Question

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

(i) Relation $R$ in the set $A=\{1,2,3,...,13,14\}$ defined as

$R=\left\{\right(x,y):3x-y=0\}$

(ii) Relative $R$ in the set $N$ of natural numbers defined as

$R=\left\{\right(x,y):y=x+5\text{and}x<4\}$

(iii) Relation $R$ in the set $A=\{1,2,3,4,5,6\}$ as

$R=\left\{\right(x,y):y\text{is divisible by}x\}$

(iv) Relative $R$ in the set $Z$ of all integers defined as

$R=\left\{\right(x,y):x-y\text{is an integer}\}$

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

$\left(a\right)R=\left\{\right(x,y):x\text{and}y\text{work at the same place}\}$

$\left(b\right)R=\left\{\right(x,y):x\text{and}y\text{live in the same locality}\}$

$\left(c\right)R=\left\{\right(x,y):x\text{is exactly}7\text{cm taller than}y\}$

$\left(d\right)R=\left\{\right(x,y):x\text{is wife of}y\}$

$\left(e\right)R=\left\{\right(x,y):x\text{is father of}y\}$

Answer 1

(i)

$A=\{1,2,3,....13,14\}$
$R=\left\{\right(x,y):3x-y=0\}$
$\therefore R=\left\{\right(1,3),(2,6),(3,9),(4,12\left)\right\}$
$R$ is not reflexive since $(1,1),(2,2)......(14,14)\notin R$
Also, $R$ is not symmetric as $(1,3)\in R$, but $(3,1)\notin R.$
Also, $R$ is not transitive as $(1,3),(3,9)\in R$, but $(1,9)\notin R$.
Hence, $R$ is neither reflexive, nor symmetric, nor transitive.

(ii)

$R=\left\{\right(x,y):y=x+5\text{and}x<4\}=\left\{\right(1,6),(2,7),(3,8\left)\right\}$
It is seen that $(1,1)\notin R\Rightarrow R$ is not reflexive.
Also $(1,6)\in R$.

But, $(6,1)\notin R$. $\therefore R$ is not symmetric.
Now, since there is no pair in $R$ such that $(x,y)$ and $(y,z)\in R$, then $(x,z)$ cannot belong to $R$.
$\therefore R$ is not transitive.
Hence, $R$ is neither reflexive, nor symmetric, nor transitive.

(iii)

$A=\{1,2,3,4,5,6\}$
$R=\left\{\right(x,y):\text{y is divisible by x}\}$
We know that any number $\left(x\right)$ is divisible by itself.
$\Rightarrow (x,x)\in R$,
$\therefore R$ is reflexive.
Now, $(2,4)\in R$, [ as $4$ is divisible by $2$]
But,
$(4,2)\notin R.$, [ as $2$ is not divisible by $4$]
$\therefore R$ is not symmetric.
Let $(x,y),(y,z)\in R$. Then, $y$ is divisible by $x$ and $z$ is divisible by $y$.
$\therefore z$ is divisible by $x$.
$\Rightarrow (x,z)\in R$
$\therefore R$ is transitive.
Hence, $R$ is reflexive and transitive but not symmetric.

(iv)

$R=\left\{\right(x,y):x-y\text{is an integer}\}$
Now, for every $x\in Z,(x,x)\in R$ as $x-x=0$ is an integer.
$\therefore R$ is reflexive.
Now, for every $x,y\in Z$ if $(x,y)\in R$, then $x-y$ is an integer.
$\Rightarrow -(x-y)$ is also an integer.
$\Rightarrow (y-x)$ is an integer.
$\therefore (y,x)\in R$
$\Rightarrow R$ is symmetric.
Now,
Let $(x,y)$ and $(y,z)\in R$, where $x,y,z\in Z$.
$\Rightarrow (x-y)$ and $(y-z)$ are integers.
$\Rightarrow x-z=(x-y)+(y-z)$ is an integer.
$\therefore (x,z)\in R$
$\therefore R$ is transitive.
Hence, $R$ is reflexive, symmetric, and transitive.

(v) (a)

$R=\left\{\right(x,y):xandyworkatthesameplace\}$
$\Rightarrow (x,x)\in R$
$\therefore R$ is reflexive.
If $(x,y)\in R$, then $x$ and $y$ work at the same place.
$\Rightarrow y$ and $x$ work at the same place.
$\Rightarrow (y,x)\in R$
$\therefore R$ is symmetric.
Now, let $(x,y),(y,z)\in R$
$\Rightarrow x$ and $y$ work at the same place and $y$ and $z$ work at the same place.
$\Rightarrow x$ and $z$ work at the same place.
$\Rightarrow (x,z)\in R$
$\therefore R$ is transitive.
Hence, $R$ is reflexive, symmetric and transitive.

(v) (b)

$R=\left\{\right(x,y):\text{x and y live in the same locality}\}$
Clearly $(x,y)\in R$ as $x$ and $x$ is the same human being.
$\therefore R$ is reflexive.
If $(x,y)\in R$, then $x$ and $y$ live in the same locality.
$\Rightarrow y$ and $x$ live in the same locality.
$\Rightarrow (y,x)\in R$
$\therefore R$ is symmetric.
Now, let $(x,y)\in R$ and $(y,z)\in R$.
$\Rightarrow$ $x$ and $y$ live in the same locality and $y$ and $z$ live in the same locality.
$\Rightarrow$ $x$ and $z$ live in the same locality.
$\Rightarrow (x,z)\in R$
$\therefore R$ is transitive.
Hence, $R$ is reflexive, symmetric and transitive.

(v)(c)

$R=\left\{\right(x,y):\text{x is exactly 7 cm taller than y}\}$
Now,
$(x,x)\notin R$
Since human being $x$ cannot be taller than himself.
$\therefore R$ is not reflexive.
Now, let $(x,y)\in R$.
$\Rightarrow x$ is exactly $7$cm taller than $y$
Then, $y$ is not taller than $x$.
$\therefore (y,x)\notin R$
Indeed if $x$ is exactly $7$cm taller than $y$, then $y$ is exactly $7$cm shorter than $x$.
$\therefore R$ is not symmetric.
Now, let $(x,y),(y,z)\in R$
$\Rightarrow$ $x$ is exactly $7$cm taller than $y$ and $y$ is exactly $7$cm taller than$z$.
$\Rightarrow$ $x$ is exactly $14$ taller than $z$.
$\therefore (x,z)\notin R$
$\therefore R$ is not transitive.
Hence, $R$ is neither reflexive, nor symmetric, nor transitive.

(v)(d)

$R=\left\{\right(x,y):\text{x is the wife of y}\}$
Now, $(x,x)\notin R$
Since $x$ cannot be the wife of herself.
$\therefore R$ is not reflexive.
Now, let $(x,y)\in R$
$\Rightarrow x$ is the wife of $y$.
Clearly $y$ is not wife of $x$.
$\therefore (y,x)\notin R$
Indeed if $x$ is the wife of $y$, then $y$ is the husband of $x$.
$\therefore R$ is not transitive.
Let $(x,y),(y,z)\in R$
$\Rightarrow$ $x$ is the wife of $y$ and $y$ is the wife of $z$.
This case is not possible. Also, this does not imply that $x$ is the wife of $z$.
$\therefore (x,z)\notin R$
$\therefore R$ is not transitive.
Hence, $R$ is neither reflexive, nor symmetric, nor transitive.

(v)(e)

$R=\left\{\right(x,y):\text{x is the father of y}\}$
$(x,x)\notin R$
As $x$ cannot be the father of himself.
$\therefore R$ is not reflexive.
Now, let $(x,y)\in R$
$\Rightarrow x$ is the father of $y$.
$\Rightarrow y$ cannot be the father of $y$.
Indeed, $y$ is the son or the daughter of $y$.
$\therefore (y,x)\notin R$
$\therefore R$ is not symmetric.
Now, let $(x,y)\in R$ and $(y,z)\in R$.
$\Rightarrow x$ is the father of $y$ and $y$ is the father of $z$.
$\Rightarrow x$ is not the father of $z$.
Indeed $x$ is the grandfather of $z$.
$\therefore (x,z)\notin R$
$\therefore R$ is not transitive.
Hence, $R$ is neither reflexive, nor symmetric, nor transitive.