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## Here's the Solution to this Question

a.)i

$x$ is related to $y$ if $x$ and $y$ begins with the same letter.

Thus, the relation is reflexive since every English words will be related to itself.

Also, if $xRy$ , that is they both have the same first letter, then $yRx.$ Thus the relation is symmetric.

Lastly, if $xRy$ and $yRz$ that is, $x$ and $y$ have the same letter and $y$ and $z$ have the same letter. Definitely, $x$ and $z$ will have the same letter, that is $xRz$. Thus, the relation is transitive.

Since the relation is reflexive, symmetric and transitive, then the relation is an equivalence relation.

ii)

$C(quadratic)=\text{Set of all English words that start with letter } 'q'\\ C(rhombus)=\text{Set of all English words that start with letter } 'r'$

iii)

Since there are 26 English alphabets, then will we have 26 equivalence class

iv)

The English words will be partition into words with same first alphabet

b)

$xRy$ if $x-y>0$ $, x,y \in \Z$

To check if it is reflexive, let $x \in \Z$. We want to check if $xRx$

$x-x=0 \not>3$

$\implies x \cancel R x$

Thus, it is not reflexive.

To check if it is symmetric. Let $x,y \in \Z$ and $xRy$

$xRy \implies x-y >3\\ \implies -x+y <-3\\ \implies y-x < -3 \not> 3\\ \implies y \cancel R x$

Thus, the relation is not symmetric.

To check if it is transitive. Let $x,y,z \in \Z, xRy \text{ and } yRz$

$xRy\implies x-y >3\\ yRz \implies y-z>3$

Add the two together, we have that:

$x-y>3 \implies xRz$

Thus the relation is transitive.