Examples (Revised) - Chapter 1 Relations And Functions class 12 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

NCERT Solutions Class 12 Maths: Relations and Functions Revised Syllabus

Example 1

Let $A$ be the set of all students of a boys school. Show that the relation $R$ in A given by $\mathrm{R}=\{(a, b): a$ is sister of $b\}$ is the empty relation and $\mathrm{R}^{\prime}=\{(a, b)$ : the difference between heights of $a$ and $b$ is less than 3 meters $\}$ is the universal relation.

Solution

Since the school is boys school, no student of the school can be sister of any student of the school. Hence, $\mathrm{R}=\phi$, showing that $\mathrm{R}$ is the empty relation. It is also obvious that the difference between heights of any two students of the school has to be less than 3 meters. This shows that $\mathrm{R}^{\prime}=\mathrm{A} \times \mathrm{A}$ is the universal relation.

Remark In Class XI, we have seen two ways of representing a relation, namely raster method and set builder method. However, a relation $R$ in the set $\{1,2,3,4\}$ defined by $R$ $=\{(a, b): b=a+1\}$ is also expressed as $a \mathrm{R} b$ if and only if $b=a+1$ by many authors. We may also use this notation, as and when convenient.
If $(a, b) \in \mathrm{R}$, we say that $a$ is related to $b$ and we denote it as $a \mathrm{R} b$.
One of the most important relation, which plays a significant role in Mathematics, is an equivalence relation. To study equivalence relation, we first consider three types of relations, namely reflexive, symmetric and transitive.

Example 2

Let $\mathrm{T}$ be the set of all triangles in a plane with $\mathrm{R}$ a relation in $\mathrm{T}$ given by $R=\left\{\left(T_1, T_2\right): T_1\right.$ is congruent to $\left.T_2\right\}$. Show that $R$ is an equivalence relation.

Solution

$\mathrm{R}$ is reflexive, since every triangle is congruent to itself. Further, $\left(T_1, T_2\right) \in R \Rightarrow T_1$ is congruent to $T_2 \Rightarrow T_2$ is congruent to $T_1 \Rightarrow\left(T_2, T_1\right) \in R$. Hence, $R$ is symmetric. Moreover, $\left(T_1, T_2\right),\left(T_2, T_3\right) \in R \Rightarrow T_1$ is congruent to $T_2$ and $T_2$ is congruent to $\mathrm{T}_3 \Rightarrow \mathrm{T}_1$ is congruent to $\mathrm{T}_3 \Rightarrow\left(\mathrm{T}_1, \mathrm{~T}_3\right) \in \mathrm{R}$. Therefore, $\mathrm{R}$ is an equivalence relation.

Example 3

Let $L$ be the set of all lines in a plane and $R$ be the relation in $L$ defined as $\mathrm{R}=\left\{\left(\mathrm{L}_1, \mathrm{~L}_2\right): \mathrm{L}_1\right.$ is perpendicular to $\left.\mathrm{L}_2\right\}$. Show that $\mathrm{R}$ is symmetric but neither reflexive nor transitive.

Solution

$\mathrm{R}$ is not reflexive, as a line $\mathrm{L}_1$ can not be perpendicular to itself, i.e., $\left(\mathrm{L}_1, \mathrm{~L}_1\right)$ $\notin \mathrm{R} . \mathrm{R}$ is symmetric as $\left(\mathrm{L}_1, \mathrm{~L}_2\right) \in \mathrm{R}$

$\notin R . R$ is symmetric as $\left(L_1, L_2\right) \in R$
$\Rightarrow \quad \mathrm{L}_1$ is perpendicular to $\mathrm{L}_2$
$\Rightarrow \quad \mathrm{L}_2$ is perpendicular to $\mathrm{L}_1$
$\Rightarrow \quad\left(\mathrm{L}_2, \mathrm{~L}_1\right) \in \mathrm{R}$.
$R$ is not transitive. Indeed, if $L_1$ is perpendicular to $L_2$ and
$\mathrm{L}_2$ is perpendicular to $\mathrm{L}_3$, then $\mathrm{L}_1$ can never be perpendicular to
$\mathrm{L}_3$. In fact, $\mathrm{L}_1$ is parallel to $\mathrm{L}_3$, i.e., $\left(\mathrm{L}_1, \mathrm{~L}_2\right) \in \mathrm{R},\left(\mathrm{L}_2, \mathrm{~L}_3\right) \in \mathrm{R}$ but $\left(\mathrm{L}_1, \mathrm{~L}_3\right) \notin \mathrm{R}$.

Example 4

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

Solution

$\mathrm{R}$ is reflexive, since $(1,1),(2,2)$ and $(3,3)$ lie in $\mathrm{R}$. Also, $\mathrm{R}$ is not symmetric, as $(1,2) \in R$ but $(2,1) \notin R$. Similarly, $R$ is not transitive, as $(1,2) \in R$ and $(2,3) \in R$ but $(1,3) \notin \mathrm{R}$.

Example 5

Show that the relation $R$ in the set $\mathbf{Z}$ of integers given by
$
\mathrm{R}=\{(a, b): 2 \text { divides } a-b\}
$
is an equivalence relation.
Solution

$\mathrm{R}$ is reflexive, as 2 divides $(a-a)$ for all $a \in \mathbf{Z}$. Further, if $(a, b) \in \mathbf{R}$, then 2 divides $a-b$. Therefore, 2 divides $b-a$. Hence, $(b, a) \in \mathrm{R}$, which shows that $\mathrm{R}$ is symmetric. Similarly, if $(a, b) \in \mathrm{R}$ and $(b, c) \in \mathrm{R}$, then $a-b$ and $b-c$ are divisible by 2. Now, $a-c=(a-b)+(b-c)$ is even (Why?). So, $(a-c)$ is divisible by 2 . This shows that $\mathrm{R}$ is transitive. Thus, $\mathrm{R}$ is an equivalence relation in $\mathbf{Z}$.

In Example 5, note that all even integers are related to zero, as $(0, \pm 2),(0, \pm 4)$ etc., lie in $\mathrm{R}$ and no odd integer is related to 0 , as $(0, \pm 1),(0, \pm 3)$ etc., do not lie in $\mathrm{R}$. Similarly, all odd integers are related to one and no even integer is related to one. Therefore, the set $\mathrm{E}$ of all even integers and the set $\mathrm{O}$ of all odd integers are subsets of $\mathbf{Z}$ satisfying following conditions:
(i) All elements of $\mathrm{E}$ are related to each other and all elements of $\mathrm{O}$ are related to each other.
(ii) No element of $\mathrm{E}$ is related to any element of $\mathrm{O}$ and vice-versa.
(iii) $\mathrm{E}$ and $\mathrm{O}$ are disjoint and $\mathrm{Z}=\mathrm{E} \cup \mathrm{O}$.

The subset $\mathrm{E}$ is called the equivalence class containing zero and is denoted by [0]. Similarly, $O$ is the equivalence class containing 1 and is denoted by [1]. Note that $[0] \neq[1],[0]=[2 r]$ and $[1]=[2 r+1], r \in \mathbf{Z}$. Infact, what we have seen above is true for an arbitrary equivalence relation $\mathrm{R}$ in a set $\mathrm{X}$. Given an arbitrary equivalence relation $\mathrm{R}$ in an arbitrary set $\mathrm{X}, \mathrm{R}$ divides $\mathrm{X}$ into mutually disjoint subsets $\mathrm{A}_i$ called partitions or subdivisions of $\mathrm{X}$ satisfying:
(i) all elements of $\mathrm{A}_i$ are related to each other, for all $i$.
(ii) no element of $\mathrm{A}_i$ is related to any element of $\mathrm{A}_j, i \neq j$.
(iii) $\cup \mathrm{A}_j=\mathrm{X}$ and $\mathrm{A}_i \cap \mathrm{A}_j=\phi, i \neq j$.

The subsets $\mathrm{A}_i$ are called equivalence classes. The interesting part of the situation is that we can go reverse also. For example, consider a subdivision of the set $\mathbf{Z}$ given by three mutually disjoint subsets $A_1, A_2$ and $A_3$ whose union is $\mathbf{Z}$ with
$
\begin{aligned}
& \mathbf{A}_1=\{x \in \mathbf{Z}: x \text { is a multiple of } 3\}=\{\ldots,-6,-3,0,3,6, \ldots\} \\
& A_2=\{x \in \mathbf{Z}: x-1 \text { is a multiple of } 3\}=\{\ldots,-5,-2,1,4,7, \ldots\} \\
& A_3=\{x \in \mathbf{Z}: x-2 \text { is a multiple of } 3\}=\{\ldots,-4,-1,2,5,8, \ldots\}
\end{aligned}
$

Define a relation $\mathbf{R}$ in $\mathbf{Z}$ given by $\mathrm{R}=\{(a, b): 3$ divides $a-b\}$. Following the arguments similar to those used in Example 5, we can show that $R$ is an equivalence relation. Also, $\mathrm{A}_1$ coincides with the set of all integers in $\mathbf{Z}$ which are related to zero, $\mathrm{A}_2$ coincides with the set of all integers which are related to 1 and $\mathrm{A}_3$ coincides with the set of all integers in $\mathbf{Z}$ which are related to 2. Thus, $A_1=[0], A_2=[1]$ and $A_3=[2]$. In fact, $\mathrm{A}_1=[3 r], \mathrm{A}_2=[3 r+1]$ and $\mathrm{A}_3=[3 r+2]$, for all $r \in \mathbf{Z}$.

Example 6

Let $\mathrm{R}$ be the relation defined in the set $\mathrm{A}=\{1,2,3,4,5,6,7\}$ by $\mathrm{R}=\{(a, b)$ : both $a$ and $b$ are either odd or even $\}$. Show that $\mathrm{R}$ is an equivalence relation. Further, show that all the elements of the subset $\{1,3,5,7\}$ are related to each other and all the elements of the subset $\{2,4,6\}$ are related to each other, but no element of the subset $\{1,3,5,7\}$ is related to any element of the subset $\{2,4,6\}$.

Solution

Given any element $a$ in A, both $a$ and $a$ must be either odd or even, so that $(a, a) \in \mathrm{R}$. Further, $(a, b) \in \mathrm{R} \Rightarrow$ both $a$ and $b$ must be either odd or even $\Rightarrow(b, a) \in \mathrm{R}$. Similarly, $(a, b) \in \mathrm{R}$ and $(b, c) \in \mathrm{R} \Rightarrow$ all elements $a, b, c$, must be either even or odd simultaneously $\Rightarrow(a, c) \in \mathrm{R}$. Hence, $\mathrm{R}$ is an equivalence relation. Further, all the elements of $\{1,3,5,7\}$ are related to each other, as all the elements of this subset are odd. Similarly, all the elements of the subset $\{2,4,6\}$ are related to each other, as all of them are even. Also, no element of the subset $\{1,3,5,7\}$ can be related to any element of $\{2,4,6\}$, as elements of $\{1,3,5,7\}$ are odd, while elements of $\{2,4,6\}$ are even.

Example 7

Let $\mathrm{A}$ be the set of all 50 students of Class $\mathrm{X}$ in a school. Let $f: \mathrm{A} \rightarrow \mathrm{N}$ be function defined by $f(x)=$ roll number of the student $x$. Show that $f$ is one-one but not onto.

Solution

No two different students of the class can have same roll number. Therefore, $f$ must be one-one. We can assume without any loss of generality that roll numbers of students are from 1 to 50 . This implies that 51 in $\mathbf{N}$ is not roll number of any student of the class, so that 51 can not be image of any element of $\mathrm{X}$ under $f$. Hence, $f$ is not onto.

Example 8 

Show that the function $f: \mathbf{N} \rightarrow \mathbf{N}$, given by $f(x)=2 x$, is one-one but not onto.

Solution

The function $f$ is one-one, for $f\left(x_1\right)=f\left(x_2\right) \Rightarrow 2 x_1=2 x_2 \Rightarrow x_1=x_2$. Further, $f$ is not onto, as for $1 \in \mathbf{N}$, there does not exist any $x$ in $\mathbf{N}$ such that $f(x)=2 x=1$.

Example 9

Prove that the function $f: \mathbf{R} \rightarrow \mathbf{R}$, given by $f(x)=2 x$, is one-one and onto.

Solution

$f$ is one-one, as $f\left(x_1\right)=f\left(x_2\right) \Rightarrow 2 x_1=2 x_2 \Rightarrow x_1=x_2$. Also, given any real number $y$ in $\mathrm{R}$, there exists $\frac{y}{2}$ in $\mathrm{R}$ such that $f\left(\frac{y}{2}\right)=2 \cdot\left(\frac{y}{2}\right)=y$. Hence, $f$ is onto.

Example 10

Show that the function $f: \mathbf{N} \rightarrow \mathbf{N}$, given by $f(1)=f(2)=1$ and $f(x)=x-1$, for every $x>2$, is onto but not one-one.

Solution

$f$ is not one-one, as $f(1)=f(2)=1$. But $f$ is onto, as given any $y \in \mathbf{N}, y \neq 1$, we can choose $x$ as $y+1$ such that $f(y+1)=y+1-1=y$. Also for $1 \in \mathbf{N}$, we have $f(1)=1$.

Example 11

Show that the function $f: \mathbf{R} \rightarrow \mathbf{R}$, defined as $f(x)=x^2$, is neither one-one nor onto.

Solution

Since $f(-1)=1=f(1), f$ is not oneone. Also, the element -2 in the co-domain $\mathbf{R}$ is not image of any element $x$ in the domain $\mathbf{R}$ (Why?). Therefore $f$ is not onto.

Example 12

Show that $f: \mathbf{N} \rightarrow \mathbf{N}$, given by
$
f(x)=\begin{aligned}
& x+1, \text { if } x \text { is odd, } \\
& x-1, \text { if } x \text { is even }
\end{aligned}
$
is both one-one and onto.

Solution

Suppose $f\left(x_1\right)=f\left(x_2\right)$. Note that if $x_1$ is odd and $x_2$ is even, then we will have $x_1+1=x_2-1$, i.e., $x_2-x_1=2$ which is impossible. Similarly, the possibility of $x_1$ being even and $x_2$ being odd can also be ruled out, using the similar argument. Therefore, both $x_1$ and $x_2$ must be either odd or even. Suppose both $x_1$ and $x_2$ are odd. Then $f\left(x_1\right)=f\left(x_2\right) \Rightarrow x_1+1=x_2+1 \Rightarrow x_1=x_2$. Similarly, if both $x_1$ and $x_2$ are even, then also $f\left(x_1\right)=f\left(x_2\right) \Rightarrow x_1-1=x_2-1 \Rightarrow x_1=x_2$. Thus, $f$ is one-one. Also, any odd number $2 r+1$ in the co-domain $\mathbf{N}$ is the image of $2 r+2$ in the domain $\mathbf{N}$ and any even number $2 r$ in the co-domain $\mathbf{N}$ is the image of $2 r-1$ in the domain $\mathbf{N}$. Thus, $f$ is onto.

Example 13

Show that an onto function $f:\{1,2,3\} \rightarrow\{1,2,3\}$ is always one-one.
Solution

Suppose $f$ is not one-one. Then there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same. Also, the image of 3 under $f$ can be only one element. Therefore, the range set can have at the most two elements of the co-domain $\{1,2,3\}$, showing that $f$ is not onto, a contradiction. Hence, $f$ must be one-one.
Example 14

Show that a one-one function $f:\{1,2,3\} \rightarrow\{1,2,3\}$ must be onto.
Solution

Since $f$ is one-one, three elements of $\{1,2,3\}$ must be taken to 3 different elements of the co-domain $\{1,2,3\}$ under $f$. Hence, $f$ has to be onto.

Remark The results mentioned in Examples 13 and 14 are also true for an arbitrary finite set $\mathrm{X}$, i.e., a one-one function $f: \mathrm{X} \rightarrow \mathrm{X}$ is necessarily onto and an onto map $f: \mathrm{X} \rightarrow \mathrm{X}$ is necessarily one-one, for every finite set $\mathrm{X}$. In contrast to this, Examples 8 and 10 show that for an infinite set, this may not be true. In fact, this is a characteristic difference between a finite and an infinite set.

Example 15

Let $f:\{2,3,4,5\} \rightarrow\{3,4,5,9\}$ and $g:\{3,4,5,9\} \rightarrow\{7,11,15\}$ be functions defined as $f(2)=3, f(3)=4, f(4)=f(5)=5$ and $g(3)=g(4)=7$ and $g(5)=g(9)=11$. Find $g \circ f$.
Solution

We have $g \circ f(2)=g(f(2))=g(3)=7, g \circ f(3)=g(f(3))=g(4)=7$, $g \circ f(4)=g(f(4))=g(5)=11$ and $g \circ f(5)=g(5)=11$.
Example 16

Find gof and $f \circ g$, if $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ are given by $f(x)=\cos x$ and $g(x)=3 x^2$. Show that $g \circ f \neq f \circ g$.
Solution

We have $g \circ f(x)=g(f(x))=g(\cos x)=3(\cos x)^2=3 \cos ^2 x$. Similarly, $f \circ g(x)=f(g(x))=f\left(3 x^2\right)=\cos \left(3 x^2\right)$. Note that $3 \cos ^2 x \neq \cos 3 x^2$, for $x=0$. Hence, $g \circ f \neq f \circ g$.

Example 17

Let $f: \mathbf{N} \rightarrow \mathrm{Y}$ be a function defined as $f(x)=4 x+3$, where, $\mathbf{Y}=\{y \in \mathbf{N}: y=4 x+3$ for some $x \in \mathbf{N}\}$. Show that $f$ is invertible. Find the inverse.

Solution

Consider an arbitrary element $y$ of Y. By the definition of $\mathrm{Y}, y=4 x+3$, for some $x$ in the domain $\mathbf{N}$. This shows that $x=\frac{(y-3)}{4}$. Define $g: \mathrm{Y} \rightarrow \mathbf{N}$ by

$g(y)=\frac{(y-3)}{4}$. Now, $g \circ f(x)=g(f(x))=g(4 x+3)=\frac{(4 x+3-3)}{4}=x$ and
$f \circ g(y)=f(g(y))=f\left(\frac{(y-3)}{4}\right)=\frac{4(y-3)}{4}+3=y-3+3=y$. This shows that $g \circ f=\mathrm{I}_{\mathrm{N}}$ and $f o g=\mathrm{I}_{\mathrm{Y}}$, which implies that $f$ is invertible and $g$ is the inverse of $f$.

Example 18

If $R_1$ and $R_2$ are equivalence relations in a set $A$, show that $R_1 \cap R_2$ is also an equivalence relation.

Solution

Since $\mathrm{R}_1$ and $\mathrm{R}_2$ are equivalence relations, $(a, a) \in \mathrm{R}_1$, and $(a, a) \in \mathrm{R}_2 \forall a \in \mathrm{A}$. This implies that $(a, a) \in \mathrm{R}_1 \cap \mathrm{R}_2, \forall a$, showing $\mathrm{R}_1 \cap \mathrm{R}_2$ is reflexive. Further, $(a, b) \in \mathrm{R}_1 \cap \mathrm{R}_2 \Rightarrow(a, b) \in \mathrm{R}_1$ and $(a, b) \in \mathrm{R}_2 \Rightarrow(b, a) \in \mathrm{R}_1$ and $(b, a) \in \mathrm{R}_2 \Rightarrow$ $(b, a) \in \mathrm{R}_1 \cap \mathrm{R}_2$, hence, $\mathrm{R}_1 \cap \mathrm{R}_2$ is symmetric. Similarly, $(a, b) \in \mathrm{R}_1 \cap \mathrm{R}_2$ and $(b, c) \in \mathrm{R}_1 \cap \mathrm{R}_2 \Rightarrow(a, c) \in \mathrm{R}_1$ and $(a, c) \in \mathrm{R}_2 \Rightarrow(a, c) \in \mathrm{R}_1 \cap \mathrm{R}_2$. This shows that $\mathrm{R}_1 \cap \mathrm{R}_2$ is transitive. Thus, $\mathrm{R}_1 \cap \mathrm{R}_2$ is an equivalence relation.

Example 19

Let $\mathrm{R}$ be a relation on the set $\mathrm{A}$ of ordered pairs of positive integers defined by $(x, y) \mathrm{R}(u, v)$ if and only if $x v=y u$. Show that $\mathrm{R}$ is an equivalence relation.

Solution

Clearly, $(x, y) \mathrm{R}(x, y), \forall(x, y) \in \mathrm{A}$, since $x y=y x$. This shows that $\mathrm{R}$ is reflexive. Further, $(x, y) \mathrm{R}(u, v) \Rightarrow x v=y u \Rightarrow u y=v x$ and hence $(u, v) \mathrm{R}(x, y)$. This shows that $\mathrm{R}$ is symmetric. Similarly, $(x, y) \mathrm{R}(u, v)$ and $(u, v) \mathrm{R}(a, b) \Rightarrow x v=y u$ and $u b=v a \Rightarrow x v \frac{a}{u}=y u \frac{a}{u} \Rightarrow x v \frac{b}{v}=y u \frac{a}{u} \Rightarrow x b=y a$ and hence $(x, y) \mathrm{R}(a, b)$. Thus, $\mathrm{R}$ is transitive. Thus, $\mathrm{R}$ is an equivalence relation.

Example 20

Let $\mathrm{X}=\{1,2,3,4,5,6,7,8,9\}$. Let $\mathrm{R}_1$ be a relation in $\mathrm{X}$ given by $\mathrm{R}_1=\{(x, y): x-y$ is divisible by 3$\}$ and $\mathrm{R}_2$ be another relation on $\mathrm{X}$ given by $\mathrm{R}_2=\{(x, y):\{x, y\} \subset\{1,4,7\}\}$ or $\{x, y\} \subset\{2,5,8\}$ or $\left.\{x, y\} \subset\{3,6,9\}\right\}$. Show that $\mathrm{R}_1=\mathrm{R}_2$.
Solution

Note that the characteristic of sets $\{1,4,7\},\{2,5,8\}$ and $\{3,6,9\}$ is that difference between any two elements of these sets is a multiple of 3 . Therefore, $(x, y) \in \mathrm{R}_1 \Rightarrow x-y$ is a multiple of $3 \Rightarrow\{x, y\} \subset\{1,4,7\}$ or $\{x, y\} \subset\{2,5,8\}$ or $\{x, y\} \subset\{3,6,9\} \Rightarrow(x, y) \in \mathbf{R}_2$. Hence, $\mathbf{R}_1 \subset \mathbf{R}_2$. Similarly, $\{x, y\} \in \mathbf{R}_2 \Rightarrow\{x, y\}$

$\subset\{1,4,7\}$ or $\{x, y\} \subset\{2,5,8\}$ or $\{x, y\} \subset\{3,6,9\} \Rightarrow x-y$ is divisible by $3 \Rightarrow\{x, y\} \in \mathrm{R}_1$. This shows that $\mathrm{R}_2 \subset \mathrm{R}_1$. Hence, $\mathrm{R}_1=\mathrm{R}_2$.
Example 21

Let $f: \mathrm{X} \rightarrow \mathrm{Y}$ be a function. Define a relation $\mathrm{R}$ in $\mathrm{X}$ given by $\mathrm{R}=\{(a, b): f(a)=f(b)\}$. Examine whether $\mathrm{R}$ is an equivalence relation or not.

Solution

For every $a \in \mathrm{X},(a, a) \in \mathrm{R}$, since $f(a)=f(a)$, showing that $\mathrm{R}$ is reflexive. Similarly, $(a, b) \in \mathrm{R} \Rightarrow f(a)=f(b) \Rightarrow f(b)=f(a) \Rightarrow(b, a) \in \mathrm{R}$. Therefore, $\mathrm{R}$ is symmetric. Further, $(a, b) \in \mathrm{R}$ and $(b, c) \in \mathrm{R} \Rightarrow f(a)=f(b)$ and $f(b)=f(c) \Rightarrow f(a)$ $=f(c) \Rightarrow(a, c) \in \mathrm{R}$, which implies that $\mathrm{R}$ is transitive. Hence, $\mathrm{R}$ is an equivalence relation.

Example 22

Find the number of all one-one functions from set $A=\{1,2,3\}$ to itself.
Solution

One-one function from $\{1,2,3\}$ to itself is simply a permutation on three symbols $1,2,3$. Therefore, total number of one-one maps from $\{1,2,3\}$ to itself is same as total number of permutations on three symbols $1,2,3$ which is $3 !=6$.

Example 23

Let $A=\{1,2,3\}$. Then show that the number of relations containing $(1,2)$ and $(2,3)$ which are reflexive and transitive but not symmetric is three.

Solution

The smallest relation $\mathrm{R}_1$ containing $(1,2)$ and $(2,3)$ which is reflexive and transitive but not symmetric is $\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$. Now, if we add the pair $(2,1)$ to $R_1$ to get $R_2$, then the relation $R_2$ will be reflexive, transitive but not symmetric. Similarly, we can obtain $\mathrm{R}_3$ by adding $(3,2)$ to $\mathrm{R}_1$ to get the desired relation. However, we can not add two pairs $(2,1),(3,2)$ or single pair $(3,1)$ to $\mathrm{R}_1$ at a time, as by doing so, we will be forced to add the remaining pair in order to maintain transitivity and in the process, the relation will become symmetric also which is not required. Thus, the total number of desired relations is three.

Example 24

Show that the number of equivalence relation in the set $\{1,2,3\}$ containing $(1,2)$ and $(2,1)$ is two.

Solution

The smallest equivalence relation $\mathrm{R}_1$ containing $(1,2)$ and $(2,1)$ is $\{(1,1)$, $(2,2),(3,3),(1,2),(2,1)\}$. Now we are left with only 4 pairs namely $(2,3),(3,2)$, $(1,3)$ and $(3,1)$. If we add any one, say $(2,3)$ to $R_1$, then for symmetry we must add $(3,2)$ also and now for transitivity we are forced to add $(1,3)$ and $(3,1)$. Thus, the only equivalence relation bigger than $R_1$ is the universal relation. This shows that the total number of equivalence relations containing $(1,2)$ and $(2,1)$ is two.

Example 25

Consider the identity function $\mathrm{I}_{\mathrm{N}}: \mathbf{N} \rightarrow \mathbf{N}$ defined as $\mathrm{I}_{\mathrm{N}}(x)=x \forall x \in \mathbf{N}$. Show that although $\mathrm{I}_{\mathrm{N}}$ is onto but $\mathrm{I}_{\mathrm{N}}+\mathrm{I}_{\mathrm{N}}: \mathrm{N} \rightarrow \mathrm{N}$ defined as
$
\left(\mathrm{I}_{\mathrm{N}}+\mathrm{I}_{\mathrm{N}}\right)(x)=\mathrm{I}_{\mathrm{N}}(x)+\mathrm{I}_{\mathrm{N}}(x)=x+x=2 x \text { is not onto. }
$

Solution

Clearly $I_N$ is onto. But $I_N+I_N$ is not onto, as we can find an element 3 in the co-domain $\mathbf{N}$ such that there does not exist any $x$ in the domain $\mathbf{N}$ with $\left(\mathrm{I}_{\mathrm{N}}+\mathrm{I}_{\mathrm{N}}\right)(x)=2 x=3$.
Example 26

Consider a function $f:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbf{R}$ given by $f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbf{R}$ given by $g(x)=\cos x$. Show that $f$ and $g$ are one-one, but $f+g$ is not one-one.

Solution

Since for any two distinct elements $x_1$ and $x_2$ in $\left[0, \frac{\pi}{2}\right], \sin x_1 \neq \sin x_2$ and $\cos x_1 \neq \cos x_2$, both $f$ and $g$ must be one-one. But $(f+g)(0)=\sin 0+\cos 0=1$ and $(f+g)\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}+\cos \frac{\pi}{2}=1$. Therefore, $f+g$ is not one-one.