Question

Which of the following defined on Z is not an equivalence relation?

• A. $(x,y)\in S\iff x\ge y$
• B. $(x,y)\in S\iff x=y$
• C. $(x,y)\in S\iff x-y$ is a multiple of $3$
• D. $(x,y)\in S$ if $\left|x-y\right|$ is even

Answer 1

The correct option is A

$(x,y)\in S\iff x\ge y$

Explanation for the correct answer

For option (a)

$(x,y)\in S\iff x\ge y$

Reflexive: Let $x\in S$

$x\ge x$ is always hold then $(x,x)\in S$

Therefore, $S$ is a reflexive relation.

Symmetric: Let $x,y\in S$

$(x,y)\in S\iff x\ge y$ it does not imply $y\ge x$

Therefore, $S$ is not a symmetric relation.

Hence, $(x,y)\in S\iff x\ge y$ is not an equivalence relation.

The explanation for incorrect options

For option (b)

$(x,y)\in S\iff x=y$

Reflexive: Let $x\in S$

$x=x$ is always hold then $(x,x)\in S$

Therefore, $S$ is a reflexive relation.

Symmetric: Let $x,y\in S$

$$\begin{gathered} x=y \\ \Rightarrow y=x \\ \Rightarrow (y,x)\in S \end{gathered}$$

Therefore, $S$ is a symmetric relation.

Transitive: Let $x,y,z\in S$

$(x,y)\in S\iff x=y$ and $(y,z)\in S\iff y=z$

$$\begin{gathered} \Rightarrow x=z \\ \Rightarrow (x,z)\in S \end{gathered}$$

Therefore, $S$ is a transitive relation.

Hence, $(x,y)\in S\iff x=y$ is an equivalence relation.

For option (c)

$(x,y)\in S\iff x-y$ is a multiple of $3$

Reflexive: Let $x\in S$

$x-x$ is always multiple of $3$ then $(x,x)\in S$

Therefore, $S$ is a reflexive relation.

Symmetric: Let $x,y\in S$

$x-y$ is a multiple of $3$

$=-\left(y-x\right)$ is a multiple of $3$

$\Rightarrow (y,x)\in S$

Therefore, $S$ is a symmetric relation.

Transitive: Let $x,y,z\in S$

$(x,y)\in S\iff x-y$ is a multiple of $3$ and $(y,z)\in S\iff y-z$ is a multiple of $3$

$=x-y+y-z$

$=x-z$ is a multiple of $3$

$\Rightarrow (x,z)\in S$

Therefore, $S$ is a transitive relation.

Hence, $S$ is an equivalence relation.

For option (d)

$(x,y)\in S$ if $\left|x-y\right|$ is even

Reflexive: Let $x\in S$

$\left|x-x\right|$ is always even then $(x,x)\in S$

Therefore, $S$ is a reflexive relation.

Symmetric: Let $x,y\in S$

$\left|x-y\right|$

$\Rightarrow \left|y-x\right|$ is even

$\Rightarrow (y,x)\in S$

Therefore, $S$ is a symmetric relation.

Transitive: Let $x,y,z\in S$

$(x,y)\in S$ if $\left|x-y\right|$ is even and $(y,z)\in S$ if $\left|y-z\right|$ is even

$\Rightarrow \left|x-y+y-z\right|$ is even

$=\left|x-z\right|$ is even

$\Rightarrow (x,z)\in S$

Therefore, $S$ is a transitive relation.

Hence, $S$ is an equivalence relation.

Hence, the correct option is option (a).