Question

Let O be the origin. We define a relation between two points P and Q in a plane if OP=OQ. Show that the relation, so defined is an equivalence relation.

Answer 1

We have to prove the given relation is reflexive, symmetric and transitive to prove the relation is symmetric. We show $\left( P,P \right)\in R$for reflexive, $\left( P,Q \right)\in R\Rightarrow \left( Q,P \right)\in R$ for symmetric and $\left( P,Q \right)\in R,\left( Q,S \right)\in R\Rightarrow \left( P,S \right)\in R$ for transitive where $S$is another point on the plane.

Complete step-by-step solution:
We know that a relation $R$ from set $A$ to set $B$ is a set of order pairs written as $R:A\to B$ which takes its elements from the set $A\times B$. A relation $R$ defined on the set A which means $R:A\to A$ is called reflexive if $aRa\Rightarrow \left( a,a \right)\in R$. The relation $R$ is symmetric if $aRB\Rightarrow bRa$ in other words $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$. The relation $R$ is transitive if $aRB,bRc\Rightarrow aRc$ in other words $\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$. Here $a,b,c\in A$. The relation is equivalent when $R$ is reflexive, symmetric and transitive. $$$$

The given relationship is the existence of a pair of points such that their distances from the origin are equal. We write it in symbols, R=\left\\{ \left( P,Q \right):OP=OQ \right\\}where $O$ is the origin. Then $OP$ is the distance of $P$ from the origin and $OQ$ is the distance of $Q$ from the origin. We have to prove that $R$ is equivalence then we have to prove $R$ is reflexive, symmetric, and transitive. Checking Reflexive: $R$ is reflexive when $P$ is related to $P$. We can say OP=OP as P is a fixed not a varying point in the plane. So we have $\left( P,P \right)\in R$. So $R$ is reflexive. Checking Reflexive: $R$ is symmetric when $P$ is related to $Q$ implies $Q$ is related to P. We know that if $OP=OQ$ then also $OQ=OP$. So we have $\left( P,Q \right)\in R\Rightarrow \left( Q,P \right)\in R$. $R$ is symmetric.
Checking Transitive: We have to take another point say $S$ on the plane. Let the distance from the origin is $OS$. We know that if $OP=OQ$ and $OQ=OS$ then $OQ=OS$. So we have $\left( P,Q \right)\in R,\left( Q,S \right)\in R\Rightarrow \left( P,S \right)\in R$. So $R$ is transitive. $$$$
Since $R$ is reflexive, symmetric, and transitive, so $R$ is an equivalence.

Note: We have to be careful of confusion of symmetric from anti-symmetric relation which is defined as $\left( a,b \right)\in R,\left( b,a \right)\in R\Rightarrow a=b$. Equivalence relation is different from equivalence class which is defined for $a\in X$ as \left[ a \right]=\left\\{ x\in X:aRx \right\\} on some set $X$.