$p\&q$ are equidistance from origin. Then$q~p$ that is $q\&p$are equidistance from origin. So, the relation is $~$symmetric.

For transitive,

Let$p,q,r\in P$ such that $p~q$and$q~r$ which means $p\&q$and $q\&r$are equidistance from origin. Then$p\&r$ are also equidistance from origin, that is$p~r$ . So, the relation $~$is transitive.

Therefore, the given relation$~$ is equivalence.