Question

In the set $\mathbb{Z}$ of all integers which of the following relation is an equivalence relation.
[i] $xRy\Leftrightarrow x\le y$
[ii] $xRy\Leftrightarrow x=-y$
[iii] $xRy\Leftrightarrow x-y\text{ is even}$
[iv] $xRy\Leftrightarrow x\equiv y\left( \bmod 3 \right)$

Answer 1

Hint: We will check for each of the relations whether they are symmetric, reflexive and transitive or not and hence we determine which of the given relations are equivalence relations.

Complete step-by-step answer:
Reflexive relation: A relation R on a set “A” is said to be reflexive if $\forall a\in A$, we have $aRa$.
Symmetric relation: A relation R on a set “A” is said to be reflexive if $aRb\Rightarrow bRa$.
Transitive relation: A relation R on a set “A” is said to be transitive if $aRb,bRc\Rightarrow aRc$.
Equivalence relation: A relation R on a set “A” is said to be an equivalence relation if it is reflexive, symmetric and transitive.
[i] $xRy\Leftrightarrow x\le y$
Reflexivity: Since $\forall x\in \mathbb{Z},$ we have $x\le x$. Hence, we have xRx , and hence R is reflexive
Symmetricity: Since $\exists 2,3\in \mathbb{Z}$ such that $2R3$ as $2\le 3$ but $3{R}2$ as $3>2$. Hence R is not symmetric
Transitivity: Since $x\le y,y\le z\Rightarrow x\le z$, we have $xRy,yRz\Rightarrow xRz$ , and hence R is transitive.
Equivalence: Since R is not symmetric, R is not an equivalence relation

[ii] $xRy\Leftrightarrow x=-y$
Reflexivity: Since $\exists 1\in \mathbb{Z}$ such that $1\ne -1$, we have $1{R}1$ , and hence R is not reflexive.
Symmetricity: Since $\forall x,y\in \mathbb{Z},$ we have $x=-y\Rightarrow y=-x$. Hence, we have $xRy\Rightarrow yRx$ , and hence R is symmetric
Transitivity: Since $1R-1,-1R1$ but $1{R}-1$, we have R is not transitive
Equivalence: Since R is not reflexive and transitive, R is not an equivalence relation

[iii] $xRy\Leftrightarrow x-y\text{ is even}$
Reflexivity: Since $\forall x\in \mathbb{Z},$ we have $x-x=0$ which is even, we have xRx , and hence R is reflexive.
Symmetricity: Since if $2|x-y\Rightarrow 2|y-x,$ we have if x-y is even, then y-x is also even and hence $xRy\Rightarrow yRx$ and hence R is symmetric.
Transitivity: Since the sum of two integers is even, we have if x-y is even and y-z is even then x-y+y-z = x-z is also even. Hence, we have $xRy,yRz\Rightarrow xRz$ , and hence R is transitive.
Equivalence: Since R is symmetric, reflexive as well as transitive, R is an equivalence relation.

[iv] $xRy\Leftrightarrow x\equiv y\left( \bmod 3 \right)$
Before solving this part, we need to understand what $x\equiv y\left( \bmod m \right)$ means. We have if $m|x-y,$ then $x\equiv y\left( \bmod m \right)$
Reflexivity: Now since $3|x-x=0$ we have $x\equiv x\left( \bmod 3 \right)$ and hence $xRx,\forall x\in \mathbb{Z}$. Hence R is reflexive.
Symmetricity: Since $3|x-y\Rightarrow 3|y-x,$ we have $x\equiv y\left( \bmod 3 \right)\Rightarrow y\equiv x\left( \bmod 3 \right)$ and hence $xRy\Rightarrow yRx$. Hence R is a symmetric relation
Transitivity: Since $3|x-y,3|y-z\Rightarrow 3|x-y+y-z$, we have $3|x-y,3|y-z\Rightarrow 3|x-z$ and hence $x\equiv y\left( \bmod 3 \right),y\equiv z\left( \bmod 3 \right)\Rightarrow x\equiv z\left( \bmod 3 \right)$. Hence $xRy,yRz\Rightarrow xRz$ and hence R is transitive.
Equivalence: Since R is reflexive, symmetric and transitive, we have R as an equivalence relation.
Hence options [iii] and [iv] are correct.

Note: In the questions of the above type, we need to understand and remember the definitions of various types of relations. Sometimes students forget to prove that xRx has to be true for all elements and not just for few and hence arrive at incorrect results.