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

Solution:

(i) A = {1, 2, 3 … 13, 14}

R = {(x, y): 3x − y = 0}

∴R = {(1, 3), (2, 6), (3, 9), (4, 12)}

$R$ is not reflexive since $(1,1),(2,2) \ldots(14,14) \notin R$.

Also, $R$ is not symmetric as $(1,3) \in R$, but $(3,1) \notin R .[3(3)-1 \neq 0]$

Also, $R$ is not transitive as $(1,3),(3,9) \in R$, but $(1,9) \notin R$.

$[3(1)-9 \neq 0]$

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

(ii) $R=\{(x, y): y=x+5$ and $x<4\}=\{(1,6),(2,7),(3,8)\}$

It is seen that $(1,1) \notin R$.

∴R is not reflexive.

$(1,6) \in R$

But,

 

$(6,1) \notin \mathrm{R} .$

∴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 $R$.

∴ R is not transitive.

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

(iii) A = {1, 2, 3, 4, 5, 6}

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

We know that any number (x) is divisible by itself.

$\Rightarrow(x, x) \in R$

∴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$]$

∴R is not symmetric.

Let $(x, y),(y, z) \in R$. Then, $y$ is divisible by $x$ and $z$ is divisible by $y$.

∴z is divisible by x.

$\Rightarrow(x, z) \in R$

∴R is transitive.

Hence, R is reflexive and transitive but not symmetric.

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

Now, for every $x \in \mathbf{Z},(x, x) \in R$ as $x-x=0$ is an integer.

∴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.

$\therefore(y, x) \in R$

∴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.

$\therefore(x, z) \in R$

∴R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

$\Rightarrow(x, x) \in \mathrm{R}$

∴ 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.

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

∴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.

$\Rightarrow(x, z) \in R$

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

Clearly $(x, x) \in R$ as $x$ and $x$ is the same human being.

∴ 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.

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

∴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.

$\Rightarrow(x, z) \in R$

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

Now,

$(x, x) \notin R$

Since human being x cannot be taller than himself.

∴R is not reflexive.

Now, let $(x, y) \in R$.

$\Rightarrow x$ is exactly $7 \mathrm{~cm}$ taller than $y$.

Then, y is not taller than x.

$\therefore(y, x) \notin \mathrm{R}$

Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.

∴R is not symmetric.

Now,

Let $(x, y),(y, z) \in R$.

$\Rightarrow x$ is exactly $7 \mathrm{~cm}$ taller thany and $y$ is exactly $7 \mathrm{~cm}$ taller than $z$.

$\Rightarrow x$ is exactly $14 \mathrm{~cm}$ taller than $z$.

$\therefore(x, z) \notin \mathrm{R}$

∴ R is not transitive.

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

(d) R = {(x, y): x is the wife of y}

Now,

$(x, x) \notin R$

Since x cannot be the wife of herself.

∴R is not reflexive.

Now, let $(x, y) \in R$

$\Rightarrow x$ is the wife of $y$

Clearly y is not the wife of x.

$\therefore(y, x) \notin \mathrm{R}$

Indeed if x is the wife of y, then y is the husband of x.

∴ R is not symmetric.

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.

$\therefore(x, z) \notin \mathrm{R}$

∴R is not transitive.

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

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

       $(x, x) \notin \mathrm{R}$

As x cannot be the father of himself.

∴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.

$\therefore(y, x) \notin \mathbf{R}$

∴ 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.

$\therefore(x, z) \notin \mathrm{R}$

∴R is not transitive.

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