Q: Determine whether each of the following relations are reflexive, symmetric and transitive. (i) Relation R in the set $A=\{1,2,3, \ldots, 13,14\}$ defined as $$ R=\{(x, y): 3 x-y=0\} $$ (ii) Relation R in the set of N natural numbers defined as $$ R=\{(x, y): y=x+5 \text { and } x<4\} $$ (iii) Relation R in the set $A=\{1,2,3,4,5,6\}$ defined as $$ R=\{(x, y): y \text { is divisible by }

(i) $\quad R=\{(1,3),(2,6),(3,9),(4,12)\}$ R is not reflexive because $(1,1),(2,2) \ldots$ and $(14,14) \notin R$. R is not symmetric because $(1,3) \in R$, but $(3,1) \notin R$. [since $3(3) \neq 0$ ]. R is not transitive because $(1,3),(3,9) \in R$, but $(1,9) \notin R .[3(1)-9 \neq 0]$. Hence, R is neither reflexive nor symmetric nor transitive. (ii) $R=\{(1,6),(2,7),(3,8)\}$ R is not reflexive because $(1,1) \notin R$. R is not symmetric because $(1,6) \in R$ but $(6,1) \notin R$. R is not transitive because there isn't any ordered pair in R such that $(x, y),(y, z) \in R$, so $(x, z) \notin R$. Hence, R is neither reflexive nor symmetric nor transitive. (iii) $R=\{(x, y): y$ is divisible by $x\}$ We know that any number other than 0 is divisible by itself. Thus, $(x, x) \in R$ So, R is reflexive. $(2,4) \in R \quad$ [because 4 is divisible by 2 ] But $(4,2) \notin R$ [since 2 is not divisible by 4 ] So, R is not symmetric. Let $(x, y)$ and $(y, z) \in R$. So, y is divisible by x and z is divisible by y . So, z is divisible by $\mathrm{x} \Rightarrow(x, z) \in R$ So, R is transitive. So, $R$ is reflexive and transitive but not symmetric. (iv) $R=\{(x, y): x-y$ is an integer $\}$ For $x \in \mathrm{Z},(x, x) \notin R$ because $x-x=0$ is an integer. So, R is reflexive. For, $x, y \in Z$, if $x, y \in \mathrm{R}$, then $x-y$ is an integer $\Rightarrow(y-x)$ is an integer. So, $(y, x) \in R$ So, R is symmetric. Let $(x, y)$ and $(y, z) \in R$, where $x, y, z \in \mathrm{Z}$. $\Rightarrow(x-y)$ and $(y-z)$ are integers. $\Rightarrow x-z=(x-y)+(y-z)$ is an integer. So, R is transitive. So, $R$ is reflexive, symmetric and transitive. (v) a) $\quad R=\{(x, y): x$ and $y$ work at the same place $\}$ R is reflexive because $(x, x) \in R$ R is symmetric because, If $(x, y) \in R$, then $x$ and y work at the same place and y and $x$ also work at the same place. $(y, x) \in R$. R is transitive because, Let $(x, y),(y, z) \in R$ $x$ and

Q: Show that the relation R in the set R of real numbers, defined as $R=\left\{(a, b): a \leq b^{2}\right\}$ is neither reflexive nor symmetric nor transitive.

$R=\left\{(a, b): a \leq b^{2}\right\}$ $\left(\frac{1}{2}, \frac{1}{2}\right) \notin R$ because $\frac{1}{2}>\left(\frac{1}{2}\right)^{2}$ R is not reflexive. $(1,4) \in R$ as $1(1.5)^{2}=2.25$ $\therefore(3,1.5) \notin R$ R is not transitive. R is neither reflective nor symmetric nor transitive.

Q: Check whether the relation R in R defined as $R=\left\{(a, b): a \leq b^{3}\right\}$ is reflexive, symmetric or transitive.

$R=\left\{(a, b): a \leq b^{3}\right\}$ $\left(\frac{1}{2}, \frac{1}{2}\right) \notin R$, since $\frac{1}{2}>\left(\frac{1}{2}\right)^{3}$ R is not reflexive. $(1,2) \in R\left(\right.$ as $\left.11=8\right)$ R is not symmetric. $\left(3, \frac{3}{2}\right),\left(\frac{3}{2}, \frac{6}{5}\right) \in R$, since $3\left(\frac{6}{5}\right)^{3}$ R is not transitive. R is neither reflexive nor symmetric nor transitive.

Q: Show that the relation R in the set $\{1,2,3\}$ given by $R=\{(1,2),(2,1)\}$ is symmetric but neither reflexive nor transitive.

$A=\{1,2,3\}$ $R=\{(1,2),(2,1)\}$ $(1,1),(2,2),(3,3) \notin R$ R is not reflexive. $(1,2) \in R$ and $(2,1) \in R$ R is symmetric. $(1,2) \in R$ and $(2,1) \in R$ $(1,1) \in R$ R is not transitive. R is symmetric, but not reflexive or transitive.

Question 1

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

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

$$

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

$$

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

$$

R=\{(x, y): y=x+5 	ext { and } x<4\}

$$

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

$$

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

Accepted Answer

(i) $\quad R=\{(1,3),(2,6),(3,9),(4,12)\}$

R is not reflexive because $(1,1),(2,2) \ldots$ and $(14,14) 
otin R$.

R is not symmetric because $(1,3) \in R$, but $(3,1) 
otin R$. [since $3(3) 
eq 0$ ].

R is not transitive because $(1,3),(3,9) \in R$, but $(1,9) 
otin R .[3(1)-9 
eq 0]$.

Hence, R is neither reflexive nor symmetric nor transitive.

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

R is not reflexive because $(1,1) 
otin R$.

R is not symmetric because $(1,6) \in R$ but $(6,1) 
otin R$.

R is not transitive because there isn't any ordered pair in R such that $(x, y),(y, z) \in R$, so $(x, z) 
otin R$.

Hence, R is neither reflexive nor symmetric nor transitive.

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

We know that any number other than 0 is divisible by itself.

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

So, R is reflexive.

$(2,4) \in R \quad$ [because 4 is divisible by 2 ]

But $(4,2) 
otin R$ [since 2 is not divisible by 4 ]

So, R is not symmetric.

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

So, z is divisible by $\mathrm{x} \Rightarrow(x, z) \in R$

So, R is transitive.

So, $R$ is reflexive and transitive but not symmetric.

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

For $x \in \mathrm{Z},(x, x) 
otin R$ because $x-x=0$ is an integer.

So, R is reflexive.

For, $x, y \in Z$, if $x, y \in \mathrm{R}$, then $x-y$ is an integer $\Rightarrow(y-x)$ is an integer.

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

So, R is symmetric.

Let $(x, y)$ and $(y, z) \in R$, where $x, y, z \in \mathrm{Z}$.

$\Rightarrow(x-y)$ and $(y-z)$ are integers.

$\Rightarrow x-z=(x-y)+(y-z)$ is an integer.

So, R is transitive.

So, $R$ is reflexive, symmetric and transitive.

(v)

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

R is reflexive because $(x, x) \in R$

R is symmetric because,

If $(x, y) \in R$, then $x$ and y work at the same place and y and $x$ also work at the same place. $(y, x) \in R$.

R is transitive because,

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

$x$ and

Question 2

Show that the relation R in the set R of real numbers, defined as $R=\left\{(a, b): a \leq b^{2}\right\}$ is neither reflexive nor symmetric nor transitive.

Accepted Answer

$R=\left\{(a, b): a \leq b^{2}ight\}$

$\left(\frac{1}{2}, \frac{1}{2}ight) 
otin R$ because $\frac{1}{2}>\left(\frac{1}{2}ight)^{2}$

R is not reflexive.

$(1,4) \in R$ as $1(1.5)^{2}=2.25$

$	herefore(3,1.5) 
otin R$

R is not transitive.

R is neither reflective nor symmetric nor transitive.

Question 3

Check whether the relation R defined in the set $\{1,2,3,4,5,6\}$ as $R=\{(a, b): b=a+1\}$ is reflexive, symmetric or transitive.

Accepted Answer

$A=\{1,2,3,4,5,6\}$

$R=\{(a, b): b=a+1\}$

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

$(a, a) 
otin R, a \in A$

$(1,1),(2,2),(3,3),(4,4),(5,5) 
otin R$

R is not reflexive.

$(1,2) \in R$, but $(2,1) 
otin R$

R is not symmetric.

$$

\begin{aligned}

& (1,2),(2,3) \in R \

& (1,3) 
otin R \

& \quad \mathrm{R} 	ext { is not transitive. }

\end{aligned}

$$

R is neither reflective nor symmetric nor transitive.

Question 4

Show that the relation R in R defined as $R=\{(a, b): a \leq b\}$ is reflexive and transitive, but not symmetric.

Accepted Answer

$R=\{(a, b): a \leq b\}$

$(a, a) \in R$

R is reflexive.

$$

\begin{aligned}

& (2,4) \in R(	ext { as } 22) \

& \quad R 	ext { is not symmetric. }

\end{aligned}

$$

\begin{aligned}

& (a, b),(b, c) \in R \

& a \leq b 	ext { and } b \leq c \

& \Rightarrow a \leq c \

& \Rightarrow(a, c) \in R \

& \quad \mathrm{R} 	ext { is transitive. }

\end{aligned}

$$

$R$ is reflexive and transitive but not symmetric.

Question 5

Check whether the relation R in R defined as $R=\left\{(a, b): a \leq b^{3}\right\}$ is reflexive, symmetric or transitive.

Accepted Answer

$R=\left\{(a, b): a \leq b^{3}ight\}$

$\left(\frac{1}{2}, \frac{1}{2}ight) 
otin R$, since $\frac{1}{2}>\left(\frac{1}{2}ight)^{3}$

R is not reflexive.

$(1,2) \in R\left(ight.$ as $\left.11=8ight)$

R is not symmetric.

$\left(3, \frac{3}{2}ight),\left(\frac{3}{2}, \frac{6}{5}ight) \in R$, since $3\left(\frac{6}{5}ight)^{3}$

R is not transitive.

R is neither reflexive nor symmetric nor transitive.

Question 6

Show that the relation R in the set $\{1,2,3\}$ given by $R=\{(1,2),(2,1)\}$ is symmetric but neither reflexive nor transitive.

Accepted Answer

$A=\{1,2,3\}$

$R=\{(1,2),(2,1)\}$

$(1,1),(2,2),(3,3) 
otin R$

R is not reflexive.

$(1,2) \in R$ and $(2,1) \in R$

R is symmetric.

$(1,2) \in R$ and $(2,1) \in R$

$(1,1) \in R$

R is not transitive.

R is symmetric, but not reflexive or transitive.

Question 7

Show that the relation R in the set A of all books in a library of a college, given by $R=\{(x, y): x$ and $y$ have same number of pages $\}$ is an equivalence relation.

Accepted Answer

$R=\{(x, y): x$ and $y$ have same number of pages $\}$

R is reflexive since $(x, x) \in R$ as $x$ and $x$ have same number of pages.

R is reflexive.

$(x, y) \in R$

$x$ and y have same number of pages and y and $x$ have same number of pages $(y, x) \in R$ R is symmetric.

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

$x$ and $y$ have same number of pages, $y$ and $z$ have same number of pages.

Then $x$ and $z$ have same number of pages.

$(x, z) \in R$

R is transitive.

$R$ is an equivalence relation.

Question 8

Show that the relation R in the set $A=\{1,2,3,4,5\}$ given by $R=\{(a, b):|a-b|$ is even $\}$ is an equivalence relation. Show that all the elements of $\{1,3,5\}$ are related to each other and all the elements of $\{2,4\}$ are related to each other. But no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$.

Accepted Answer

$a \in A$

$|a-a|=0$ (which is even)

R is reflective.

$(a, b) \in R$

$\Rightarrow|a-b|$ [is even]

$\Rightarrow|-(a-b)|=|b-a|$ [is even]

$(b, a) \in R$

R is symmetric.

$(a, b) \in R$ and $(b, c) \in \mathrm{R}$

$\Rightarrow|a-b|$ is even and $|b-c|$ is even

$\Rightarrow(a-b)$ is even and $(b-c)$ is even

$\Rightarrow(a-c)=(a+b)+(b-c)$ is even

$\Rightarrow|a-b|$ is even

$\Rightarrow(a, c) \in R$

R is transitive.

$R$ is an equivalence relation.

All elements of $\{1,3,5\}$ are related to each other because they are all odd. So, the modulus of the difference between any two elements is even.

Similarly, all elements $\{2,4\}$ are related to each other because they are all even.

No element of $\{1,3,5\}$ is related to any elements of $\{2,4\}$ as all elements of $\{1,3,5\}$ are odd and all elements of $\{2,4\}$ are even. So, the modulus of the difference between the two elements will not be even.

1. Relations and Functions

1.1

1.2

1.3

1.4

Miscellaneous