Q1 (i): Relation R in the set A=1,2,3, …, 13,14 defined as R=(x, y): 3 x-y=0

(i) Relation R in the set A=1,2,3, …, 13,14 defined as R=(x, y): 3 x-y=0 Given: Set A=1,2,3, …, 13,14 Relation R=(x, y): 3 x-y=0, which can be written as R=(x, y): y=3x. Solution: First, let's list the elements of the relation R. The pairs (x, y) must be in A × A. If x=1, y=3(1)=3. So, (1, 3) in R.…

Q1 (ii): Relation R in the set N of natural numbers defined as R=(x, y): y=x+5 and x<4

(i) Relation R in the set A=1,2,3, …, 13,14 defined as R=(x, y): 3 x-y=0 Given: Set A=1,2,3, …, 13,14 Relation R=(x, y): 3 x-y=0, which can be written as R=(x, y): y=3x. Solution: First, let's list the elements of the relation R. The pairs (x, y) must be in A × A. If x=1, y=3(1)=3. So, (1, 3) in R.…

Q1 (iii): Relation R in the set A=1,2,3,4,5,6 as R=(x, y): y is divisible by x

(i) Relation R in the set A=1,2,3, …, 13,14 defined as R=(x, y): 3 x-y=0 Given: Set A=1,2,3, …, 13,14 Relation R=(x, y): 3 x-y=0, which can be written as R=(x, y): y=3x. Solution: First, let's list the elements of the relation R. The pairs (x, y) must be in A × A. If x=1, y=3(1)=3. So, (1, 3) in R.…

Q1 (iv): Relation R in the set Z of all integers defined as R=(x, y): x-y is an integer

(i) Relation R in the set A=1,2,3, …, 13,14 defined as R=(x, y): 3 x-y=0 Given: Set A=1,2,3, …, 13,14 Relation R=(x, y): 3 x-y=0, which can be written as R=(x, y): y=3x. Solution: First, let's list the elements of the relation R. The pairs (x, y) must be in A × A. If x=1, y=3(1)=3. So, (1, 3) in R.…

Q1 (v): Relation R in the set A of human beings in a 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…

(i) Relation R in the set A=1,2,3, …, 13,14 defined as R=(x, y): 3 x-y=0 Given: Set A=1,2,3, …, 13,14 Relation R=(x, y): 3 x-y=0, which can be written as R=(x, y): y=3x. Solution: First, let's list the elements of the relation R. The pairs (x, y) must be in A × A. If x=1, y=3(1)=3. So, (1, 3) in R.…

Relations and Functions Class 12 Mathematics NCERT Solutions

Q1EXERCISE 1.1

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

(i)

Relation R in the set $\mathrm{A}=\{1,2,3, \ldots, 13,14\}$ defined as $\mathrm{R}=\{(x, y): 3 x-y=0\}$

(ii)

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

(iii)

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

(iv)

Relation R in the set $\mathbf{Z}$ of all integers defined as $\mathrm{R}=\{(x, y): x-y \text { is an integer }\}$

(v)

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

(a)

\mathrm{R}=\{(x, y): x

and

y

work at the same place

\}

(b)

\mathrm{R}=\{(x, y): x

and

y

live in the same locality

\}

(c)

\mathrm{R}=\{(x, y): x

is exactly 7 cm taller than

y\}

(d)

\mathrm{R}=\{(x, y): x

is wife of

y\}

(e)

\mathrm{R}=\{(x, y): x

is father of

y\}

Solution

(i) Relation R in the set $\mathrm{A}=\{1,2,3, \ldots, 13,14\}$ defined as $\mathrm{R}=\{(x, y): 3 x-y=0\}$

Given: Set

\mathrm{A}=\{1,2,3, \ldots, 13,14\}

Relation

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

, which can be written as

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

.

Solution: First, let's list the elements of the relation R. The pairs

(x, y)

must be in A

\times

A.

If $x=1$ , $y=3(1)=3$ . So, $(1, 3) \in \mathrm{R}$ .
If $x=2$ , $y=3(2)=6$ . So, $(2, 6) \in \mathrm{R}$ .
If $x=3$ , $y=3(3)=9$ . So, $(3, 9) \in \mathrm{R}$ .
If $x=4$ , $y=3(4)=12$ . So, $(4, 12) \in \mathrm{R}$ .
If $x=5$ , $y=3(5)=15$ . Since $15 \notin \mathrm{A}$ , this pair is not in R. We stop here. So, $\mathrm{R} = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$ .

Checking the properties:

Reflexivity: A relation is reflexive if $(x, x) \in \mathrm{R}$ for every $x \in \mathrm{A}$ . For $x=1$ , $(1, 1)$ should be in R. This means $3(1)-1=0$ , which implies $2=0$ . This is false. So, $(1, 1) \notin \mathrm{R}$ . Therefore, R is not reflexive.
Symmetry: A relation is symmetric if $(x, y) \in \mathrm{R}$ implies $(y, x) \in \mathrm{R}$ . We have $(1, 3) \in \mathrm{R}$ . For R to be symmetric, $(3, 1)$ must also be in R. This would require $3(3)-1=0$ , which implies $8=0$ . This is false. So, $(3, 1) \notin \mathrm{R}$ . Therefore, R is not symmetric.
Transitivity: A relation is transitive if $(x, y) \in \mathrm{R}$ and $(y, z) \in \mathrm{R}$ implies $(x, z) \in \mathrm{R}$ . We have $(1, 3) \in \mathrm{R}$ and $(3, 9) \in \mathrm{R}$ . For R to be transitive, $(1, 9)$ must also be in R. This would require $3(1)-9=0$ , which implies $-6=0$ . This is false. So, $(1, 9) \notin \mathrm{R}$ . Therefore, R is not transitive.

Final Answer for (i): The relation R is neither reflexive, nor symmetric, nor transitive.

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

Given: Set is

\mathbf{N}

(natural numbers). Relation

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

.

Solution: Since

x

is a natural number and

x<4

, the possible values for

x

are 1, 2, and 3.

If $x=1$ , $y=1+5=6$ . So, $(1, 6) \in \mathrm{R}$ .
If $x=2$ , $y=2+5=7$ . So, $(2, 7) \in \mathrm{R}$ .
If $x=3$ , $y=3+5=8$ . So, $(3, 8) \in \mathrm{R}$ . So, $\mathrm{R} = \{(1, 6), (2, 7), (3, 8)\}$ .

Checking the properties:

Reflexivity: For R to be reflexive, $(x, x) \in \mathrm{R}$ for every $x \in \mathbf{N}$ . For $x=1$ , $(1, 1)$ should be in R. This requires $1=1+5$ , which is false. So, $(1, 1) \notin \mathrm{R}$ . Therefore, R is not reflexive.
Symmetry: For R to be symmetric, if $(x, y) \in \mathrm{R}$ , then $(y, x) \in \mathrm{R}$ . We have $(1, 6) \in \mathrm{R}$ . For R to be symmetric, $(6, 1)$ must be in R. For $(6, 1) \in \mathrm{R}$ , we need $x=6$ . But the condition for R is $x<4$ . Since $6$ is not less than $4$ , $(6, 1) \notin \mathrm{R}$ . Therefore, R is not symmetric.
Transitivity: For R to be transitive, if $(x, y) \in \mathrm{R}$ and $(y, z) \in \mathrm{R}$ , then $(x, z) \in \mathrm{R}$ . The elements of R are $(1, 6), (2, 7), (3, 8)$ . For any pair $(x, y) \in \mathrm{R}$ , the second element $y$ is 6, 7, or 8. There is no pair in R whose first element is 6, 7, or 8. Thus, the condition $(x, y) \in \mathrm{R}$ and $(y, z) \in \mathrm{R}$ is never satisfied. A relation is considered non-transitive only if we find a case where $(x, y) \in \mathrm{R}$ and $(y, z) \in \mathrm{R}$ but $(x, z) \notin \mathrm{R}$ . Since we cannot find such a case, the condition for transitivity is vacuously true. Therefore, R is transitive.

Final Answer for (ii): The relation R is not reflexive and not symmetric, but it is transitive.

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

Given: Set

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

. Relation

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

.

Solution: Checking the properties:

Reflexivity: For any $x \in \mathrm{A}$ , $x$ is divisible by $x$ . So, $(x, x) \in \mathrm{R}$ for all $x \in \mathrm{A}$ . For example, $(1,1), (2,2), ..., (6,6)$ are all in R. Therefore, R is reflexive.
Symmetry: If $(x, y) \in \mathrm{R}$ , this means $y$ is divisible by $x$ . This does not imply that $x$ is divisible by $y$ . For example, $(2, 4) \in \mathrm{R}$ because 4 is divisible by 2. However, 2 is not divisible by 4, so $(4, 2) \notin \mathrm{R}$ . Therefore, R is not symmetric.
Transitivity: Let $(x, y) \in \mathrm{R}$ and $(y, z) \in \mathrm{R}$ . $(x, y) \in \mathrm{R} \Rightarrow y$ is divisible by $x \Rightarrow y = kx$ for some integer $k$ . $(y, z) \in \mathrm{R} \Rightarrow z$ is divisible by $y \Rightarrow z = my$ for some integer $m$ . Substituting the value of $y$ , we get $z = m(kx) = (mk)x$ . Since $m$ and $k$ are integers, their product $mk$ is also an integer. This shows that $z$ is divisible by $x$ . So, $(x, z) \in \mathrm{R}$ . Therefore, R is transitive.

Final Answer for (iii): The relation R is reflexive and transitive, but not symmetric.

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

Given: Set is

\mathbf{Z}

(all integers). Relation

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

.

Solution: Checking the properties:

Reflexivity: For any integer $x \in \mathbf{Z}$ , $x-x = 0$ . Since 0 is an integer, $(x, x) \in \mathrm{R}$ for all $x \in \mathbf{Z}$ . Therefore, R is reflexive.
Symmetry: Let $(x, y) \in \mathrm{R}$ . This means $x-y$ is an integer. Let $x-y = k$ , where $k \in \mathbf{Z}$ . Then $y-x = -(x-y) = -k$ . Since $k$ is an integer, $-k$ is also an integer. So, $y-x$ is an integer, which means $(y, x) \in \mathrm{R}$ . Therefore, R is symmetric.
Transitivity: Let $(x, y) \in \mathrm{R}$ and $(y, z) \in \mathrm{R}$ . $(x, y) \in \mathrm{R} \Rightarrow x-y$ is an integer. Let $x-y = k$ , where $k \in \mathbf{Z}$ . $(y, z) \in \mathrm{R} \Rightarrow y-z$ is an integer. Let $y-z = m$ , where $m \in \mathbf{Z}$ . Adding these two equations: $(x-y) + (y-z) = k+m$ . This simplifies to $x-z = k+m$ . Since $k$ and $m$ are integers, their sum $k+m$ is also an integer. So, $x-z$ is an integer, which means $(x, z) \in \mathrm{R}$ . Therefore, R is transitive.

Final Answer for (iv): The relation R is reflexive, symmetric, and transitive. It is an equivalence relation.

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

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

Reflexive: Any person $x$ works at the same place as themself. So $(x,x) \in R$ . R is reflexive.
Symmetric: If $x$ and $y$ work at the same place, then $y$ and $x$ also work at the same place. So if $(x,y) \in R$ , then $(y,x) \in R$ . R is symmetric.
Transitive: If $x$ and $y$ work at the same place, and $y$ and $z$ work at the same place, then $x$ and $z$ must also work at that same place. So if $(x,y) \in R$ and $(y,z) \in R$ , then $(x,z) \in R$ . R is transitive. Conclusion (a): R is reflexive, symmetric, and transitive.

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

Reflexive: Any person $x$ lives in the same locality as themself. So $(x,x) \in R$ . R is reflexive.
Symmetric: If $x$ and $y$ live in the same locality, then $y$ and $x$ also live in the same locality. So if $(x,y) \in R$ , then $(y,x) \in R$ . R is symmetric.
Transitive: If $x$ and $y$ live in the same locality, and $y$ and $z$ live in the same locality, then $x$ and $z$ must also live in that same locality. So if $(x,y) \in R$ and $(y,z) \in R$ , then $(x,z) \in R$ . R is transitive. Conclusion (b): R is reflexive, symmetric, and transitive.

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

Reflexive: A person $x$ cannot be exactly 7 cm taller than themself. So $(x,x) \notin R$ . R is not reflexive.
Symmetric: If $x$ is exactly 7 cm taller than $y$ , then $y$ is exactly 7 cm shorter than $x$ , not taller. So if $(x,y) \in R$ , then $(y,x) \notin R$ . R is not symmetric.
Transitive: If $x$ is exactly 7 cm taller than $y$ , and $y$ is exactly 7 cm taller than $z$ , then $x$ is exactly $7+7=14$ cm taller than $z$ . $x$ is not 7 cm taller than $z$ . So if $(x,y) \in R$ and $(y,z) \in R$ , then $(x,z) \notin R$ . R is not transitive. Conclusion (c): R is neither reflexive, nor symmetric, nor transitive.

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

Reflexive: A person $x$ cannot be their own wife. So $(x,x) \notin R$ . R is not reflexive.
Symmetric: If $x$ is the wife of $y$ , then $y$ is the husband of $x$ , not the wife. So if $(x,y) \in R$ , then $(y,x) \notin R$ . R is not symmetric.
Transitive: If $x$ is the wife of $y$ , then $y$ is a male. $y$ cannot be the wife of anyone. So the condition $(x,y) \in R$ and $(y,z) \in R$ can never be satisfied. Thus, we cannot find a counterexample to transitivity. The relation is vacuously transitive. Conclusion (d): R is not reflexive, not symmetric, but is transitive.

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

Reflexive: A person $x$ cannot be their own father. So $(x,x) \notin R$ . R is not reflexive.
Symmetric: If $x$ is the father of $y$ , then $y$ is the son or daughter of $x$ , not the father. So if $(x,y) \in R$ , then $(y,x) \notin R$ . R is not symmetric.
Transitive: If $x$ is the father of $y$ , and $y$ is the father of $z$ , then $x$ is the grandfather of $z$ , not the father. So if $(x,y) \in R$ and $(y,z) \in R$ , then $(x,z) \notin R$ . R is not transitive. Conclusion (e): R is neither reflexive, nor symmetric, nor transitive.

Q2EXERCISE 1.1

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

Solution

Given: The relation R in the set

\mathbf{R}

of real numbers, defined as

\mathrm{R}=\left\{(a, b): a \leq b^{2}\right\}

.

To Show: The relation R is neither reflexive, nor symmetric, nor transitive.

Proof:

1. Checking for Reflexivity A relation is reflexive if

(a, a) \in \mathrm{R}

for every

a \in \mathbf{R}

. This means

a \leq a^2

must be true for all real numbers

a

. Let's consider a counterexample. Let

a = \frac{1}{2}

. Then

a^2 = (\frac{1}{2})^2 = \frac{1}{4}

. The condition

a \leq a^2

becomes

\frac{1}{2} \leq \frac{1}{4}

, which is false. Since the condition fails for

a = \frac{1}{2}

,

(\frac{1}{2}, \frac{1}{2}) \notin \mathrm{R}

. Hence, R is not reflexive.

2. Checking for Symmetry A relation is symmetric if

(a, b) \in \mathrm{R}

implies

(b, a) \in \mathrm{R}

. This means if

a \leq b^2

, then it must follow that

b \leq a^2

. Let's consider a counterexample. Let

a = 1

and

b = 2

. For

(1, 2)

:

a \leq b^2 \Rightarrow 1 \leq 2^2 \Rightarrow 1 \leq 4

, which is true. So,

(1, 2) \in \mathrm{R}

. Now let's check for

(2, 1)

. The condition is

b \leq a^2 \Rightarrow 2 \leq 1^2 \Rightarrow 2 \leq 1

, which is false. So,

(2, 1) \notin \mathrm{R}

. Since

(1, 2) \in \mathrm{R}

but

(2, 1) \notin \mathrm{R}

. Hence, R is not symmetric.

3. Checking for Transitivity A relation is transitive if

(a, b) \in \mathrm{R}

and

(b, c) \in \mathrm{R}

implies

(a, c) \in \mathrm{R}

. This means if

a \leq b^2

and

b \leq c^2

, then it must follow that

a \leq c^2

. Let's consider a counterexample. Let

a = 2

,

b = -2

, and

c = -1

. For

(a, b) = (2, -2)

: The condition is

a \leq b^2 \Rightarrow 2 \leq (-2)^2 \Rightarrow 2 \leq 4

, which is true. So,

(2, -2) \in \mathrm{R}

. For

(b, c) = (-2, -1)

: The condition is

b \leq c^2 \Rightarrow -2 \leq (-1)^2 \Rightarrow -2 \leq 1

, which is true. So,

(-2, -1) \in \mathrm{R}

. Now let's check for

(a, c) = (2, -1)

. The condition is

a \leq c^2 \Rightarrow 2 \leq (-1)^2 \Rightarrow 2 \leq 1

, which is false. So,

(2, -1) \notin \mathrm{R}

. Since

(2, -2) \in \mathrm{R}

and

(-2, -1) \in \mathrm{R}

, but

(2, -1) \notin \mathrm{R}

. Hence, R is not transitive.

Conclusion: The relation R is neither reflexive, nor symmetric, nor transitive. Hence Proved.

Q3EXERCISE 1.1

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

Solution

Given: Set

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

. Relation R defined as

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

.

To Check: Whether the relation R is reflexive, symmetric, or transitive.

Solution: First, let's list the elements of the relation R.

If $a=1$ , $b=1+1=2$ . So, $(1, 2) \in \mathrm{R}$ .
If $a=2$ , $b=2+1=3$ . So, $(2, 3) \in \mathrm{R}$ .
If $a=3$ , $b=3+1=4$ . So, $(3, 4) \in \mathrm{R}$ .
If $a=4$ , $b=4+1=5$ . So, $(4, 5) \in \mathrm{R}$ .
If $a=5$ , $b=5+1=6$ . So, $(5, 6) \in \mathrm{R}$ .
If $a=6$ , $b=6+1=7$ . Since $7 \notin \mathrm{A}$ , this pair is not in R. So, $\mathrm{R} = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$ .

1. Checking for Reflexivity A relation is reflexive if

(a, a) \in \mathrm{R}

for every

a \in \mathrm{A}

. Let's check for

a=1

. For

(1, 1)

to be in R, we must have

1 = 1+1

, which means

1=2

. This is false. So,

(1, 1) \notin \mathrm{R}

. Hence, R is not reflexive.

2. Checking for Symmetry A relation is symmetric if

(a, b) \in \mathrm{R}

implies

(b, a) \in \mathrm{R}

. We have

(1, 2) \in \mathrm{R}

. For R to be symmetric,

(2, 1)

must also be in R. For

(2, 1)

to be in R, we must have

1 = 2+1

, which means

1=3

. This is false. Also, looking at the elements of R,

(2,1)

is not present. Hence, R is not symmetric.

3. Checking for Transitivity A relation is transitive if

(a, b) \in \mathrm{R}

and

(b, c) \in \mathrm{R}

implies

(a, c) \in \mathrm{R}

. Let's take an example. We have

(1, 2) \in \mathrm{R}

and

(2, 3) \in \mathrm{R}

. For R to be transitive,

(1, 3)

must also be in R. For

(1, 3)

to be in R, we must have

3 = 1+1

, which means

3=2

. This is false. Also, looking at the elements of R,

(1,3)

is not present. Hence, R is not transitive.

Final Answer: The relation R is neither reflexive, nor symmetric, nor transitive.

Q4EXERCISE 1.1

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

Solution

Given: The relation R in the set

\mathbf{R}

of real numbers, defined as

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

.

To Show: The relation R is reflexive and transitive but not symmetric.

Proof:

1. Checking for Reflexivity A relation is reflexive if

(a, a) \in \mathrm{R}

for every

a \in \mathbf{R}

. This means we need to check if

a \leq a

for all real numbers

a

. Since any real number

a

is equal to itself (

a=a

), the condition

a \leq a

is always true. Therefore,

(a, a) \in \mathrm{R}

for all

a \in \mathbf{R}

. Hence, R is reflexive.

2. Checking for Symmetry A relation is symmetric if

(a, b) \in \mathrm{R}

implies

(b, a) \in \mathrm{R}

. This means if

a \leq b

, then it must follow that

b \leq a

. This is not always true. For example, let

a=2

and

b=3

. We have

2 \leq 3

, so

(2, 3) \in \mathrm{R}

. However,

3 \leq 2

is false, so

(3, 2) \notin \mathrm{R}

. Since

(2, 3) \in \mathrm{R}

but

(3, 2) \notin \mathrm{R}

. Hence, R is not symmetric.

3. Checking for Transitivity A relation is transitive if

(a, b) \in \mathrm{R}

and

(b, c) \in \mathrm{R}

implies

(a, c) \in \mathrm{R}

. Let

(a, b) \in \mathrm{R}

and

(b, c) \in \mathrm{R}

. This means

a \leq b

and

b \leq c

. From the properties of inequalities, if

a

is less than or equal to

b

, and

b

is less than or equal to

c

, it follows that

a

must be less than or equal to

c

. So,

a \leq c

, which implies

(a, c) \in \mathrm{R}

. Hence, R is transitive.

Conclusion: The relation R is reflexive and transitive, but not symmetric. Hence Proved.

Q5EXERCISE 1.1

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

Solution

Given: The relation R in the set

\mathbf{R}

of real numbers, defined by

\mathrm{R}=\left\{(a, b): a \leq b^{3}\right\}

.

To Check: Whether the relation R is reflexive, symmetric, or transitive.

Solution:

1. Checking for Reflexivity A relation is reflexive if

(a, a) \in \mathrm{R}

for every

a \in \mathbf{R}

. This means

a \leq a^3

must be true for all real numbers

a

. Let's consider a counterexample. Let

a = \frac{1}{2}

. Then

a^3 = (\frac{1}{2})^3 = \frac{1}{8}

. The condition

a \leq a^3

becomes

\frac{1}{2} \leq \frac{1}{8}

, which is false. Since the condition fails for

a = \frac{1}{2}

,

(\frac{1}{2}, \frac{1}{2}) \notin \mathrm{R}

. Hence, R is not reflexive.

2. Checking for Symmetry A relation is symmetric if

(a, b) \in \mathrm{R}

implies

(b, a) \in \mathrm{R}

. This means if

a \leq b^3

, then it must follow that

b \leq a^3

. Let's consider a counterexample. Let

a = 1

and

b = 2

. For

(1, 2)

: The condition is

a \leq b^3 \Rightarrow 1 \leq 2^3 \Rightarrow 1 \leq 8

, which is true. So,

(1, 2) \in \mathrm{R}

. Now let's check for

(2, 1)

. The condition is

b \leq a^3 \Rightarrow 2 \leq 1^3 \Rightarrow 2 \leq 1

, which is false. So,

(2, 1) \notin \mathrm{R}

. Since

(1, 2) \in \mathrm{R}

but

(2, 1) \notin \mathrm{R}

. Hence, R is not symmetric.

3. Checking for Transitivity A relation is transitive if

(a, b) \in \mathrm{R}

and

(b, c) \in \mathrm{R}

implies

(a, c) \in \mathrm{R}

. This means if

a \leq b^3

and

b \leq c^3

, then it must follow that

a \leq c^3

. Let's consider a counterexample. Let

a = 100

,

b = 5

, and

c = 2

. For

(a, b) = (100, 5)

: The condition is

a \leq b^3 \Rightarrow 100 \leq 5^3 \Rightarrow 100 \leq 125

, which is true. So,

(100, 5) \in \mathrm{R}

. For

(b, c) = (5, 2)

: The condition is

b \leq c^3 \Rightarrow 5 \leq 2^3 \Rightarrow 5 \leq 8

, which is true. So,

(5, 2) \in \mathrm{R}

. Now let's check for

(a, c) = (100, 2)

. The condition is

a \leq c^3 \Rightarrow 100 \leq 2^3 \Rightarrow 100 \leq 8

, which is false. So,

(100, 2) \notin \mathrm{R}

. Since

(100, 5) \in \mathrm{R}

and

(5, 2) \in \mathrm{R}

, but

(100, 2) \notin \mathrm{R}

. Hence, R is not transitive.

Final Answer: The relation R is neither reflexive, nor symmetric, nor transitive.

Q6EXERCISE 1.1

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

Solution

Given: The set is

A = \{1, 2, 3\}

. The relation R on A is given by

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

.

To Show: The relation R is symmetric but neither reflexive nor transitive.

Solution:

1. Check for Reflexivity: A relation R on a set A is reflexive if

(a, a) \in R

for every

a \in A

. In this case, the set is

A = \{1, 2, 3\}

. We check if

(1, 1), (2, 2), (3, 3)

are in R.

$(1, 1) \notin R$
$(2, 2) \notin R$
$(3, 3) \notin R$ Since R does not contain $(a, a)$ for any $a \in A$ , the relation R is not reflexive.

2. Check for Symmetry: A relation R on a set A is symmetric if

(a, b) \in R

implies that

(b, a) \in R

for all

a, b \in A

. Here,

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

.

For the pair $(1, 2) \in R$ , we check if $(2, 1)$ is in R. Yes, $(2, 1) \in R$ .
For the pair $(2, 1) \in R$ , we check if $(1, 2)$ is in R. Yes, $(1, 2) \in R$ . Since for every pair in R, its reverse is also in R, the relation R is symmetric.

3. Check for Transitivity: A relation R on a set A is transitive if

(a, b) \in R

and

(b, c) \in R

implies that

(a, c) \in R

for all

a, b, c \in A

. Let's consider the pairs from R. We have

(1, 2) \in R

and

(2, 1) \in R

. According to the condition for transitivity, this should imply that

(1, 1) \in R

. However, we can see that

(1, 1) \notin R

. Therefore, the relation R is not transitive.

Conclusion: The relation R is symmetric, but it is neither reflexive nor transitive.

Hence Shown.

Q7EXERCISE 1.1

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

Solution

Given: Let A be the set of all books in a library of a college. The relation R on A is given by

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

.

To Show: The relation R is an equivalence relation.

Solution: For R to be an equivalence relation, it must be reflexive, symmetric, and transitive.

1. Check for Reflexivity: A relation R on a set A is reflexive if

(x, x) \in R

for every

x \in A

. Let

x

be any book in the library. The pair

(x, x)

would be in R if book

x

and book

x

have the same number of pages. A book always has the same number of pages as itself. So,

(x, x) \in R

for all

x \in A

. Therefore, R is reflexive.

2. Check for Symmetry: A relation R on a set A is symmetric if

(x, y) \in R

implies that

(y, x) \in R

for all

x, y \in A

. Let

(x, y) \in R

. This means that book

x

and book

y

have the same number of pages. If book

x

and book

y

have the same number of pages, then it is also true that book

y

and book

x

have the same number of pages. This implies that

(y, x) \in R

. So,

(x, y) \in R \implies (y, x) \in R

. Therefore, R is symmetric.

3. Check for Transitivity: A relation R on a set A is transitive if

(x, y) \in R

and

(y, z) \in R

implies that

(x, z) \in R

for all

x, y, z \in A

. Let

(x, y) \in R

and

(y, z) \in R

.

$(x, y) \in R$ means that book $x$ and book $y$ have the same number of pages. Let this number be $p$ .
$(y, z) \in R$ means that book $y$ and book $z$ have the same number of pages. Since book $y$ has $p$ pages, book $z$ must also have $p$ pages. From these two statements, we can conclude that book $x$ and book $z$ both have $p$ pages. This means they have the same number of pages. This implies that $(x, z) \in R$ . So, $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$ . Therefore, R is transitive.

Conclusion: Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

Hence Shown.

Q8EXERCISE 1.1

Show that the relation R in the set $\mathrm{A}=\{1,2,3,4,5\}$ given by $\mathrm{R}=\{(a, b):|a-b|$ 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 $\{2,4\}$ .

Solution

Given: The set is

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

. The relation R on A is given by

R = \{(a, b) : |a - b| \text{ is even}\}

.

Part 1: Show that R is an equivalence relation. For R to be an equivalence relation, it must be reflexive, symmetric, and transitive.

Reflexivity: For any element

a \in A

, we consider

|a - a| = |0| = 0

. Since 0 is an even number (

0 = 2 \times 0

),

|a - a|

is even. Thus,

(a, a) \in R

for all

a \in A

. So, R is reflexive.

Symmetry: Let

(a, b) \in R

. This means

|a - b|

is an even number. We know that

|b - a| = |-(a - b)| = |a - b|

. Since

|a - b|

is even,

|b - a|

is also even. Thus,

(b, a) \in R

. So, R is symmetric.

Transitivity: Let

(a, b) \in R

and

(b, c) \in R

.

$(a, b) \in R \implies |a - b|$ is even. This means $a - b$ is an even integer. Let $a - b = 2k$ for some integer $k$ .
$(b, c) \in R \implies |b - c|$ is even. This means $b - c$ is an even integer. Let $b - c = 2m$ for some integer $m$ . Now, let's consider $a - c$ : $a - c = (a - b) + (b - c) = 2k + 2m = 2(k + m)$ . Since $k$ and $m$ are integers, $k + m$ is also an integer. So, $a - c$ is an even integer. This implies that $|a - c|$ is even. Thus, $(a, c) \in R$ . So, R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Part 2: Show relationships between subsets.

Show that all the elements of $\{1, 3, 5\}$ are related to each other. The elements of this set are all odd integers. The difference between any two odd integers is always an even integer.

$|1 - 3| = |-2| = 2$ (even)
$|1 - 5| = |-4| = 4$ (even)
$|3 - 5| = |-2| = 2$ (even) Since the absolute difference between any pair of elements from $\{1, 3, 5\}$ is even, all elements of $\{1, 3, 5\}$ are related to each other.

Show that all the elements of $\{2, 4\}$ are related to each other. The elements of this set are all even integers. The difference between any two even integers is always an even integer.

$|2 - 4| = |-2| = 2$ (even) Since the absolute difference between any pair of elements from $\{2, 4\}$ is even, all elements of $\{2, 4\}$ are related to each other.

Show that no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$ . Let's take one element from

\{1, 3, 5\}

(an odd number) and one element from

\{2, 4\}

(an even number). The difference between an odd integer and an even integer is always an odd integer.

$|1 - 2| = |-1| = 1$ (odd)
$|1 - 4| = |-3| = 3$ (odd)
$|3 - 2| = |1| = 1$ (odd)
$|3 - 4| = |-1| = 1$ (odd)
$|5 - 2| = |3| = 3$ (odd)
$|5 - 4| = |1| = 1$ (odd) In all cases, the absolute difference is an odd number. Therefore, no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$ .

Hence Shown.

Q9EXERCISE 1.1

Show that each of the relation R in the set $\mathrm{A}=\{x \in \mathbf{Z}: 0 \leq x \leq 12\}$ , given by

(i)

$\mathrm{R}=\{(a, b):|a-b|$ is a multiple of 4 $\}$

(ii)

$\mathrm{R}=\{(a, b): a=b\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution

Given: The set is

A = \{x \in \mathbf{Z} : 0 \leq x \leq 12\} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}

.

(i) $R = \{(a, b) : |a - b| \text{ is a multiple of 4}\}$

To Show: R is an equivalence relation. Reflexivity: For any

a \in A

, we have

|a - a| = 0

. Since

0 = 4 \times 0

, 0 is a multiple of 4. So,

(a, a) \in R

. Thus, R is reflexive.

Symmetry: Let

(a, b) \in R

. Then

|a - b|

is a multiple of 4. Let

|a - b| = 4k

for some non-negative integer

k

. Since

|b - a| = |-(a - b)| = |a - b|

, we have

|b - a| = 4k

, which is also a multiple of 4. So,

(b, a) \in R

. Thus, R is symmetric.

Transitivity: Let

(a, b) \in R

and

(b, c) \in R

.

(a, b) \in R \implies |a - b|

is a multiple of 4

\implies a - b

is a multiple of 4. Let

a - b = 4k

for some integer

k

.

(b, c) \in R \implies |b - c|

is a multiple of 4

\implies b - c

is a multiple of 4. Let

b - c = 4m

for some integer

m

. Then

a - c = (a - b) + (b - c) = 4k + 4m = 4(k + m)

. Since

k+m

is an integer,

a-c

is a multiple of 4. This implies

|a - c|

is a multiple of 4. So,

(a, c) \in R

. Thus, R is transitive. Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Find the set of all elements related to 1: We need to find all elements

x \in A

such that

(x, 1) \in R

. This means

|x - 1|

must be a multiple of 4. Since

x \in \{0, 1, ..., 12\}

, we check the possible values of

|x - 1|

:

If $|x - 1| = 0 \implies x - 1 = 0 \implies x = 1$ .
If $|x - 1| = 4 \implies x - 1 = 4$ or $x - 1 = -4$ . This gives $x = 5$ or $x = -3$ . Since $x \in A$ , we have $x = 5$ .
If $|x - 1| = 8 \implies x - 1 = 8$ or $x - 1 = -8$ . This gives $x = 9$ or $x = -7$ . Since $x \in A$ , we have $x = 9$ .
If $|x - 1| = 12 \implies x - 1 = 12$ or $x - 1 = -12$ . This gives $x = 13$ or $x = -11$ . Neither is in A. The next multiple of 4 is 16, which will also give values of $x$ outside A. So, the elements related to 1 are 1, 5, and 9. Final Answer for (i): The set of all elements related to 1 is $\{1, 5, 9\}$ .

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

To Show: R is an equivalence relation. Reflexivity: For any

a \in A

, we have

a = a

, which is true. So,

(a, a) \in R

. Thus, R is reflexive.

Symmetry: Let

(a, b) \in R

. This means

a = b

. If

a = b

, then it is also true that

b = a

. So,

(b, a) \in R

. Thus, R is symmetric.

Transitivity: Let

(a, b) \in R

and

(b, c) \in R

.

(a, b) \in R \implies a = b

.

(b, c) \in R \implies b = c

. From

a = b

and

b = c

, it follows that

a = c

. So,

(a, c) \in R

. Thus, R is transitive. Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Find the set of all elements related to 1: We need to find all elements

x \in A

such that

(x, 1) \in R

. The condition is

x = 1

. The only element in A that is equal to 1 is 1 itself. Final Answer for (ii): The set of all elements related to 1 is

\{1\}

.

Q10EXERCISE 1.1

Give an example of a relation. Which is

(i)

Symmetric but neither reflexive nor transitive.

(ii)

Transitive but neither reflexive nor symmetric.

(iii)

Reflexive and symmetric but not transitive.

(iv)

Reflexive and transitive but not symmetric.

(v)

Symmetric and transitive but not reflexive.

Solution

To Do: Give an example of a relation for each of the five given conditions. Let's consider the set

A = \{1, 2, 3\}

for constructing the examples.

(i) Symmetric but neither reflexive nor transitive. Example: Let the relation R on A be defined as

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

.

Symmetric: For $(1, 2) \in R$ , we have $(2, 1) \in R$ . So, it is symmetric.
Not Reflexive: $(1, 1) \notin R$ , $(2, 2) \notin R$ , $(3, 3) \notin R$ . So, it is not reflexive.
Not Transitive: We have $(1, 2) \in R$ and $(2, 1) \in R$ , but $(1, 1) \notin R$ . So, it is not transitive.

(ii) Transitive but neither reflexive nor symmetric. Example: Let the relation R on A be defined as

R = \{(1, 2)\}

.

Transitive: The condition for transitivity is that if $(a, b) \in R$ and $(b, c) \in R$ , then $(a, c) \in R$ . Here, we have $(1, 2) \in R$ , but there is no pair of the form $(2, c)$ . So, the premise of the implication is never met, which makes the relation vacuously transitive.
Not Reflexive: $(1, 1) \notin R$ . So, it is not reflexive.
Not Symmetric: $(1, 2) \in R$ , but $(2, 1) \notin R$ . So, it is not symmetric. (Another example: The relation "is less than" on the set of integers $\mathbf{Z}$ .)

(iii) Reflexive and symmetric but not transitive. Example: Let the relation R on A be defined as

R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\}

.

Reflexive: $(1, 1), (2, 2), (3, 3)$ are all in R. So, it is reflexive.
Symmetric: For $(1, 2) \in R$ , we have $(2, 1) \in R$ . For $(2, 3) \in R$ , we have $(3, 2) \in R$ . So, it is symmetric.
Not Transitive: We have $(1, 2) \in R$ and $(2, 3) \in R$ , but $(1, 3) \notin R$ . So, it is not transitive.

(iv) Reflexive and transitive but not symmetric. Example: Let R be the relation "is less than or equal to" (

\"$$\leq$$\"") on the set

A = {1, 2, 3}

.

R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)}$.

Reflexive: $(1, 1), (2, 2), (3, 3)$ are in R because $a \leq a$ . So, it is reflexive.
Transitive: If $a \leq b$ and $b \leq c$ , then $a \leq c$ . For example, $(1, 2) \in R$ and $(2, 3) \in R$ implies $(1, 3) \in R$ . This holds for all pairs. So, it is transitive.
Not Symmetric: $(1, 2) \in R$ (since $1 \leq 2$ ), but $(2, 1) \notin R$ (since $2 \not\leq 1$ ). So, it is not symmetric.

(v) Symmetric and transitive but not reflexive. Example: Let the relation R on A be defined as

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

.

Symmetric: For $(1, 2) \in R$ , we have $(2, 1) \in R$ . So, it is symmetric.
Transitive: We can check all cases:
- $(1, 2) \in R$ and $(2, 1) \in R \implies (1, 1) \in R$ . (True)
- $(1, 2) \in R$ and $(2, 2) \in R \implies (1, 2) \in R$ . (True)
- $(2, 1) \in R$ and $(1, 1) \in R \implies (2, 1) \in R$ . (True)
- $(2, 1) \in R$ and $(1, 2) \in R \implies (2, 2) \in R$ . (True) All other cases involve reflexive pairs and are trivial. So, it is transitive.
Not Reflexive: The element $3 \in A$ , but $(3, 3) \notin R$ . So, it is not reflexive. (Another example: The empty relation $\emptyset$ on a non-empty set is symmetric and transitive, but not reflexive.)

Q11EXERCISE 1.1

Show that the relation R in the set A of points in a plane given by $\mathrm{R}=\{(\mathrm{P}, \mathrm{Q})$ : distance of the point P from the origin is same as the distance of the point Q from the origin} $, is an equivalence relation. Further, show that the set of all points related to a point$ \mathrm{P} \neq(0,0)$ is the circle passing through P with origin as centre.

Solution

Given: The relation R in the set A of points in a plane is defined as:

R = \{ (P, Q) : \text{distance of the point P from the origin is same as the distance of the point Q from the origin} \}

To Show (Part 1): R is an equivalence relation.

Proof: Let O be the origin (0, 0). The distance of a point P from the origin is denoted by OP. So,

R = \{ (P, Q) : OP = OQ \}

.

Reflexivity: For any point P ∈ A, we have OP = OP. This is always true. Therefore, (P, P) ∈ R for all P ∈ A. Hence, R is reflexive.
Symmetry: Let (P, Q) ∈ R. This implies that OP = OQ. By the property of equality, if OP = OQ, then OQ = OP. This implies that (Q, P) ∈ R. Hence, R is symmetric.
Transitivity: Let (P, Q) ∈ R and (Q, S) ∈ R for points P, Q, S ∈ A. (P, Q) ∈ R implies OP = OQ. (Q, S) ∈ R implies OQ = OS. From these two equations, we get OP = OS. This implies that (P, S) ∈ R. Hence, R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation. Hence Proved.

To Show (Part 2): The set of all points related to a point P ≠ (0,0) is the circle passing through P with origin as centre.

Proof: Let P be a point with coordinates

(x_1, y_1)

, where P ≠ (0,0). The distance of P from the origin O(0,0) is

OP = \sqrt{x_1^2 + y_1^2}

. Let this distance be

k

. Since P ≠ (0,0),

k > 0

.

Let Q(x, y) be any point in the set of all points related to P. By the definition of the relation R, (P, Q) ∈ R, which means OQ = OP.

So, the distance of Q from the origin must be

k

.

OQ = \sqrt{x^2 + y^2} = k

Squaring both sides, we get:

x^2 + y^2 = k^2

This is the equation of a circle with its centre at the origin (0,0) and radius

k

.

Since the point P

(x_1, y_1)

satisfies this equation (as

x_1^2 + y_1^2 = (OP)^2 = k^2

), the circle passes through the point P.

Final Answer: The set of all points related to P is the circle with centre at the origin and radius equal to the distance of P from the origin, which is the circle passing through P with the origin as its centre. Hence Proved.

Q12EXERCISE 1.1

Show that the relation R defined in the set A of all triangles as $\mathrm{R}=\left\{\left(\mathrm{T}_{1}, \mathrm{T}_{2}\right): \mathrm{T}_{1}\right.$ is similar to $\left.\mathrm{T}_{2}\right\}$ , is equivalence relation. Consider three right angle triangles $\mathrm{T}_{1}$ with sides 3, 4, 5, $\mathrm{T}_{2}$ with sides 5, 12, 13 and $\mathrm{T}_{3}$ with sides 6, 8, 10. Which triangles among $\mathrm{T}_{1}, \mathrm{T}_{2}$ and $\mathrm{T}_{3}$ are related?

Solution

Given: The relation R defined in the set A of all triangles as:

R = \{ (T_1, T_2) : T_1 \text{ is similar to } T_2 \}

To Show: R is an equivalence relation.

Proof:

Reflexivity: Every triangle is similar to itself. So, for any triangle $T_1 \in A$ , $T_1$ is similar to $T_1$ . Therefore, $(T_1, T_1) \in R$ . Hence, R is reflexive.
Symmetry: Let $(T_1, T_2) \in R$ . This implies that $T_1$ is similar to $T_2$ . If $T_1$ is similar to $T_2$ , then $T_2$ is also similar to $T_1$ . This implies that $(T_2, T_1) \in R$ . Hence, R is symmetric.
Transitivity: Let $(T_1, T_2) \in R$ and $(T_2, T_3) \in R$ . $(T_1, T_2) \in R$ implies $T_1$ is similar to $T_2$ . $(T_2, T_3) \in R$ implies $T_2$ is similar to $T_3$ . By the property of similarity, if one triangle is similar to a second, and the second is similar to a third, then the first triangle is similar to the third. So, $T_1$ is similar to $T_3$ . This implies that $(T_1, T_3) \in R$ . Hence, R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation. Hence Proved.

Further Investigation: Consider three right-angled triangles:

$T_1$ with sides 3, 4, 5.
$T_2$ with sides 5, 12, 13.
$T_3$ with sides 6, 8, 10.

Two triangles are related if they are similar. Two triangles are similar if their corresponding sides are in proportion.

Comparing $T_1$ and $T_2$ : The ratios of their corresponding sides are $\frac{3}{5}$ , $\frac{4}{12} = \frac{1}{3}$ , and $\frac{5}{13}$ . Since $\frac{3}{5} \neq \frac{1}{3} \neq \frac{5}{13}$ , $T_1$ and $T_2$ are not similar. Thus, they are not related.
Comparing $T_1$ and $T_3$ : The ratios of their corresponding sides are $\frac{3}{6} = \frac{1}{2}$ , $\frac{4}{8} = \frac{1}{2}$ , and $\frac{5}{10} = \frac{1}{2}$ . Since the ratios of the corresponding sides are equal, $T_1$ and $T_3$ are similar. Thus, they are related.
Comparing $T_2$ and $T_3$ : The ratios of their corresponding sides are $\frac{5}{6}$ , $\frac{12}{8} = \frac{3}{2}$ , and $\frac{13}{10}$ . Since the ratios are not equal, $T_2$ and $T_3$ are not similar. Thus, they are not related.

Final Answer: Among the given triangles,

T_1

and

T_3

are related.

Q13EXERCISE 1.1

Show that the relation R defined in the set A of all polygons as R = {(P_{1}, P_{2}): P_{1} and P_{2} have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3,4 and 5 ?

Solution

Given: The relation R defined in the set A of all polygons as:

R = \{ (P_1, P_2) : P_1 \text{ and } P_2 \text{ have same number of sides} \}

To Show: R is an equivalence relation.

Proof:

Reflexivity: For any polygon $P_1 \in A$ , $P_1$ has the same number of sides as itself. Therefore, $(P_1, P_1) \in R$ . Hence, R is reflexive.
Symmetry: Let $(P_1, P_2) \in R$ . This implies that $P_1$ and $P_2$ have the same number of sides. This also means that $P_2$ and $P_1$ have the same number of sides. This implies that $(P_2, P_1) \in R$ . Hence, R is symmetric.
Transitivity: Let $(P_1, P_2) \in R$ and $(P_2, P_3) \in R$ . $(P_1, P_2) \in R$ implies $P_1$ and $P_2$ have the same number of sides. $(P_2, P_3) \in R$ implies $P_2$ and $P_3$ have the same number of sides. Therefore, $P_1$ and $P_3$ must have the same number of sides. This implies that $(P_1, P_3) \in R$ . Hence, R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation. Hence Proved.

Further Question: What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Solution: The given element is a right-angled triangle T. A triangle is a polygon with 3 sides. An element (polygon) P in A is related to T if and only if P and T have the same number of sides. Therefore, any polygon related to T must have 3 sides. The set of all polygons with 3 sides is the set of all triangles.

Final Answer: The set of all elements in A related to the right-angled triangle T with sides 3, 4, and 5 is the set of all triangles.

Q14EXERCISE 1.1

Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L_{1}, L_{2}): L_{1} is parallel to L_{2}}. Show that R is an equivalence relation. Find the set of all lines related to the line $y=2 x+4$ .

Solution

Given: L is the set of all lines in the XY plane and R is the relation in L defined as:

R = \{ (L_1, L_2) : L_1 \text{ is parallel to } L_2 \}

To Show: R is an equivalence relation.

Proof:

Reflexivity: Any line $L_1$ is considered parallel to itself (as it has the same slope as itself). Therefore, $(L_1, L_1) \in R$ for all $L_1 \in L$ . Hence, R is reflexive.
Symmetry: Let $(L_1, L_2) \in R$ . This implies that line $L_1$ is parallel to line $L_2$ . If $L_1$ is parallel to $L_2$ , then $L_2$ is also parallel to $L_1$ . This implies that $(L_2, L_1) \in R$ . Hence, R is symmetric.
Transitivity: Let $(L_1, L_2) \in R$ and $(L_2, L_3) \in R$ . $(L_1, L_2) \in R$ implies $L_1$ is parallel to $L_2$ . $(L_2, L_3) \in R$ implies $L_2$ is parallel to $L_3$ . By a fundamental property of parallel lines, if a line is parallel to a second line, and the second is parallel to a third, then the first line is parallel to the third. So, $L_1$ is parallel to $L_3$ . This implies that $(L_1, L_3) \in R$ . Hence, R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation. Hence Proved.

Further Question: Find the set of all lines related to the line

y = 2x + 4

.

Solution: A line is related to

y = 2x + 4

if it is parallel to it. Two lines are parallel if and only if they have the same slope. The given line is in the slope-intercept form

y = mx + c

, where the slope

m

is 2. Therefore, any line parallel to

y = 2x + 4

must also have a slope of 2. The general equation of a line with slope 2 is given by

y = 2x + k

, where

k

is any real constant representing the y-intercept.

Final Answer: The set of all lines related to the line

y = 2x + 4

is the set of all lines of the form

y = 2x + k

, where

k

is a real number.

Q15EXERCISE 1.1

Let R be the relation in the set {1,2,3,4} given by R = {(1,2),(2,2),(1,1),(4,4), (1,3),(3,3),(3,2)}. Choose the correct answer.

(A)

R is reflexive and symmetric but not transitive.

(B)

R is reflexive and transitive but not symmetric.

(C)

R is symmetric and transitive but not reflexive.

(D)

R is an equivalence relation.

Solution

Given: The set A = {1, 2, 3, 4} and the relation R on A is given by:

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

To Find: Choose the correct description of the relation R.

Solution: We check the properties of an equivalence relation (reflexivity, symmetry, and transitivity) for the given relation R.

Reflexivity: A relation is reflexive if $(a, a) \in R$ for every $a \in A$ . Here, A = {1, 2, 3, 4}.
- Is (1, 1) ∈ R? Yes.
- Is (2, 2) ∈ R? Yes.
- Is (3, 3) ∈ R? Yes.
- Is (4, 4) ∈ R? Yes. Since $(a, a) \in R$ for all $a \in \{1, 2, 3, 4\}$ , the relation R is reflexive. This eliminates option (C).
Symmetry: A relation is symmetric if for every $(a, b) \in R$ , we also have $(b, a) \in R$ .
- We have (1, 2) ∈ R. But (2, 1) ∉ R.
- We have (1, 3) ∈ R. But (3, 1) ∉ R.
- We have (3, 2) ∈ R. But (2, 3) ∉ R. Since we found counterexamples, the relation R is not symmetric. This eliminates options (A) and (D).
Transitivity: A relation is transitive if for every $(a, b) \in R$ and $(b, c) \in R$ , we also have $(a, c) \in R$ . Let's check all such pairs:
- We have (1, 3) ∈ R and (3, 2) ∈ R. For transitivity, we need (1, 2) ∈ R. Yes, (1, 2) is in R.
- Let's check other combinations to be sure:
  - (1, 2) ∈ R and (2, 2) ∈ R implies (1, 2) ∈ R. (True)
  - (1, 3) ∈ R and (3, 3) ∈ R implies (1, 3) ∈ R. (True) There are no other pairs of the form $(a, b)$ and $(b, c)$ to check. Therefore, the relation R is transitive.

Conclusion: The relation R is reflexive and transitive, but not symmetric. This corresponds to option (B).

Final Answer: (B) R is reflexive and transitive but not symmetric.

Q16EXERCISE 1.1

Let R be the relation in the set N given by R = {(a, b): $a=b-2, b>6$ }. Choose the correct answer.

(A)

(2,4) ∈ Р

(B)

(3,8) ∈ Р

(C)

(6,8) ∈ Р

(D)

(8,7) ∈ Р

Solution

Given: The relation R in the set of natural numbers

\mathbf{N}

is given by R = {(a, b):

a = b - 2, b > 6

}.

To Find: Which of the given options represents an element of R.

Solution: We need to find an ordered pair (a, b) that satisfies both conditions:

b > 6

and

a = b - 2

.

Let's check each option:

(A) (2, 4) ∈ R Here,

a = 2

and

b = 4

. The condition

b > 6

is not satisfied, as

4

is not greater than

6

. So,

(2, 4) \notin

R.

(B) (3, 8) ∈ R Here,

a = 3

and

b = 8

. The condition

b > 6

is satisfied, as

8 > 6

. Now, let's check the condition

a = b - 2

.

3 = 8 - 2

3 = 6

, which is false. So,

(3, 8) \notin

R.

(C) (6, 8) ∈ R Here,

a = 6

and

b = 8

. The condition

b > 6

is satisfied, as

8 > 6

. Now, let's check the condition

a = b - 2

.

6 = 8 - 2

6 = 6

, which is true. Both conditions are satisfied. So,

(6, 8) \in

R.

(D) (8, 7) ∈ R Here,

a = 8

and

b = 7

. The condition

b > 6

is satisfied, as

7 > 6

. Now, let's check the condition

a = b - 2

.

8 = 7 - 2

8 = 5

, which is false. So,

(8, 7) \notin

R.

Final Answer: The correct answer is (C) because the pair (6, 8) satisfies both conditions of the relation R.

Q1EXERCISE 1.2

Show that the function $f: \mathbf{R}_{} \rightarrow \mathbf{R}_{}$ defined by $f(x)=\frac{1}{x}$ is one-one and onto, where $\mathbf{R}_{}$ is the set of all non-zero real numbers. Is the result true, if the domain $\mathbf{R}_{}$ is replaced by $\mathbf{N}$ with co-domain being same as $\mathbf{R}_{*}$ ?

Solution

Given: A function

f: \mathbf{R}_{*} \rightarrow \mathbf{R}_{*}

defined by

f(x)=\frac{1}{x}

, where

\mathbf{R}_{*}

is the set of all non-zero real numbers.

Part 1: To show that f is one-one and onto.

One-one (Injectivity): A function

f

is one-one if

f(x_1) = f(x_2)

implies

x_1 = x_2

for all

x_1, x_2

in the domain. Let

x_1, x_2 \in \mathbf{R}_{*}

. Assume

f(x_1) = f(x_2)

.

\frac{1}{x_1} = \frac{1}{x_2}

This implies

x_1 = x_2

. Therefore,

f

is one-one.

Onto (Surjectivity): A function

f

is onto if for every element

y

in the co-domain, there exists an element

x

in the domain such that

f(x) = y

. Let

y \in \mathbf{R}_{*}

(the co-domain). We need to find an

x \in \mathbf{R}_{*}

(the domain) such that

f(x) = y

.

f(x) = y

\frac{1}{x} = y

x = \frac{1}{y}

Since

y \in \mathbf{R}_{*}

,

y

is a non-zero real number. Thus,

x = \frac{1}{y}

is also a well-defined non-zero real number. So,

x \in \mathbf{R}_{*}

. This means that for every

y

in the co-domain

\mathbf{R}_{*}

, there exists a pre-image

x = \frac{1}{y}

in the domain

\mathbf{R}_{*}

. Therefore,

f

is onto.

Since

f

is both one-one and onto, it is a bijective function.

Part 2: Is the result true if the domain $\mathbf{R}_{*}$ is replaced by $\mathbf{N}$ with co-domain being same as $\mathbf{R}_{*}$ ?

The new function is

g: \mathbf{N} \rightarrow \mathbf{R}_{*}

defined by

g(x) = \frac{1}{x}

.

One-one (Injectivity): Let

x_1, x_2 \in \mathbf{N}

. Assume

g(x_1) = g(x_2)

.

\frac{1}{x_1} = \frac{1}{x_2}

This implies

x_1 = x_2

. Therefore, the function

g

is one-one.

Onto (Surjectivity): Let

y \in \mathbf{R}_{*}

. We need to find an

x \in \mathbf{N}

such that

g(x) = y

.

g(x) = y

\frac{1}{x} = y

x = \frac{1}{y}

For

g

to be onto, for every

y \in \mathbf{R}_{*}

, the corresponding

x = \frac{1}{y}

must be a natural number. Let's take an element from the co-domain, say

y = 2 \in \mathbf{R}_{*}

. Then

x = \frac{1}{2}

. But

x = \frac{1}{2}

is not a natural number (

x \notin \mathbf{N}

). So, the element

2

in the co-domain

\mathbf{R}_{*}

has no pre-image in the domain

\mathbf{N}

. Therefore, the function

g

is not onto.

Final Answer: The function

f: \mathbf{R}_{*} \rightarrow \mathbf{R}_{*}

is one-one and onto. However, if the domain is replaced by

\mathbf{N}

, the new function

g: \mathbf{N} \rightarrow \mathbf{R}_{*}

is one-one but not onto.

Q2EXERCISE 1.2

Check the injectivity and surjectivity of the following functions:

(i)

$f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{2}$

(ii)

$f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{2}$

(iii)

$f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=x^{2}$

(iv)

$f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{3}$

(v)

$f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{3}$

Solution

To check the injectivity and surjectivity of the following functions:

(i) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{2}$ Injectivity (One-one): Let

x_1, x_2 \in \mathbf{N}

such that

f(x_1) = f(x_2)

.

x_1^2 = x_2^2

Since

x_1, x_2

are natural numbers, they are positive. Taking the square root, we get

x_1 = x_2

. Hence,

f

is injective. Surjectivity (Onto): Let

y \in \mathbf{N}

(co-domain). We need to find an

x \in \mathbf{N}

(domain) such that

f(x) = y

, i.e.,

x^2 = y

, which means

x = \sqrt{y}

. For

y

to have a pre-image in

\mathbf{N}

,

\sqrt{y}

must be a natural number. This is not true for all

y \in \mathbf{N}

. For example, if we take

y=2 \in \mathbf{N}

, then

x = \sqrt{2}

, which is not a natural number. Thus,

2

has no pre-image in the domain

\mathbf{N}

. Hence,

f

is not surjective. Conclusion: The function is injective but not surjective.

(ii) $f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{2}$ Injectivity (One-one): Let

x_1, x_2 \in \mathbf{Z}

such that

f(x_1) = f(x_2)

. Consider

x_1 = -1

and

x_2 = 1

. Both belong to

\mathbf{Z}

.

f(-1) = (-1)^2 = 1

f(1) = (1)^2 = 1

So,

f(-1) = f(1)

but

-1 \neq 1

. Hence,

f

is not injective. Surjectivity (Onto): Let

y \in \mathbf{Z}

. We need to find an

x \in \mathbf{Z}

such that

f(x) = y

, i.e.,

x^2 = y

. The square of any integer is always non-negative. So, no negative integer in the co-domain has a pre-image. For example, if we take

y=-2 \in \mathbf{Z}

, there is no integer

x

such that

x^2 = -2

. Hence,

f

is not surjective. Conclusion: The function is neither injective nor surjective.

(iii) $f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=x^{2}$ Injectivity (One-one): Similar to the case of integers, consider

x_1 = -1

and

x_2 = 1

. Both belong to

\mathbf{R}

.

f(-1) = (-1)^2 = 1

f(1) = (1)^2 = 1

So,

f(-1) = f(1)

but

-1 \neq 1

. Hence,

f

is not injective. Surjectivity (Onto): The square of any real number is non-negative. The range of the function is

[0, \infty)

. The co-domain is

\mathbf{R}

. Since the range is a proper subset of the co-domain, the function is not onto. For example, if we take

y=-1 \in \mathbf{R}

, there is no real number

x

such that

x^2 = -1

. Hence,

f

is not surjective. Conclusion: The function is neither injective nor surjective.

(iv) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{3}$ Injectivity (One-one): Let

x_1, x_2 \in \mathbf{N}

such that

f(x_1) = f(x_2)

.

x_1^3 = x_2^3

Taking the cube root on both sides, we get

x_1 = x_2

. Hence,

f

is injective. Surjectivity (Onto): Let

y \in \mathbf{N}

. We need an

x \in \mathbf{N}

such that

f(x) = y

, i.e.,

x^3 = y

, which means

x = \sqrt[3]{y}

. For

y

to have a pre-image in

\mathbf{N}

,

\sqrt[3]{y}

must be a natural number. This is not true for all

y \in \mathbf{N}

. For example, if we take

y=2 \in \mathbf{N}

, then

x = \sqrt[3]{2}

, which is not a natural number. Thus,

2

has no pre-image in the domain

\mathbf{N}

. Hence,

f

is not surjective. Conclusion: The function is injective but not surjective.

(v) $f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{3}$ Injectivity (One-one): Let

x_1, x_2 \in \mathbf{Z}

such that

f(x_1) = f(x_2)

.

x_1^3 = x_2^3

Taking the cube root on both sides, we get

x_1 = x_2

. (Unlike squares, the cube of a negative number is negative, so

x_1^3 = x_2^3

implies

x_1=x_2

for integers). Hence,

f

is injective. Surjectivity (Onto): Let

y \in \mathbf{Z}

. We need an

x \in \mathbf{Z}

such that

f(x) = y

, i.e.,

x^3 = y

, which means

x = \sqrt[3]{y}

. For

y

to have a pre-image in

\mathbf{Z}

,

\sqrt[3]{y}

must be an integer. This is not true for all

y \in \mathbf{Z}

. For example, if we take

y=2 \in \mathbf{Z}

, then

x = \sqrt[3]{2}

, which is not an integer. Thus,

2

has no pre-image in the domain

\mathbf{Z}

. Hence,

f

is not surjective. Conclusion: The function is injective but not surjective.

Q3EXERCISE 1.2

Prove that the Greatest Integer Function $f: \mathbf{R} \rightarrow \mathbf{R}$ , given by $f(x)=[x]$ , is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$ .

Solution

Given: The Greatest Integer Function

f: \mathbf{R} \rightarrow \mathbf{R}

, given by

f(x)=[x]

, where

[x]

denotes the greatest integer less than or equal to

x

.

To Prove: The function

f

is neither one-one nor onto.

Proof:

Checking for One-one (Injectivity): A function is one-one if distinct elements in the domain have distinct images in the co-domain. To show

f

is not one-one, we need to find at least two distinct elements in the domain

\mathbf{R}

that have the same image.

Let's consider two distinct real numbers,

x_1 = 1.2

and

x_2 = 1.8

. Both are in the domain

\mathbf{R}

.

f(x_1) = f(1.2) = [1.2] = 1

.

f(x_2) = f(1.8) = [1.8] = 1

. Here, we have

f(x_1) = f(x_2)

but

x_1 \neq x_2

. Since different elements (1.2 and 1.8) have the same image (1), the function

f

is not one-one.

Checking for Onto (Surjectivity): A function is onto if every element in the co-domain has at least one pre-image in the domain. The co-domain of

f

is

\mathbf{R}

. The range of

f(x) = [x]

is the set of all integers,

\mathbf{Z}

, because the output of the greatest integer function is always an integer.

To show

f

is not onto, we need to find an element in the co-domain

\mathbf{R}

that has no pre-image in the domain

\mathbf{R}

.

Let's consider a non-integer value from the co-domain, for example,

y = 0.5 \in \mathbf{R}

. For

f

to be onto, there must exist some

x \in \mathbf{R}

such that

f(x) = y

.

[x] = 0.5

By the definition of the greatest integer function,

[x]

must be an integer. Since

0.5

is not an integer, there is no real number

x

for which

[x] = 0.5

. Thus, the element

0.5

in the co-domain has no pre-image in the domain. Since the range

\mathbf{Z}

is a proper subset of the co-domain

\mathbf{R}

, the function

f

is not onto.

Conclusion: The Greatest Integer Function

f(x) = [x]

is neither one-one nor onto.

Hence Proved.

Q4EXERCISE 1.2

Show that the Modulus Function $f: \mathbf{R} \rightarrow \mathbf{R}$ , given by $f(x)=|x|$ , is neither oneone nor onto, where $|x|$ is $x$ , if $x$ is positive or 0 and $|x|$ is $-x$ , if $x$ is negative.

Solution

Given: The Modulus Function

f: \mathbf{R} \rightarrow \mathbf{R}

, given by

f(x)=|x|

. The definition of

|x|

is:

|x| = x

, if

x \ge 0

|x| = -x

, if

x < 0

To Show: The function

f

is neither one-one nor onto.

Solution:

Checking for One-one (Injectivity): A function is one-one if distinct elements in the domain have distinct images. To show

f

is not one-one, we need to find at least two distinct elements in the domain

\mathbf{R}

that map to the same image.

Let's consider two distinct real numbers,

x_1 = 1

and

x_2 = -1

. Both are in the domain

\mathbf{R}

.

f(x_1) = f(1) = |1| = 1

.

f(x_2) = f(-1) = |-1| = 1

. Here, we have

f(x_1) = f(x_2)

but

x_1 \neq x_2

. Since different elements (1 and -1) have the same image (1), the function

f

is not one-one.

Checking for Onto (Surjectivity): A function is onto if its range is equal to its co-domain. The co-domain of

f

is

\mathbf{R}

. The range of

f(x) = |x|

is the set of all non-negative real numbers, i.e.,

[0, \infty)

, because the modulus of any real number is always non-negative.

To show

f

is not onto, we need to find an element in the co-domain

\mathbf{R}

that has no pre-image in the domain

\mathbf{R}

.

Let's consider a negative number from the co-domain, for example,

y = -1 \in \mathbf{R}

. For

f

to be onto, there must exist some

x \in \mathbf{R}

such that

f(x) = y

.

|x| = -1

By the definition of the modulus function,

|x|

is always greater than or equal to 0 for all

x \in \mathbf{R}

. Therefore, there is no real number

x

for which

|x| = -1

. Thus, the element

-1

in the co-domain has no pre-image in the domain. Since the range

[0, \infty)

is a proper subset of the co-domain

\mathbf{R}

, the function

f

is not onto.

Conclusion: The Modulus Function

f(x) = |x|

is neither one-one nor onto.

Q5EXERCISE 1.2

Show that the Signum Function $f: \mathbf{R} \rightarrow \mathbf{R}$ , given by $f(x)=\begin{array}{r} 1, \text { if } x>0 \ 0, \text { if } x=0 \ 1, \text { if } x<0 \end{array}$ is neither one-one nor onto.

Solution

Given: The Signum Function

f: \mathbf{R} \rightarrow \mathbf{R}

, given by

f(x)=\begin{cases} 1, & \text{if } x>0 \ 0, & \text{if } x=0 \ -1, & \text{if } x<0 \end{cases}

To Show: The function

f

is neither one-one nor onto.

Solution:

1. Checking for One-one (Injectivity): A function is one-one if for every

x_1, x_2

in the domain,

f(x_1) = f(x_2)

implies

x_1 = x_2

. In other words, distinct elements in the domain must have distinct images.

Let us consider two distinct elements from the domain

\mathbf{R}

, for example,

x_1 = 1

and

x_2 = 2

. Since

1 > 0

,

f(1) = 1

. Since

2 > 0

,

f(2) = 1

. Here, we have

f(1) = f(2)

, but

1 \neq 2

. Since different elements in the domain have the same image, the function

f

is not one-one.

2. Checking for Onto (Surjectivity): A function is onto if its range is equal to its codomain. The codomain of the function is given as

\mathbf{R}

.

From the definition of the function

f(x)

, the value of

f(x)

is always either

1

,

0

, or

-1

, for any real number

x

. Therefore, the range of the function

f

is the set

\{-1, 0, 1\}

.

The codomain of the function is

\mathbf{R}

. Since Range

\{-1, 0, 1\} \neq

Codomain

\mathbf{R}

, the function is not onto. For instance, the real number

2

is in the codomain

\mathbf{R}

, but there is no

x

in the domain

\mathbf{R}

for which

f(x) = 2

.

Conclusion: Since the function

f

is neither one-one nor onto, it is not a bijective function.

Q6EXERCISE 1.2

Let A = {1,2,3}, B = {4,5,6,7} and let $f=\{(1,4),(2,5),(3,6)\}$ be a function from A to B . Show that $f$ is one-one.

Solution

Given: Sets A = {1, 2, 3} and B = {4, 5, 6, 7}. A function

f: A \rightarrow B

is defined as

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

.

To Show: The function

f

is one-one.

Solution: A function is one-one (injective) if distinct elements in the domain have distinct images in the codomain.

The domain of the function

f

is the set A = {1, 2, 3}. The codomain of the function

f

is the set B = {4, 5, 6, 7}.

Let us find the images of the elements of the domain A under the function

f

. From the definition of

f

, we have:

f(1) = 4

f(2) = 5

f(3) = 6

The elements in the domain are 1, 2, and 3. These are all distinct. Their corresponding images are 4, 5, and 6. These are also all distinct.

Since for any two distinct elements

x_1, x_2 \in A

, their images

f(x_1)

and

f(x_2)

are also distinct, the function

f

is one-one.

Final Answer: The function

f

is one-one.

Q7EXERCISE 1.2

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i)

$f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=3-4 x$

(ii)

$f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=1+x^{2}$

Solution

(i) $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=3-4x$

To state: Whether the function is one-one, onto or bijective.

Solution:

1. Checking for One-one (Injectivity): Let

x_1, x_2 \in \mathbf{R}

such that

f(x_1) = f(x_2)

.

3 - 4x_1 = 3 - 4x_2

-4x_1 = -4x_2

x_1 = x_2

Since

f(x_1) = f(x_2)

implies

x_1 = x_2

, the function

f

is one-one.

2. Checking for Onto (Surjectivity): Let

y

be an arbitrary element in the codomain

\mathbf{R}

. We need to find an element

x

in the domain

\mathbf{R}

such that

f(x) = y

.

f(x) = y

3 - 4x = y

4x = 3 - y

x = \frac{3 - y}{4}

Since

y

is a real number,

3-y

is also a real number, and so is

x = \frac{3 - y}{4}

. Thus, for any

y \in \mathbf{R}

(codomain), there exists an

x = \frac{3 - y}{4} \in \mathbf{R}

(domain) such that

f(x) = f(\frac{3 - y}{4}) = 3 - 4(\frac{3-y}{4}) = 3 - (3-y) = y

. Therefore, the function

f

is onto.

3. Checking for Bijectivity: Since the function

f

is both one-one and onto, it is a bijective function.

Final Answer for (i): The function is one-one, onto, and hence bijective.

(ii) $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=1+x^{2}$

To state: Whether the function is one-one, onto or bijective.

Solution:

1. Checking for One-one (Injectivity): Let us take two distinct elements from the domain

\mathbf{R}

, for example,

x_1 = 1

and

x_2 = -1

.

f(1) = 1 + (1)^2 = 1 + 1 = 2

f(-1) = 1 + (-1)^2 = 1 + 1 = 2

Here, we have

f(1) = f(-1)

, but

1 \neq -1

. Since different elements in the domain have the same image, the function

f

is not one-one.

2. Checking for Onto (Surjectivity): Let

y

be an arbitrary element in the codomain

\mathbf{R}

. We need to find an element

x

in the domain

\mathbf{R}

such that

f(x) = y

.

f(x) = y

1 + x^2 = y

x^2 = y - 1

x = \pm \sqrt{y - 1}

For

x

to be a real number, the term under the square root must be non-negative, i.e.,

y - 1 \geq 0

, which implies

y \geq 1

. This means that for any real number

y < 1

in the codomain, there is no corresponding

x

in the domain. For example, if we take

y = -2

from the codomain

\mathbf{R}

, we get

x^2 = -2 - 1 = -3

, which has no real solution for

x

. The range of the function is

[1, \infty)

, which is a proper subset of the codomain

\mathbf{R}

. Therefore, the function

f

is not onto.

3. Checking for Bijectivity: Since the function

f

is neither one-one nor onto, it is not a bijective function.

Final Answer for (ii): The function is neither one-one nor onto.

Q8EXERCISE 1.2

Let A and B be sets. Show that $f: \mathrm{A} \times \mathrm{B} \rightarrow \mathrm{B} \times \mathrm{A}$ such that $f(a, b)=(b, a)$ is bijective function.

Solution

Given: Sets A and B. A function

f: A \times B \rightarrow B \times A

such that

f(a, b)=(b, a)

.

To Show: The function

f

is a bijective function.

Proof: A function is bijective if it is both one-one (injective) and onto (surjective).

1. Checking for One-one (Injectivity): Let

(a_1, b_1)

and

(a_2, b_2)

be two arbitrary elements in the domain

A \times B

. Assume that

f(a_1, b_1) = f(a_2, b_2)

. By the definition of the function

f

, this means:

(b_1, a_1) = (b_2, a_2)

By the property of equality of ordered pairs, two ordered pairs are equal if and only if their corresponding components are equal. Therefore, we have:

b_1 = b_2 \quad \text{and} \quad a_1 = a_2

This implies that the original elements are equal:

(a_1, b_1) = (a_2, b_2)

Since

f(a_1, b_1) = f(a_2, b_2)

implies

(a_1, b_1) = (a_2, b_2)

, the function

f

is one-one.

2. Checking for Onto (Surjectivity): Let

(b, a)

be an arbitrary element in the codomain

B \times A

. This means that

b \in B

and

a \in A

. We need to find a pre-image

(x, y)

in the domain

A \times B

such that

f(x, y) = (b, a)

. By the definition of

f

, we have

f(x, y) = (y, x)

. So, we need to find

(x, y)

such that:

(y, x) = (b, a)

By the equality of ordered pairs, this gives us:

y = b \quad \text{and} \quad x = a

So, the pre-image is

(a, b)

. Since

a \in A

and

b \in B

, the element

(a, b)

belongs to the domain

A \times B

. Thus, for any element

(b, a)

in the codomain

B \times A

, there exists a pre-image

(a, b)

in the domain

A \times B

such that

f(a, b) = (b, a)

. Therefore, the function

f

is onto.

Conclusion: Since the function

f

is both one-one and onto, it is a bijective function.

Hence Proved.

Q9EXERCISE 1.2

Let $f: \mathbf{N} \rightarrow \mathbf{N}$ be defined by $f(n)=\begin{array}{l} \frac{n+1}{2}, \text { if } n \text { is odd } \ \frac{n}{2}, \text { if } n \text { is even } \end{array}$ for all $n \in \mathbf{N}$ . State whether the function $f$ is bijective. Justify your answer.

Solution

Given: A function

f: \mathbf{N} \rightarrow \mathbf{N}

defined by

f(n)=\begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}

for all

n \in \mathbf{N}

.

To state: Whether the function

f

is bijective and justify the answer.

Solution: A function is bijective if it is both one-one (injective) and onto (surjective).

1. Checking for One-one (Injectivity): A function is one-one if distinct elements in the domain have distinct images. Let us test some values from the domain

\mathbf{N}

. Consider the element

n=1

(which is odd):

f(1) = \frac{1+1}{2} = \frac{2}{2} = 1

Consider the element

n=2

(which is even):

f(2) = \frac{2}{2} = 1

Here, we have two distinct elements in the domain,

1

and

2

, but their images are the same, i.e.,

f(1) = f(2) = 1

. Since

1 \neq 2

, the function is not one-one. It is a many-one function.

2. Checking for Onto (Surjectivity): A function is onto if its range is equal to its codomain. The codomain is

\mathbf{N}

. Let

y

be an arbitrary natural number in the codomain

\mathbf{N}

. We need to find if there is a pre-image

n \in \mathbf{N}

in the domain such that

f(n)=y

.

Case 1: The pre-image $n$ is odd. If

n

is odd, then

f(n) = \frac{n+1}{2} = y

. This gives

n+1 = 2y

, so

n = 2y - 1

. If

y

is a natural number (

y \geq 1

), then

2y \geq 2

, and

n=2y-1 \geq 1

. Also,

2y-1

is always an odd number. So, for any

y \in \mathbf{N}

, we can find an odd pre-image

n=2y-1 \in \mathbf{N}

. For example, if

y=3

, the pre-image is

n = 2(3)-1=5

. And

f(5) = (5+1)/2 = 3

.

Case 2: The pre-image $n$ is even. If

n

is even, then

f(n) = \frac{n}{2} = y

. This gives

n = 2y

. If

y

is a natural number (

y \geq 1

), then

n=2y \geq 2

. Also,

2y

is always an even number. So, for any

y \in \mathbf{N}

, we can find an even pre-image

n=2y \in \mathbf{N}

. For example, if

y=3

, the pre-image is

n=2(3)=6

. And

f(6) = 6/2 = 3

.

Since for every natural number

y

in the codomain, we can find at least one pre-image

n

in the domain, the function is onto.

Conclusion: The function

f

is onto but not one-one. Since a function must be both one-one and onto to be bijective, the function

f

is not bijective.

Final Answer: The function

f

is not bijective because it is not one-one.

Q10EXERCISE 1.2

Let $\mathrm{A}=\mathbf{R}-\{3\}$ and $\mathrm{B}=\mathbf{R}-\{1\}$ . Consider the function $f: \mathrm{A} \rightarrow \mathrm{B}$ defined by $f(x)=\left(\frac{x-2}{x-3}\right)$ . Is $f$ one-one and onto? Justify your answer.

Solution

Given: A function

f: A \rightarrow B

defined by

f(x)=\left(\frac{x-2}{x-3}\right)

, where

A=\mathbf{R}-\{3\}

and

B=\mathbf{R}-\{1\}

.

To check: If

f

is one-one and onto.

Solution:

1. Check for one-one (injective): A function

f

is one-one if

f(x_1) = f(x_2)

implies

x_1 = x_2

for all

x_1, x_2

in the domain.

Let

x_1, x_2 \in A = \mathbf{R}-\{3\}

such that

f(x_1) = f(x_2)

.

f(x_1) = f(x_2)

\implies \frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}

On cross-multiplication, we get:

(x_1-2)(x_2-3) = (x_2-2)(x_1-3)

x_1x_2 - 3x_1 - 2x_2 + 6 = x_2x_1 - 3x_2 - 2x_1 + 6

-3x_1 - 2x_2 = -3x_2 - 2x_1

-3x_1 + 2x_1 = -3x_2 + 2x_2

-x_1 = -x_2

x_1 = x_2

Since

f(x_1) = f(x_2)

implies

x_1 = x_2

, the function

f

is one-one.

2. Check for onto (surjective): A function

f: A \rightarrow B

is onto if for every element

y

in the codomain

B

, there exists at least one element

x

in the domain

A

such that

f(x) = y

.

Let

y \in B = \mathbf{R}-\{1\}

. We need to find an

x \in A

such that

f(x) = y

.

y = f(x) = \frac{x-2}{x-3}

y(x-3) = x-2

xy - 3y = x-2

xy - x = 3y - 2

x(y-1) = 3y-2

x = \frac{3y-2}{y-1}

Since

y \in B = \mathbf{R}-\{1\}

, we have

y \neq 1

. Therefore,

y-1 \neq 0

and

x

is a well-defined real number. Now, we must check if this

x

belongs to the domain

A = \mathbf{R}-\{3\}

. This means

x

cannot be equal to 3. Let us check if

x

can be 3.

3 = \frac{3y-2}{y-1}

3(y-1) = 3y-2

3y - 3 = 3y-2

-3 = -2

This is a contradiction, which means our assumption that

x=3

is false. So,

x \neq 3

. Thus, for any

y \in B

, there exists an

x = \frac{3y-2}{y-1} \in A

such that

f(x)=y

. Therefore, the function

f

is onto.

Final Answer: The function

f

is both one-one and onto.

Q11EXERCISE 1.2

Let $f: \boldsymbol{\mathrm { R }} \rightarrow \boldsymbol{\mathrm { R }}$ be defined as $f(x)=x^{4}$ . Choose the correct answer.

(A)

f

is one-one onto

(B)

f

is many-one onto

(C)

f

is one-one but not onto

(D)

f

is neither one-one nor onto.

Solution

Given: A function

f: \mathbf{R} \rightarrow \mathbf{R}

defined as

f(x)=x^{4}

.

To determine: Whether

f

is one-one, onto, or neither.

Solution:

1. Check for one-one (injective): A function is one-one if distinct elements in the domain map to distinct elements in the codomain. Let's take two distinct elements from the domain

\mathbf{R}

, for example,

x_1 = -1

and

x_2 = 1

.

f(x_1) = f(-1) = (-1)^4 = 1

f(x_2) = f(1) = (1)^4 = 1

Here,

x_1 \neq x_2

, but

f(x_1) = f(x_2)

. Since different elements in the domain have the same image, the function

f

is not one-one. It is a many-one function.

2. Check for onto (surjective): A function is onto if its range is equal to its codomain. The codomain of

f

is

\mathbf{R}

. The range of

f

is the set of all possible values of

f(x) = x^4

. For any real number

x

,

x^4

is always non-negative. So, the range of

f

is

[0, \infty)

. Since the range

[0, \infty)

is a proper subset of the codomain

\mathbf{R}

, the function is not onto. For example, there is no

x \in \mathbf{R}

such that

f(x) = x^4 = -2

, because

-2

is not in the range.

Conclusion: The function

f

is neither one-one nor onto.

Final Answer: The correct option is (D)

f

is neither one-one nor onto.

Q12EXERCISE 1.2

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $f(x)=3 x$ . Choose the correct answer.

(A)

f

is one-one onto

(B)

f

is many-one onto

(C)

f

is one-one but not onto

(D)

f

is neither one-one nor onto.

Solution

Given: A function

f: \mathbf{R} \rightarrow \mathbf{R}

defined as

f(x)=3x

.

To determine: Whether

f

is one-one, onto, or neither.

Solution:

1. Check for one-one (injective): A function is one-one if

f(x_1) = f(x_2)

implies

x_1 = x_2

. Let

x_1, x_2 \in \mathbf{R}

such that

f(x_1) = f(x_2)

.

f(x_1) = f(x_2)

3x_1 = 3x_2

Dividing both sides by 3, we get:

x_1 = x_2

Since

f(x_1) = f(x_2)

implies

x_1 = x_2

, the function

f

is one-one.

2. Check for onto (surjective): A function is onto if for any element

y

in the codomain, there is a pre-image

x

in the domain such that

f(x)=y

. The codomain of

f

is

\mathbf{R}

. Let

y

be an arbitrary element in the codomain

\mathbf{R}

. We need to find an

x

in the domain

\mathbf{R}

such that

f(x) = y

.

3x = y

x = \frac{y}{3}

Since

y

is a real number,

x = \frac{y}{3}

is also a real number. Thus, for any

y \in \mathbf{R}

(codomain), there exists a pre-image

x = \frac{y}{3} \in \mathbf{R}

(domain). Therefore, the function

f

is onto.

Conclusion: The function

f

is both one-one and onto.

Final Answer: The correct option is (A)

f

is one-one onto.

Q1Miscellaneous Exercise on Chapter 1

Show that the function $f: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1<x<1\}$ defined by $f(x)=\frac{x}{1+|x|}$ , $x \in \mathbf{R}$ is one one and onto function.

Solution

Given: The function

f: \mathbf{R} \rightarrow \{x \in \mathbf{R}:-1 0

. Therefore,

x = \frac{y}{1-y} \geq 0

. This means for any

y \in [0, 1)

, there exists a pre-image

x \in [0, \infty)

.

Case 2:

-1 0

. Therefore,

x = \frac{y}{1+y} < 0

. This means for any

y \in (-1, 0)

, there exists a pre-image

x \in (-\infty, 0)

.

From both cases, we see that for every

y

in the codomain

(-1, 1)

, there exists a corresponding

x

in the domain

\mathbf{R}

. Hence,

f

is onto.

Since

f

is both one-one and onto, it is a bijective function.

Hence Proved.

Q2Miscellaneous Exercise on Chapter 1

Show that the function $f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=x^{3}$ is injective.

Solution

Given: The function

f: \mathbf{R} \rightarrow \mathbf{R}

given by

f(x)=x^{3}

.

To Prove: The function

f

is injective. An injective function is another term for a one-one function.

Proof: To prove that

f

is injective (one-one), we need to show that for any

x_1, x_2 \in \mathbf{R}

, if

f(x_1) = f(x_2)

, then

x_1 = x_2

.

Let

x_1, x_2

be two arbitrary elements in the domain

\mathbf{R}

. Assume

f(x_1) = f(x_2)

. By the definition of the function

f

, this means:

x_1^3 = x_2^3

To solve for the relationship between

x_1

and

x_2

, we can move all terms to one side:

x_1^3 - x_2^3 = 0

Using the algebraic identity for the difference of cubes,

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

, we get:

(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) = 0

This equation implies that either

(x_1 - x_2) = 0

or

(x_1^2 + x_1x_2 + x_2^2) = 0

.

Case 1:

x_1 - x_2 = 0

This directly implies

x_1 = x_2

.

Case 2:

x_1^2 + x_1x_2 + x_2^2 = 0

We can rewrite the expression by completing the square:

x_1^2 + x_1x_2 + \frac{x_2^2}{4} - \frac{x_2^2}{4} + x_2^2 = 0

\left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3x_2^2}{4} = 0

Since

x_1

and

x_2

are real numbers,

\left(x_1 + \frac{x_2}{2}\right)^2 \geq 0

and

\frac{3x_2^2}{4} \geq 0

. The sum of two non-negative numbers can be zero only if both numbers are zero. So, we must have:

\frac{3x_2^2}{4} = 0 \implies x_2^2 = 0 \implies x_2 = 0

And,

\left(x_1 + \frac{x_2}{2}\right)^2 = 0 \implies x_1 + \frac{x_2}{2} = 0

Substituting

x_2 = 0

into this equation gives:

x_1 + 0 = 0 \implies x_1 = 0

So, this case holds only if

x_1 = x_2 = 0

. This is a specific instance of the conclusion from Case 1 (

x_1=x_2

).

In all possible scenarios, the condition

f(x_1) = f(x_2)

leads to the conclusion that

x_1 = x_2

.

Therefore, the function

f(x) = x^3

is injective.

Hence Proved.

Q3Miscellaneous Exercise on Chapter 1

Given a non empty set X , consider $\mathrm{P}(\mathrm{X})$ which is the set of all subsets of X . Define the relation R in $\mathrm{P}(\mathrm{X})$ as follows: For subsets $\mathrm{A}, \mathrm{B}$ in $\mathrm{P}(\mathrm{X}), \mathrm{ARB}$ if and only if $\mathrm{A} \subset \mathrm{B}$ . Is R an equivalence relation on $\mathrm{P}(\mathrm{X})$ ? Justify your answer.

Solution

Given: A non-empty set X and its power set P(X). A relation R on P(X) is defined as: For subsets A, B in P(X), ARB if and only if A ⊂ B.

To determine: Whether R is an equivalence relation on P(X).

Solution: For a relation to be an equivalence relation, it must be reflexive, symmetric, and transitive.

1. Reflexivity: A relation R on P(X) is reflexive if ARA for every A ∈ P(X). ARA means A ⊂ A. Every set is a subset of itself. Therefore, A ⊂ A is true for all subsets A of X. So, R is reflexive.

2. Symmetry: A relation R on P(X) is symmetric if for all A, B ∈ P(X), ARB implies BRA. ARB implies A ⊂ B. BRA implies B ⊂ A. If A ⊂ B, it is not necessary that B ⊂ A. For B ⊂ A to be true, we must have A = B. But the condition A ⊂ B allows for A to be a proper subset of B.

Let's take a counterexample. Let X = {1, 2}. Then the power set is P(X) = {∅, {1}, {2}, {1, 2}}. Let A = {1} and B = {1, 2}. Both A and B are in P(X). Here, A ⊂ B since every element of A is in B. So, (A, B) ∈ R. However, B is not a subset of A, because 2 ∈ B but 2 ∉ A. So, B ⊄ A. This means (B, A) ∉ R. Since (A, B) ∈ R but (B, A) ∉ R, the relation R is not symmetric.

3. Transitivity: A relation R on P(X) is transitive if for all A, B, C ∈ P(X), (ARB and BRC) implies ARC. ARB means A ⊂ B. BRC means B ⊂ C. If A ⊂ B, then every element of A is also an element of B. If B ⊂ C, then every element of B is also an element of C. Combining these, it follows that every element of A is also an element of C. Therefore, A ⊂ C, which means ARC. So, R is transitive.

Conclusion: The relation R is reflexive and transitive, but it is not symmetric. Since an equivalence relation must satisfy all three properties, R is not an equivalence relation.

Final Answer: The relation R is not an equivalence relation because it is not symmetric.

Q4Miscellaneous Exercise on Chapter 1

Find the number of all onto functions from the set $\{1,2,3, \ldots \ldots, n\}$ to itself.

Solution

Given: A set A = {1, 2, 3, ..., n}.

To Find: The number of all onto functions from the set A to itself.

Solution: Let the given set be A = {1, 2, 3, ..., n}. The cardinality of the set is |A| = n. We need to find the number of onto (surjective) functions f: A → A.

For a function f: X → Y between two finite sets, if the sets have the same cardinality, i.e., |X| = |Y|, then the function is one-one (injective) if and only if it is onto (surjective).

In this case, the domain and codomain are the same set A, and |A| = n. So, any onto function from A to A must also be a one-one function.

A one-one function from a finite set to itself is a permutation of the elements of the set. We need to count the number of such permutations.

Let's consider the mapping of each element from the domain A to the codomain A.

The image of the first element, f(1), can be any of the n elements in the codomain. So, there are n choices for f(1).
Since the function is one-one, the image of the second element, f(2), must be different from f(1). So, there are (n-1) choices for f(2).
Similarly, there are (n-2) choices for f(3).
This continues until the last element, n. There will be only 1 choice left for f(n).

By the fundamental principle of counting (multiplication rule), the total number of such functions is the product of the number of choices for each element: Number of onto functions = n × (n-1) × (n-2) × ... × 2 × 1

This product is defined as n factorial, denoted by n!.

Final Answer: The number of all onto functions from the set {1, 2, 3, ..., n} to itself is n!.

Q5Miscellaneous Exercise on Chapter 1

Let $\mathrm{A}=\{-1,0,1,2\}, \mathrm{B}=\{-4,-2,0,2\}$ and $f, g: \mathrm{A} \rightarrow \mathrm{B}$ be functions defined by $f(x)=x^{2}-x, x \in \mathrm{A}$ and $g(x)=2\left|x-\frac{1}{2}\right|-1, x \in \mathrm{A}$ . Are $f$ and $g$ equal? Justify your answer. (Hint: One may note that two functions $f: \mathrm{A} \rightarrow \mathrm{B}$ and $g: \mathrm{A} \rightarrow \mathrm{B}$ such that $f(a)=g(a) \forall a \in \mathrm{A}$ , are called equal functions).

Solution

Given: Sets A = {-1, 0, 1, 2} and B = {-4, -2, 0, 2}. Two functions f, g: A → B defined by:

f(x) = x^2 - x

, for

x \in A

g(x) = 2\left|x - \frac{1}{2}\right| - 1

, for

x \in A

To determine: Whether the functions f and g are equal.

Condition for Equal Functions: Two functions f: A → B and g: A → B are called equal functions if

f(a) = g(a)

for all

a \in A

.

Solution: We will calculate the values of f(x) and g(x) for each element x in the domain A and check if they are equal.

1. For x = -1:

f(-1) = (-1)^2 - (-1) = 1 + 1 = 2

g(-1) = 2\left|-1 - \frac{1}{2}\right| - 1 = 2\left|-\frac{3}{2}\right| - 1 = 2\left(\frac{3}{2}\right) - 1 = 3 - 1 = 2

So,

f(-1) = g(-1) = 2

. The value 2 is in the codomain B.

2. For x = 0:

f(0) = (0)^2 - 0 = 0

g(0) = 2\left|0 - \frac{1}{2}\right| - 1 = 2\left|-\frac{1}{2}\right| - 1 = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0

So,

f(0) = g(0) = 0

. The value 0 is in the codomain B.

3. For x = 1:

f(1) = (1)^2 - 1 = 1 - 1 = 0

g(1) = 2\left|1 - \frac{1}{2}\right| - 1 = 2\left|\frac{1}{2}\right| - 1 = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0

So,

f(1) = g(1) = 0

. The value 0 is in the codomain B.

4. For x = 2:

f(2) = (2)^2 - 2 = 4 - 2 = 2

g(2) = 2\left|2 - \frac{1}{2}\right| - 1 = 2\left|\frac{3}{2}\right| - 1 = 2\left(\frac{3}{2}\right) - 1 = 3 - 1 = 2

So,

f(2) = g(2) = 2

. The value 2 is in the codomain B.

Conclusion: We have checked for all elements

x \in A

and found that

f(x) = g(x)

in every case.

f(-1) = g(-1) = 2

f(0) = g(0) = 0

f(1) = g(1) = 0

f(2) = g(2) = 2

Since the condition

f(a) = g(a)

for all

a \in A

is satisfied, the functions f and g are equal.

Final Answer: Yes, the functions f and g are equal. This is justified because

f(x) = g(x)

for all

x \in A

.

Q6Miscellaneous Exercise on Chapter 1

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

(A)

1

(B)

2

(C)

3

(D)

4

Solution

Given: Set A = {1, 2, 3}. A relation on A must contain (1, 2) and (1, 3). The relation must be reflexive and symmetric but not transitive.

To Find: The number of such relations.

Solution: Let R be a relation on the set A = {1, 2, 3}. The set of all possible ordered pairs is A × A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}.

We build the relation R based on the given conditions:

Reflexive: For R to be reflexive, it must contain all pairs (a, a) for a ∈ A. So, {(1, 1), (2, 2), (3, 3)} ⊂ R.
Contains (1, 2) and (1, 3): The pairs (1, 2) and (1, 3) must be in R.
Symmetric: For R to be symmetric, if (a, b) ∈ R, then (b, a) ∈ R. Since (1, 2) ∈ R, we must have (2, 1) ∈ R. Since (1, 3) ∈ R, we must have (3, 1) ∈ R.

Combining these necessary conditions, the smallest possible relation, let's call it

R_{min}

, must contain the following set of pairs:

R_{min}

= {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)}

Now, any valid relation R must be a superset of

R_{min}

. The pairs from A × A that are not in

R_{min}

are (2, 3) and (3, 2). Let's check the properties of

R_{min}

itself.

It is reflexive and contains (1,2) and (1,3) by construction.
It is symmetric by construction.
Is it transitive? A relation is transitive if for all (a, b) ∈ R and (b, c) ∈ R, we have (a, c) ∈ R. Let's check for pairs in $R_{min}$ that could violate this. Consider (2, 1) ∈ $R_{min}$ and (1, 3) ∈ $R_{min}$ . For transitivity, the pair (2, 3) must be in $R_{min}$ . But (2, 3) ∉ $R_{min}$ . Therefore, $R_{min}$ is not transitive.

So,

R_1 = R_{min}

is one such relation that satisfies all the given conditions.

Now, let's see if we can form any other such relations by adding the remaining pairs, which are (2, 3) and (3, 2). To maintain symmetry, if we add one, we must add the other. So, we must add the set {(2, 3), (3, 2)} to

R_{min}

.

Let's form a new relation

R_2 = R_{min} \cup

{(2, 3), (3, 2)}.

R_2

= {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}. This is the universal relation A × A. Let's check the properties of

R_2

:

It is reflexive, symmetric, and contains (1, 2) and (1, 3).
Is it transitive? Since $R_2$ is the universal relation A × A, for any (a, b) ∈ $R_2$ and (b, c) ∈ $R_2$ , the pair (a, c) will also be in $R_2$ . Thus, the universal relation is always transitive.

Since the condition is that the relation must not be transitive,

R_2

is not a valid relation.

There are no other pairs to add. The only possibilities were to form

R_1

or

R_2

. Only

R_1

meets all criteria.

Therefore, there is only one such relation.

Final Answer: The number of such relations is 1. So, the correct option is (A).

Q7Miscellaneous Exercise on Chapter 1

Let $\mathrm{A}=\{1,2,3\}$ . Then number of equivalence relations containing ( 1,2 ) is

(A)

1

(B)

2

(C)

3

(D)

4

Solution

Given: Set A = {1, 2, 3}. An equivalence relation on A that must contain the pair (1, 2).

To Find: The number of such equivalence relations.

Solution: An equivalence relation on a set partitions the set into disjoint subsets, called equivalence classes. The number of equivalence relations on a set is equal to the number of possible partitions of the set.

The set is A = {1, 2, 3}. The condition that the equivalence relation must contain the pair (1, 2) means that elements 1 and 2 must belong to the same equivalence class. In other words, in any partition of A, 1 and 2 must be in the same subset.

We need to find the number of partitions of the set {1, 2, 3} such that 1 and 2 are in the same block. Let's list the possible partitions of A = {1, 2, 3} with {1, 2} as a single unit.

Case 1: The block containing {1, 2} also contains 3. In this case, all three elements are in the same block. The partition is:

P_1 = \{\{1, 2, 3\}\}

This is a valid partition. It corresponds to the universal relation A × A, which is an equivalence relation containing (1, 2). This gives 1 equivalence relation.

Case 2: The block containing {1, 2} does not contain 3. In this case, the element 3 must be in a separate block by itself. The partition is:

P_2 = \{\{1, 2\}, \{3\}\}

This is also a valid partition. It corresponds to the equivalence relation R = {(1,1), (2,2), (3,3), (1,2), (2,1)}. This gives 1 equivalence relation.

These are the only two possible ways to partition the set {1, 2, 3} such that 1 and 2 are in the same subset.

Let's verify. Any partition of {1, 2, 3} must distribute the elements 1, 2, and 3 into disjoint blocks. Since 1 and 2 must be together, we can treat {1, 2} as a single entity, say 'X'. Now we need to partition the set {X, 3}.

Possibility 1: X and 3 are in the same block: {X, 3}. This translates back to {{1, 2, 3}}. This is our partition $P_1$ .
Possibility 2: X and 3 are in different blocks: {X}, {3}. This translates back to {{1, 2}, {3}}. This is our partition $P_2$ .

There are no other ways to partition {X, 3}. Thus, there are exactly two such partitions, and therefore, two such equivalence relations.

The relation corresponding to $P_1 = \{\{1, 2, 3\}\}$ is $R_1 = \{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)\}$ .
The relation corresponding to $P_2 = \{\{1, 2\}, \{3\}\}$ is $R_2 = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ .

Both

R_1

and

R_2

are equivalence relations containing (1, 2).

Final Answer: The number of equivalence relations containing (1, 2) is 2. So, the correct option is (B).

Relations and FunctionsClass 12 Mathematics NCERT Solutions

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

Relation R in the set A={1,2,3,…,13,14}\mathrm{A}=\{1,2,3, \ldots, 13,14\}A={1,2,3,…,13,14} defined as R={(x,y):3x−y=0}\mathrm{R}=\{(x, y): 3 x-y=0\}R={(x,y):3x−y=0}

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

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

Relation R in the set Z\mathbf{Z}Z of all integers defined as R={(x,y):x−y is an integer }\mathrm{R}=\{(x, y): x-y \text { is an integer }\}R={(x,y):x−y is an integer }

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

Solution

(i) Relation R in the set A={1,2,3,…,13,14}\mathrm{A}=\{1,2,3, \ldots, 13,14\}A={1,2,3,…,13,14} defined as R={(x,y):3x−y=0}\mathrm{R}=\{(x, y): 3 x-y=0\}R={(x,y):3x−y=0}

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

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

(iv) Relation R in the set Z\mathbf{Z}Z of all integers defined as R={(x,y):x−y is an integer }\mathrm{R}=\{(x, y): x-y \text { is an integer }\}R={(x,y):x−y is an integer }

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

Show that the relation R in the set R\mathbf{R}R of real numbers, defined as R={(a,b):a≤b2}\mathrm{R}=\left\{(a, b): a \leq b^{2}\right\}R={(a,b):a≤b2} is neither reflexive nor symmetric nor transitive.

Solution

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

Solution

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

Solution

Check whether the relation R in R\mathbf{R}R defined by R={(a,b):a≤b3}\mathrm{R}=\left\{(a, b): a \leq b^{3}\right\}R={(a,b):a≤b3} is reflexive, symmetric or transitive.

Solution

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

Solution

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

Solution

Solution

Show that each of the relation R in the set A={x∈Z:0≤x≤12}\mathrm{A}=\{x \in \mathbf{Z}: 0 \leq x \leq 12\}A={x∈Z:0≤x≤12}, given by

R={(a,b):∣a−b∣\mathrm{R}=\{(a, b):|a-b|R={(a,b):∣a−b∣ is a multiple of 4}\}}

R={(a,b):a=b}\mathrm{R}=\{(a, b): a=b\}R={(a,b):a=b} is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution

Give an example of a relation. Which is

Symmetric but neither reflexive nor transitive.

Transitive but neither reflexive nor symmetric.

Reflexive and symmetric but not transitive.

Reflexive and transitive but not symmetric.

Symmetric and transitive but not reflexive.

Solution

Solution

Solution

Show that the relation R defined in the set A of all polygons as R = {(P_{1}, P_{2}): P_{1} and P_{2} have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3,4 and 5 ?

Solution

Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L_{1}, L_{2}): L_{1} is parallel to L_{2}}. Show that R is an equivalence relation. Find the set of all lines related to the line y=2x+4y=2 x+4y=2x+4.

Solution

Let R be the relation in the set {1,2,3,4} given by R = {(1,2),(2,2),(1,1),(4,4), (1,3),(3,3),(3,2)}. Choose the correct answer.

Solution

Let R be the relation in the set N given by R = {(a, b): a=b−2,b>6a=b-2, b>6a=b−2,b>6}. Choose the correct answer.

Solution

Solution

Check the injectivity and surjectivity of the following functions:

f:N→Nf: \mathbf{N} \rightarrow \mathbf{N}f:N→N given by f(x)=x2f(x)=x^{2}f(x)=x2

f:Z→Zf: \mathbf{Z} \rightarrow \mathbf{Z}f:Z→Z given by f(x)=x2f(x)=x^{2}f(x)=x2

f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R given by f(x)=x2f(x)=x^{2}f(x)=x2

f:N→Nf: \mathbf{N} \rightarrow \mathbf{N}f:N→N given by f(x)=x3f(x)=x^{3}f(x)=x3

f:Z→Zf: \mathbf{Z} \rightarrow \mathbf{Z}f:Z→Z given by f(x)=x3f(x)=x^{3}f(x)=x3

Solution

Prove that the Greatest Integer Function f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R, given by f(x)=[x]f(x)=[x]f(x)=[x], is neither one-one nor onto, where [x][x][x] denotes the greatest integer less than or equal to xxx.

Solution

Show that the Modulus Function f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R, given by f(x)=∣x∣f(x)=|x|f(x)=∣x∣, is neither oneone nor onto, where ∣x∣|x|∣x∣ is xxx, if xxx is positive or 0 and ∣x∣|x|∣x∣ is −x-x−x, if xxx is negative.

Solution

Show that the Signum Function f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R, given by f(x)=1, if x>0 0, if x=0 1, if x<0f(x)=\begin{array}{r} 1, \text { if } x>0 \ 0, \text { if } x=0 \ 1, \text { if } x<0 \end{array}f(x)=1, if x>0 0, if x=0 1, if x<0​ is neither one-one nor onto.

Solution

Let A = {1,2,3}, B = {4,5,6,7} and let f={(1,4),(2,5),(3,6)}f=\{(1,4),(2,5),(3,6)\}f={(1,4),(2,5),(3,6)} be a function from A to B . Show that fff is one-one.

Solution

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R defined by f(x)=3−4xf(x)=3-4 xf(x)=3−4x

f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R defined by f(x)=1+x2f(x)=1+x^{2}f(x)=1+x2

Solution

Let A and B be sets. Show that f:A×B→B×Af: \mathrm{A} \times \mathrm{B} \rightarrow \mathrm{B} \times \mathrm{A}f:A×B→B×A such that f(a,b)=(b,a)f(a, b)=(b, a)f(a,b)=(b,a) is bijective function.

Solution

Solution

Solution

Let f:R→Rf: \boldsymbol{\mathrm { R }} \rightarrow \boldsymbol{\mathrm { R }}f:R→R be defined as f(x)=x4f(x)=x^{4}f(x)=x4. Choose the correct answer.

Solution

Let f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R be defined as f(x)=3xf(x)=3 xf(x)=3x. Choose the correct answer.

Solution

Show that the function f:R→{x∈R:−1<x<1}f: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1<x<1\}f:R→{x∈R:−1<x<1} defined by f(x)=x1+∣x∣f(x)=\frac{x}{1+|x|}f(x)=1+∣x∣x​, x∈Rx \in \mathbf{R}x∈R is one one and onto function.

Solution

Show that the function f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R given by f(x)=x3f(x)=x^{3}f(x)=x3 is injective.

Solution

Solution

Relation R in the set $\mathrm{A}=\{1,2,3, \ldots, 13,14\}$ defined as $\mathrm{R}=\{(x, y): 3 x-y=0\}$

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

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

Relation R in the set $\mathbf{Z}$ of all integers defined as $\mathrm{R}=\{(x, y): x-y \text { is an integer }\}$

(i) Relation R in the set $\mathrm{A}=\{1,2,3, \ldots, 13,14\}$ defined as $\mathrm{R}=\{(x, y): 3 x-y=0\}$

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

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

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

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

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

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

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

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

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

Show that each of the relation R in the set $\mathrm{A}=\{x \in \mathbf{Z}: 0 \leq x \leq 12\}$ , given by

$\mathrm{R}=\{(a, b):|a-b|$ is a multiple of 4 $\}$

$\mathrm{R}=\{(a, b): a=b\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L_{1}, L_{2}): L_{1} is parallel to L_{2}}. Show that R is an equivalence relation. Find the set of all lines related to the line $y=2 x+4$ .

Let R be the relation in the set N given by R = {(a, b): $a=b-2, b>6$ }. Choose the correct answer.

$f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{2}$

$f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{2}$

$f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=x^{2}$

$f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{3}$

$f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{3}$

Prove that the Greatest Integer Function $f: \mathbf{R} \rightarrow \mathbf{R}$ , given by $f(x)=[x]$ , is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$ .

Show that the Modulus Function $f: \mathbf{R} \rightarrow \mathbf{R}$ , given by $f(x)=|x|$ , is neither oneone nor onto, where $|x|$ is $x$ , if $x$ is positive or 0 and $|x|$ is $-x$ , if $x$ is negative.

Show that the Signum Function $f: \mathbf{R} \rightarrow \mathbf{R}$ , given by $f(x)=\begin{array}{r} 1, \text { if } x>0 \ 0, \text { if } x=0 \ 1, \text { if } x<0 \end{array}$ is neither one-one nor onto.

Let A = {1,2,3}, B = {4,5,6,7} and let $f=\{(1,4),(2,5),(3,6)\}$ be a function from A to B . Show that $f$ is one-one.

$f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=3-4 x$

$f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=1+x^{2}$

Let A and B be sets. Show that $f: \mathrm{A} \times \mathrm{B} \rightarrow \mathrm{B} \times \mathrm{A}$ such that $f(a, b)=(b, a)$ is bijective function.

Let $f: \boldsymbol{\mathrm { R }} \rightarrow \boldsymbol{\mathrm { R }}$ be defined as $f(x)=x^{4}$ . Choose the correct answer.

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $f(x)=3 x$ . Choose the correct answer.

Show that the function $f: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1<x<1\}$ defined by $f(x)=\frac{x}{1+|x|}$ , $x \in \mathbf{R}$ is one one and onto function.

Show that the function $f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=x^{3}$ is injective.

Find the number of all onto functions from the set $\{1,2,3, \ldots \ldots, n\}$ to itself.

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

Let $\mathrm{A}=\{1,2,3\}$ . Then number of equivalence relations containing ( 1,2 ) is