a) Consider the whole of the English words set. Suppose an English word x is related to another English word y if x and y begin with the same letter. i) Show that this is an equivalence relation. ii) Compute C(quadratic) and C(rhombus) iii) How many equivalence classes are there in all, and why? iv) What is the partition of the English words under this relation? b) Consider Z, the set of integers. Suppose we define the relation: x is related to y if x - y > 3, x, y \in Z. Determine whether or not the relation is i) reflexive ii) symmetric iii) transitive

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.