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

(i) A $= \{1, 2, 3, 4, 5, 6, \cdots., 13, 14 \}$ is the given set
R $= \{(x, y) : 3x-y$ = $0\} \Rightarrow$ R $= \{(1, 3), (2, 6) (3, 9), (4, 12)\}$

Reflexive

Let $x \in A$ be any element.

Since, $(x,\;x) \notin R\;\;\;\;\therefore \;\;R\;\;\;is\;\;not\;\;reflexive$

Symmetric

$x,\;y \in A,\;\;(x,\;y) \in R\;\;but\;\;(y,\;x) \notin R$

Therefore, $R$ is not symmetric.

Transitive

$x,\;y,\;z \in A\;\; \Rightarrow (x,\;y) \in R\;\;and\;\;(y,\;z) \in R\; \Rightarrow (x,\;z) \in R$

For example : (1, 3)$\in$R and (3, 9)$\in$R but (1, 9) $\notin R$ ,therefore, R is not transitive.

Hence, R is neither reflexive, nor symmetric and nor transitive.

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

(ii) N is the set of natural numbers

R $= \{(x, y) : y$ = $x + 5$ and $x < 4\}$ is the set of natural numbers.

$R = \{ (1,\;6)(2,\;7),\;(3,\;8)\}$

Reflexive

Let $x \in N$ be any element.

$(x,\;x) \notin R\;\;\;\therefore \;\;R$ is not reflexive.

Symmetric

$x,\;y \in N,\;\;(x,\;y) \in R\;\;\;but\;\;(y,\;x) \notin R$

Therefore, R is not symmetric.

Transitive

(1, 6)$\in$R and (6, 7) $\notin R$ and (1, 7) $\notin R$

Therefore, R is not transitive.

Hence, R is neither reflexive, nor symmetric and nor transitive.

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

(iii) A $=$ {1, 2, 3, 4, 5, 6} is the given set

R $= \{(x, y ) :$y is divisible by x in A $\}$

R $= \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6),(3, 3), (3, 6), (4, 4), (5, 5), (6, 6)\}$

Reflexive

Let $x \in A$ be any element.

Now, $(x,\;x) \in R\;\;\;i.e.\;\;(1,\;1) \in R,\;(2,\;2) \in R,\;(3,\;3) \in R,\;(4,\;4) \in R,\;(5,\;5) \in R,\;(6,\;6) \in R$

R is reflexive.

Symmetric

$x,\;y \in A,\;\;(x,\;y) \in R\;\;$ $(y,\;x) \in R$

i.e., $(1,\;2) \in R\;\;but\;\;(2,\;1) \notin R$

Therefore, R is not symmetric.

Transitive

$x,\;y,\;z \in A,\;\;(x,\;y) \in R,\;(y,\;z) \in R\;\; \Rightarrow (x,\;z) \in R$

i.e., $(1,\;2) \in R\;\;and\;\;(2,\;4) \in R\;\; \Rightarrow \;(1,\;4) \in R$

Thus, R is transitive.

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

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

(iv) Z is the set of all integers

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

Reflexive

Let $x \in Z$ , be any element, (x, x) i.e., (1, 1) $=$ 1 - 1 $=$ 0$\in$Z.

Therefore, R is reflexive.

Symmetric

$x,\;y \in Z,\;\;(x,\;y) \in R\;\;\;\; \Rightarrow (y,\;x) \in R$

i.e.,  $x - y$ is an integer $\Rightarrow$ $(y,\;x) \in R$

i.e., $x - y$ is an integer $\Rightarrow$ $y - x$ is also an integer.

Therefore, R is symmetric.

Transitive

$(x,\;y) \in R\;\;and\;\;(y,\;z) \in R$

i.e., $(x - y)$ is integer and $(y - z)$ is integer

$\Rightarrow$ $(x - z) = (x - y + y - z) \in \;\;\;$ integer $\Rightarrow$ $(x,\;z) \in R$

Hence, R is reflexive, symmetric and transitive.

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

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

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

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

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

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

Relation R in the set A of human beings in a town at a particular time.

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

Reflexive

$(x,\;x) \in R$ because x and x work at the same place. Thus, R is reflexive.

Symmetric

Let $(x,\;y) \in R \Rightarrow$ x and y work at the same place

$\Rightarrow$ y and x work at the same place $\Rightarrow$ $(y,\;x) \in R$

Thus, R is symmetric.

Transitive

(x, y) R and (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

Thus, R is transitive.

Hence, R is reflexive, symmetric and transitive

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

Reflexive

(x, x)$\in$R because x and x live in the same locality.

Therefore, R is reflexive.

Symmetric

Let (x, y)$\in$R $\Rightarrow$ x and y live in the same locality

$\Rightarrow$ y and x also live in the same locality $\Rightarrow$ (y, x )$\in$R

Thus, R is symmetric.

Transitive

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 (x, z)$\in$R

Thus, R is transitive.

Hence, R is reflexive, symmetric and transitive.

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

Reflexive

x is not exactly 7 cm taller than x, so (x, x) $\in$ R, thus R is not reflexive.

Symmetric

If x is exactly 7 cm taller than y, then y is not exactly 7 cm taller than x.

So, if (x, y) $\in$ R then (y, x) $\notin$ R $\Rightarrow$ R is not symmetric.

Transitive

If x is exactly 7 cm taller thany and if y is exactly 7 cm taller than z, then it does not imply that x is exactly 7 cm taller than z.Thus, R is not transitive.

Hence, R is not reflexive, not symmetric and not transitive.

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

Reflexive

x is not wife of x, therefore, (x, x) $\notin$ R and thus R is not reflexive.

Symmetric

If x is wife of y, then y is not wife of x.

If (x, y)$\in$R, then (y, x) $\notin$ R.

So, R is not symmetric.

Transitive

If x is the wife ofy, then y is not wife of z.

and R is transitive as transitivity is not contradicted in this case.

$(x,\;y) \in R$ and $(y,\;z) \notin R,$ then $(x,\;z) \notin R,$ for any z

if x is wife of y, then y is a male and male cannot be wife]

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

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

Reflexive

x is not father of x, so (x, x) $\notin$ R, so R is not reflexive.

Symmetric

If x is father of y, then y is not father of x.

If (x, y)$\in$R, then (y, x) $\notin$ R, so R is not symmetric.

Transitive

If (x, y) $\in$ R and (y, z)$\in$R, then (x, z)$\notin$ R.

i.e., x is father of y, y is father of z, then x is not father of z.

So, R is not transitive.

Hence, R is neither reflexive, nor symmetric nor transitive.