Question

Determine whether each of the following relations are reflexive, symmetric, and transitive:

$\left(i\right)$ Relation $R$ in the set $A=\{1,2,3\dots 13,14\}$ defined as $R=\left\{\right(x,y):3x-y=0\}$

$\left(ii\right)$ Relation $R$ in the set $N$ of natural numbers defined as $R=\left\{\right(x,y):y=x+5\text{and}x<4\}$

$\left(iii\right)$ Relation $R$ in the set $A=\{1,2,3,4,5,6\}$ as
$R=\left\{\right(x,y):y\text{is divisible by}x\}$

$\left(iv\right)$ Relation $R$ in the set $Z$ of all integers defined as $R=\left\{\right(x,y):x-y\text{is an integer}\}$

$\left(v\right)$ Relation $R$ in the set $A$ of human beings in a town at a particular time given by

Answer 1

$\left(i\right)$ Given : Relation $R$ in the set
$A=\{1,2,3\dots 13,14\}$ defined as
$R=\left\{\right(x,y):3x-y=0\}$
So, $3x-y=0$
$\Rightarrow 3x=y\Rightarrow y=3x,$
Where $x,y\in A$
Now, substituting $x=1$
$y=3\in A$
Substituting $x=2$
$y=6\in A$
Substituting $x=3$
$y=9\in A$
Substituting $x=4$
$y=12\in A$
Substituting $x=5$
$y=15\notin A$
For $x=5\Rightarrow y=15\notin A$
$\therefore R=\left\{\right(1,3),(2,6),(3,9),(4,12\left)\right\}$

$A=\{1,2,\mathrm{3...13},14\}$
$R=\left\{\right(1,3),(2,6),(3,9),(4,12\left)\right\}$
Checking reflexivity of a relation:
Relation is reflexive, then $(a,a)\in A$ for every $a\in A$ Since $(1,1)\notin R$
$\therefore R$ is not reflexive.

Checking given relation is symmetric or not:
Here $(1,3)\in R\text{but}(3,1)\notin R$
$\therefore R$ is not symmetric.

Checking transitivity of a relation:
Here, $(1,3)\in R\text{and}(3,9)\in R\text{but}(1,9)\notin R$.
$\therefore R$ is not transitive.


$\left(ii\right)$ Given: Relation $R$ in the set $N$ of natural numbers defined as
$R=\left\{\right(x,y):y=x+5\text{and}x<4\}$

Here $x\&y$ are natural numbers and $x<4$
Substituting $x=1$
$y=x+5=1+5=6$
Substituting $x=2$
$y=x+5=2+5=7$
Substituting $x=3$
$y=x+5=3+5=8$
$\therefore R=\left\{\right(1,6),(2,7),(3,8\left)\right\}$

Checking reflexivity of a relation:
If the relation is reflexive,
then $(x,x)\in R$ for every $x\in N$
Since $(1,1)\notin R$
$\therefore R$ is not reflexive

Checking given relation is symmetric or not:
Here $(1,6)\in R,\text{but}(6,1)\notin R$
$\therefore R$ is not symmetric.

Checking transitivity of a relation:
If $(x,y)\in R\&(y,z)\in R\text{then}(x,z)\in R$
But there is no such pair
$\therefore R$ is not transitive.


$\left(iii\right)$ Given : Relation $R$ in the set
$A=\{1,2,3,4,5,6\}$ as
$R=\left\{\right(x,y):y\text{is divisible by}x\}$.

Checking reflexivity of a relation:
$\because x$ is divisible by $x$
$\Rightarrow (x,x)\in R$
$\therefore R$ is reflexive.

Checking given relation is symmetric or not:
If $x,y\in R,\text{then}(y,x)\in R$
Here $(3,6)\in R$, as $6$ is divisible by $3$ but $(6,3)\notin R$ as $3$ is not divisible by $6$
$\therefore R$ is not symmetric.

Checking transitivity of a relation :
If $y$ is divisible by $x\&z$ is divisible by $y$, then $z$ is divisible by $x$
$\therefore \text{If}(x,y)\in R\&(y,z)\in R,\text{then}(x,z)\in R$
For example :
$(1,5)\in R\&(5,5)\in R\text{so}(1,5)\in R$
$\therefore R$ is transitive.

Hence, $R$ is reflexive, transitive but not symmetric.


$\left(iv\right)$ Given: Relation $R$ in the set $Z$ of all integers defined as
$R=\left\{\right(x,y):x-y\text{is an integer}\}$

Checking reflexivity of a relation:
Since, $x-x=0\&0$ is an integer
$\therefore x-x\text{is an integer}\Rightarrow (x,x)\in R$
$\therefore R$ is reflexive.

Checking given relation is symmetric or not:
If $x-y$ is an integer, then $-(x-y)$ also an integer.
$\Rightarrow y-x$ is also an integer.
So, $(x,y)\in R,\text{then}(y,x)\in R$
$\therefore R$ is symmetric.

Checking transitivity of a relation:
Let $(x,y,z)\in Z$
$\because x-y$ is an integer,
Assuming
$x-y=k_1$, where $k_1$ is any integer.
$y-z=k_2$, where $k_2$ is any integer.
Adding both equations,
$x-z=k_1+k_2,$
And $(k_1+k_2)\in Z$
$\Rightarrow (x,y)\in R\&(y,z)\in R,\text{then}(x,z)\in R$
$\therefore R$ is transitive.