Let A=0,1,2, ,9.

Question

Let $A={0,1,2,\ldots ,9}$. Let R be a relation on A defined by $(x,y)\in R$ if and only if $|x-y|$ is a multiple of 3.

Given below are two statements :

Statement I : $n(R)=36$.

Statement II : R is an equivalence relation.

In the light of the above statements, choose the correct answer from the options given below :

Accepted Answer

$A={0,1,2,\ldots 9}$

A relation $R$ on $A$

$R=|x-y|$ is multiple of 3

$0,3,6,9$ are $3n$ type numbers

$1,4,7$ are $3n+1$ type numbers

$2,5,8$ are $3n+2$ type numbers

Case 1 : Relation $R$ is valid if $x\&y$ both are $3n$ type.

$=4	imes 4=16$

Case 2 : Relation $R$ is valid if $x$ is $3n+1$ type $\&y$ is $3n+2$ type. $=3	imes 3=9$ pairs

Case 3 : similarly relation $R$ is valid if $x$ is $3n+2$ type $\&y$ is $3n+1$ type. $=3	imes 3=9$ pair

$n(R)=16+9+9=34
e 36$

statement I is false.

A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

Reflexive: For any $x\in A,|x-x|=0$. Since 0 is a multiple of $3,(x,x)\in R$.

Symmetric: If $(x,y)\in R$, then $|x-y|=3k$. Since $|y-x|=|x-y|$, then $|y-x|=3k$, so $(y,x)\in R$.

Transitive: If $|x-y|$ and $|y-z|$ are multiples of 3 , then $x-y=3k$ and $y-z=3m$.

Adding these gives $x-z=3(k+m)$, so $|x-z|$ is a multiple of 3.

Statement II is correct.

Solutions · 1