MATHEMATICS CLASS- 12

CHAPTER-1
(RELATIONS AND FUNCTIONS)

CBSEChapter 1EXERCISE 1.1

Exercise 1.1

Question 1 (i)

Determine whether the following relation is reflexive, symmetric and transitive.

Relation R in the set

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

is defined as

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

Solution:

The given relation is

$$ 3x-y=0 $$

or,

$$ y=3x. $$

Since both x and y must belong to the set

$$ A=\{1,2,3,\ldots,14\}, $$

the ordered pairs of the relation are

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

For

$$ x=5,\; y=15, $$

but

$$ 15\notin A, $$

so no more ordered pairs are possible.

1. Reflexive

A relation is reflexive if

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

for every

$$ x\in A. $$

Here,

$$ (x,x)\in R \Longrightarrow 3x=x $$ $$ 2x=0. $$

This is not true for any element of the set.

For example,

$$ (1,1)\notin R,\qquad (2,2)\notin R,\qquad (3,3)\notin R. $$

Therefore, the relation is not reflexive.

2. Symmetric

A relation is symmetric if

$$ (x,y)\in R \Longrightarrow (y,x)\in R. $$

Consider the ordered pair

$$ (1,3)\in R. $$

Its reverse pair is

$$ (3,1). $$

Check whether it belongs to the relation.

$$ 3(3)-1=8\ne0. $$

Hence,

$$ (3,1)\notin R. $$

Therefore, the relation is not symmetric.

3. Transitive

A relation is transitive if

$$ (x,y)\in R \quad\text{and}\quad (y,z)\in R \Longrightarrow (x,z)\in R. $$

The relation contains the pairs

$$ (1,3),\; (2,6),\; (3,9),\; (4,12). $$

Notice that

$$ (1,3)\in R \quad\text{and}\quad (3,9)\in R. $$

If the relation were transitive, then

$$ (1,9) $$

must also belong to

$$ R. $$

Check:

$$ 3(1)-9=-6\ne0. $$

Hence,

$$ (1,9)\notin R. $$

Therefore, the relation is not transitive.

Final Answer

Property	Result	Reason
Reflexive	No	No pair of the form $(x,x)$ satisfies $3x-y=0$.
Symmetric	No	$(1,3)\in R$, but $(3,1)\notin R$.
Transitive	No	$(1,3)$ and $(3,9)$ belong to $R$, but $(1,9)\notin R$.

Answer:

$$ \boxed{\text{The relation is neither reflexive, nor symmetric, nor transitive.}} $$

Question 1 (ii)

Determine whether the following relation is reflexive, symmetric and transitive.

Relation R in the set of natural numbers N is defined as

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

Solution:

The relation is defined by

$$ y=x+5,\qquad x<4. $$

Since x is a natural number and

$$ x<4, $$

the possible values of x are

$$ 1,\;2,\;3. $$

Hence, the relation is

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

1. Reflexive

A relation is reflexive if

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

for every

$$ x\in N. $$

Here, the relation contains only

$$ (1,6),\;(2,7),\;(3,8), $$

and no ordered pair of the form

$$ (x,x) $$

belongs to the relation.

For example,

$$ (1,1)\notin R,\qquad (2,2)\notin R. $$

Therefore, the relation is not reflexive.

2. Symmetric

A relation is symmetric if

$$ (x,y)\in R \Longrightarrow (y,x)\in R. $$

Consider the ordered pair

$$ (1,6)\in R. $$

Its reverse pair is

$$ (6,1). $$

Check whether it belongs to the relation.

For

$$ (6,1), $$

we have

$$ 1\ne6+5, $$

and

$$ 6\not<4. $$

Hence,

$$ (6,1)\notin R. $$

Therefore, the relation is not symmetric.

3. Transitive

A relation is transitive if

$$ (x,y)\in R \quad\text{and}\quad (y,z)\in R \Longrightarrow (x,z)\in R. $$

The relation contains the pairs

$$ (1,6),\;(2,7),\;(3,8). $$

Notice that none of the second elements

$$ 6,\;7,\;8 $$

appear as the first element of another ordered pair in the relation.

Therefore, there are no pairs of the form

$$ (x,y)\in R \quad\text{and}\quad (y,z)\in R. $$

Hence, the transitivity condition is never violated.

Therefore, the relation is transitive.

Final Answer

Property	Result	Reason
Reflexive	No	No ordered pair of the form $(x,x)$ belongs to the relation.
Symmetric	No	$(1,6)\in R$, but $(6,1)\notin R$.
Transitive	Yes	There are no two pairs $(x,y)$ and $(y,z)$ in the relation, so the transitive condition is satisfied.

Answer:

$$ \boxed{\text{The relation is not reflexive, not symmetric, but it is transitive.}} $$

Question 1 (iii)

Determine whether the following relation is reflexive, symmetric and transitive.

Relation R in the set

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

is defined as

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

Solution:

The relation contains all ordered pairs $(x,y)$ such that y is divisible by x.

Some of the ordered pairs are

$$ \begin{aligned} R=\{& (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ &(2,2),(2,4),(2,6),\\ &(3,3),(3,6),\\ &(4,4),\\ &(5,5),\\ &(6,6)\}. \end{aligned} $$

1. Reflexive

A relation is reflexive if

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

for every

$$ x\in A. $$

Since every number is divisible by itself,

$$ 1\mid1,\; 2\mid2,\; 3\mid3,\; 4\mid4,\; 5\mid5,\; 6\mid6. $$

Hence,

$$ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\in R. $$

Therefore, the relation is reflexive.

2. Symmetric

A relation is symmetric if

$$ (x,y)\in R \Longrightarrow (y,x)\in R. $$

Consider the ordered pair

$$ (2,4)\in R, $$

because

$$ 4\text{ is divisible by }2. $$

Its reverse pair is

$$ (4,2). $$

But

$$ 2 $$

is not divisible by

$$ 4. $$

Hence,

$$ (4,2)\notin R. $$

Therefore, the relation is not symmetric.

3. Transitive

A relation is transitive if

$$ (x,y)\in R \quad\text{and}\quad (y,z)\in R \Longrightarrow (x,z)\in R. $$

Suppose

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

Then

$$ y $$

is divisible by

$$ x. $$

Also, if

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

then

$$ z $$

is divisible by

$$ y. $$

Since

$$ y $$

is divisible by

$$ x, $$

it follows that

$$ z $$

is also divisible by

$$ x. $$

Hence,

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

For example,

$$ (2,4)\in R $$

and

$$ (4,4)\in R. $$

Therefore,

$$ (2,4)\in R, $$

which satisfies the transitive property.

Thus, the relation is transitive.

Final Answer

Property	Result	Reason
Reflexive	Yes	Every element is divisible by itself.
Symmetric	No	$(2,4)\in R$, but $(4,2)\notin R$.
Transitive	Yes	If $y$ is divisible by $x$ and $z$ is divisible by $y$, then $z$ is divisible by $x$.

Answer:

$$ \boxed{\text{The relation is reflexive and transitive, but not symmetric.}} $$

Question 1 (iv)

Determine whether the following relation is reflexive, symmetric and transitive.

Relation R in the set of all integers

$$ \mathbb{Z} $$

is defined as

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

Solution:

Since

$$ x,y\in\mathbb{Z}, $$

both x and y are integers.

The difference of any two integers is always an integer.

Therefore,

$$ x-y\in\mathbb{Z} $$

for every pair

$$ (x,y)\in\mathbb{Z}\times\mathbb{Z}. $$

Hence, the relation contains every possible ordered pair of integers.

1. Reflexive

A relation is reflexive if

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

for every

$$ x\in\mathbb{Z}. $$

Here,

$$ x-x=0. $$

Since

$$ 0 $$

is an integer,

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

for every integer

$$ x. $$

Therefore, the relation is reflexive.

2. Symmetric

A relation is symmetric if

$$ (x,y)\in R \Longrightarrow (y,x)\in R. $$

Suppose

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

Then

$$ x-y $$

is an integer.

Now,

$$ y-x=-(x-y). $$

The negative of an integer is also an integer.

Hence,

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

Therefore, the relation is symmetric.

3. Transitive

A relation is transitive if

$$ (x,y)\in R \quad\text{and}\quad (y,z)\in R \Longrightarrow (x,z)\in R. $$

Suppose

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

and

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

Then

$$ x-y $$

and

$$ y-z $$

are integers.

Add the two expressions.

$$ (x-y)+(y-z)=x-z. $$

The sum of two integers is an integer.

Hence,

$$ x-z $$

is an integer.

Therefore,

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

Thus, the relation is transitive.

Final Answer

Property	Result	Reason
Reflexive	Yes	$x-x=0$, which is an integer.
Symmetric	Yes	If $x-y$ is an integer, then $y-x=-(x-y)$ is also an integer.
Transitive	Yes	If $x-y$ and $y-z$ are integers, then $x-z=(x-y)+(y-z)$ is also an integer.

Answer:

$$ \boxed{\text{The relation is reflexive, symmetric and transitive.}} $$

Question 1 (v)

Determine whether the following relation is reflexive, symmetric and transitive.

Relation R in the set

$$ A=\{1,2,3,\ldots,9\} $$

is defined as

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

Solution:

The relation is defined by

$$ (a,b)\in R \iff |a-b| \text{ is even.} $$

An even difference means that both numbers have the same parity, i.e., they are either both even or both odd.

In the given set,

Odd elements are

$$ \{1,3,5,7,9\} $$

Even elements are

$$ \{2,4,6,8\}. $$

Hence, every odd number is related to every odd number, and every even number is related to every even number.

1. Reflexive

A relation is reflexive if

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

for every

$$ a\in A. $$

Since

$$ |a-a|=0, $$

and

$$ 0 $$

is an even number, every element is related to itself.

Therefore, the relation is reflexive.

2. Symmetric

A relation is symmetric if

$$ (a,b)\in R \Longrightarrow (b,a)\in R. $$

Suppose

$$ (a,b)\in R. $$

Then

$$ |a-b| $$

is even.

Since

$$ |b-a|=|a-b|, $$

it is also even.

Hence,

$$ (b,a)\in R. $$

Therefore, the relation is symmetric.

3. Transitive

A relation is transitive if

$$ (a,b)\in R \quad\text{and}\quad (b,c)\in R \Longrightarrow (a,c)\in R. $$

$$ |a-b| $$

is even, then

$$ a $$

and

$$ b $$

have the same parity.

Similarly, if

$$ |b-c| $$

is even, then

$$ b $$

and

$$ c $$

have the same parity.

Therefore,

$$ a $$

and

$$ c $$

also have the same parity.

Hence,

$$ |a-c| $$

is even.

Thus,

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

Therefore, the relation is transitive.

Final Answer

Property	Result	Reason
Reflexive	Yes	$\|a-a\|=0$, which is even.
Symmetric	Yes	$\|a-b\|=\|b-a\|$.
Transitive	Yes	If two numbers have the same parity as a third number, then they have the same parity with each other.

Answer:

$$ \boxed{\text{The relation is reflexive, symmetric and transitive.}} $$

Question 2

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

Solution:

The given relation is

$$ (a,b)\in R \iff a\le b^2. $$

We shall check each property one by one.

1. Reflexive

A relation is reflexive if

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

for every

$$ a\in\mathbb{R}. $$

For reflexivity, we must have

$$ a\le a^2 $$

for every real number.

This is not always true.

Take

$$ a=\frac12. $$

Then

$$ \frac12\le\left(\frac12\right)^2 $$ $$ \frac12\le\frac14, $$

which is false.

Hence,

$$ \left(\frac12,\frac12\right)\notin R. $$

Therefore, the relation is not reflexive.

2. Symmetric

A relation is symmetric if

$$ (a,b)\in R \Longrightarrow (b,a)\in R. $$

Take

$$ a=1,\qquad b=2. $$

Then

$$ 1\le2^2 $$ $$ 1\le4, $$

which is true.

Therefore,

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

Now check the reverse pair.

$$ 2\le1^2 $$ $$ 2\le1, $$

which is false.

Hence,

$$ (2,1)\notin R. $$

Therefore, the relation is not symmetric.

3. Transitive

A relation is transitive if

$$ (a,b)\in R \quad\text{and}\quad (b,c)\in R \Longrightarrow (a,c)\in R. $$

Take

$$ a=4,\qquad b=-2,\qquad c=1. $$

Then

$$ 4\le(-2)^2 $$ $$ 4\le4, $$

$$ (4,-2)\in R. $$

Also,

$$ -2\le1^2 $$ $$ -2\le1, $$

$$ (-2,1)\in R. $$

Now check

$$ (4,1). $$

We have

$$ 4\le1^2 $$ $$ 4\le1, $$

which is false.

Hence,

$$ (4,1)\notin R. $$

Therefore, the relation is not transitive.

Final Answer

Property	Result	Reason
Reflexive	No	For example, $\left(\frac12,\frac12\right)\notin R$.
Symmetric	No	$(1,2)\in R$, but $(2,1)\notin R$.
Transitive	No	$(4,-2)\in R$ and $(-2,1)\in R$, but $(4,1)\notin R$.

Answer:

$$ \boxed{\text{The relation is neither reflexive, nor symmetric, nor transitive.}} $$

Question 3

Check whether the relation R defined on the set $$ A=\{1,2,3,4,5,6\} $$

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

is reflexive, symmetric or transitive.

Solution:

The relation is defined by

$$ b=a+1. $$

Hence, the ordered pairs belonging to the relation are

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

1. Reflexive

A relation is reflexive if

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

for every

$$ a\in A. $$

Here, the relation contains only the pairs

$$ (1,2),\;(2,3),\;(3,4),\;(4,5),\;(5,6). $$

No ordered pair of the form

$$ (a,a) $$

belongs to the relation.

For example,

$$ (1,1)\notin R,\qquad (2,2)\notin R. $$

Therefore, the relation is not reflexive.

2. Symmetric

A relation is symmetric if

$$ (a,b)\in R \Longrightarrow (b,a)\in R. $$

Consider the ordered pair

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

Its reverse pair is

$$ (2,1). $$

Check whether it belongs to the relation.

For

$$ (2,1), $$

we require

$$ 1=2+1, $$

which is false.

Hence,

$$ (2,1)\notin R. $$

Therefore, the relation is not symmetric.

3. Transitive

A relation is transitive if

$$ (a,b)\in R \quad\text{and}\quad (b,c)\in R \Longrightarrow (a,c)\in R. $$

Consider the pairs

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

and

$$ (2,3)\in R. $$

If the relation were transitive, then

$$ (1,3) $$

must also belong to

$$ R. $$

But for

$$ (1,3), $$

we have

$$ 3\ne1+1. $$

Hence,

$$ (1,3)\notin R. $$

Therefore, the relation is not transitive.

Final Answer

Property	Result	Reason
Reflexive	No	No ordered pair of the form $(a,a)$ belongs to the relation.
Symmetric	No	$(1,2)\in R$, but $(2,1)\notin R$.
Transitive	No	$(1,2)$ and $(2,3)$ belong to $R$, but $(1,3)\notin R$.

Answer:

$$ \boxed{\text{The relation is neither reflexive, nor symmetric, nor transitive.}} $$

Question 4

Show that the relation R on the set of real numbers $$ \mathbb{R} $$

defined by

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

is reflexive and transitive but not symmetric.

Solution:

The given relation is

$$ (a,b)\in R \iff a\le b. $$

We shall verify each property one by one.

1. Reflexive

A relation is reflexive if

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

for every

$$ a\in\mathbb{R}. $$

Since every real number is always less than or equal to itself,

$$ a\le a. $$

Therefore,

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

for every

$$ a\in\mathbb{R}. $$

Hence, the relation is reflexive.

2. Symmetric

A relation is symmetric if

$$ (a,b)\in R \Longrightarrow (b,a)\in R. $$

Take

$$ a=2,\qquad b=5. $$

Since

$$ 2\le5, $$

we have

$$ (2,5)\in R. $$

Now check the reverse pair.

$$ 5\le2, $$

which is false.

Therefore,

$$ (5,2)\notin R. $$

Hence, the relation is not symmetric.

3. Transitive

A relation is transitive if

$$ (a,b)\in R \quad\text{and}\quad (b,c)\in R \Longrightarrow (a,c)\in R. $$

Suppose

$$ (a,b)\in R $$

and

$$ (b,c)\in R. $$

Then

$$ a\le b $$

and

$$ b\le c. $$

By the transitive property of inequalities,

$$ a\le c. $$

Therefore,

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

Hence, the relation is transitive.

Final Answer

Property	Result	Reason
Reflexive	Yes	Every real number satisfies $a\le a$.
Symmetric	No	$(2,5)\in R$, but $(5,2)\notin R$.
Transitive	Yes	If $a\le b$ and $b\le c$, then $a\le c$.

Answer:

$$ \boxed{\text{The relation is reflexive and transitive, but not symmetric.}} $$

Question 5

Prove that the relation R in the set $$ A=\{1,2,3,\ldots,14\} $$

defined by

$$ R=\{(a,b):3a-b=0\} $$

is neither reflexive nor symmetric nor transitive.

Solution:

The given relation is

$$ 3a-b=0 $$

or equivalently,

$$ b=3a. $$

Since both a and b belong to the set

$$ A=\{1,2,3,\ldots,14\}, $$

the ordered pairs in the relation are

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

For

$$ a=5, $$

we get

$$ b=15, $$

which does not belong to the set

$$ A. $$

Hence, there are no more ordered pairs.

1. Reflexive

A relation is reflexive if

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

for every

$$ a\in A. $$

For

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

we must have

$$ 3a-a=0 $$ $$ 2a=0. $$

This is not true for any element of the given set.

For example,

$$ (1,1)\notin R,\qquad (2,2)\notin R. $$

Therefore, the relation is not reflexive.

2. Symmetric

A relation is symmetric if

$$ (a,b)\in R \Longrightarrow (b,a)\in R. $$

Consider the ordered pair

$$ (1,3)\in R. $$

The reverse ordered pair is

$$ (3,1). $$

Check whether it belongs to the relation.

$$ 3(3)-1=8\ne0. $$

Hence,

$$ (3,1)\notin R. $$

Therefore, the relation is not symmetric.

3. Transitive

A relation is transitive if

$$ (a,b)\in R \quad\text{and}\quad (b,c)\in R \Longrightarrow (a,c)\in R. $$

Consider the ordered pairs

$$ (1,3)\in R $$

and

$$ (3,9)\in R. $$

If the relation were transitive, then

$$ (1,9) $$

must also belong to the relation.

Check:

$$ 3(1)-9=-6\ne0. $$

Hence,

$$ (1,9)\notin R. $$

Therefore, the relation is not transitive.

Final Answer

Property	Result	Reason
Reflexive	No	No ordered pair of the form $(a,a)$ satisfies $3a-b=0$.
Symmetric	No	$(1,3)\in R$, but $(3,1)\notin R$.
Transitive	No	$(1,3)$ and $(3,9)$ belong to $R$, but $(1,9)\notin R$.

Answer:

$$ \boxed{\text{The relation is neither reflexive, nor symmetric, nor transitive.}} $$

Question 6

Show that the relation R in the set $$ A=\{1,2,3\} $$

given by

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

is symmetric but neither reflexive nor transitive.

Solution:

The given set is

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

and the relation is

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

We examine each property one by one.

1. Reflexive

A relation is reflexive if

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

for every

$$ a\in A. $$

Therefore, the relation must contain the ordered pairs

$$ (1,1),\ (2,2),\ (3,3). $$

But

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

does not contain any of these ordered pairs.

Hence,

$$ (1,1)\notin R,\qquad (2,2)\notin R,\qquad (3,3)\notin R. $$

Therefore, the relation is not reflexive.

2. Symmetric

A relation is symmetric if

$$ (a,b)\in R \Longrightarrow (b,a)\in R. $$

Here,

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

and its reverse pair

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

Similarly,

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

and its reverse pair

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

Thus, whenever

$$ (a,b)\in R, $$

the pair

$$ (b,a) $$

also belongs to

$$ R. $$

Therefore, the relation is symmetric.

3. Transitive

A relation is transitive if

$$ (a,b)\in R \quad\text{and}\quad (b,c)\in R \Longrightarrow (a,c)\in R. $$

Consider the ordered pairs

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

and

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

If the relation were transitive, then

$$ (1,1) $$

must also belong to

$$ R. $$

But

$$ (1,1)\notin R. $$

Hence, the transitive property fails.

Similarly,

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

and

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

but

$$ (2,2)\notin R. $$

Therefore, the relation is not transitive.

Final Answer

Property	Result	Reason
Reflexive	No	$(1,1)$, $(2,2)$ and $(3,3)$ do not belong to the relation.
Symmetric	Yes	$(1,2)\in R$ and $(2,1)\in R$.
Transitive	No	$(1,2)$ and $(2,1)$ belong to $R$, but $(1,1)\notin R$.

Answer:

$$ \boxed{\text{The relation is symmetric but neither reflexive nor transitive.}} $$

Question 7

Show that the relation R on the set $$ A=\{\text{all the books in a library of a college}\} $$

defined by

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

is an equivalence relation.

Solution:

To prove that R is an equivalence relation, we must show that it is:

Reflexive
Symmetric
Transitive

1. Reflexive

A relation is reflexive if

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

for every

$$ x\in A. $$

Every book has the same number of pages as itself.

Hence,

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

for every book

$$ x\in A. $$

Therefore, the relation is reflexive.

2. Symmetric

A relation is symmetric if

$$ (x,y)\in R \Longrightarrow (y,x)\in R. $$

Suppose

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

Then book

$$ x $$

and book

$$ y $$

have the same number of pages.

Therefore, book

$$ y $$

also has the same number of pages as book

$$ x. $$

Hence,

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

Therefore, the relation is symmetric.

3. Transitive

A relation is transitive if

$$ (x,y)\in R \quad\text{and}\quad (y,z)\in R \Longrightarrow (x,z)\in R. $$

Suppose

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

and

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

Then

Book x and book y have the same number of pages.
Book y and book z have the same number of pages.

Therefore, book

$$ x $$

and book

$$ z $$

also have the same number of pages.

Hence,

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

Therefore, the relation is transitive.

Final Conclusion

Property	Result	Reason
Reflexive	Yes	Every book has the same number of pages as itself.
Symmetric	Yes	If two books have the same number of pages, then the reverse is also true.
Transitive	Yes	If book x and y have the same number of pages, and y and z have the same number of pages, then x and z also have the same number of pages.

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

Answer:

$$ \boxed{\text{The relation }R\text{ is an equivalence relation.}} $$

Question 8

Show that the relation R in the set $$ A=\{1,2,3,4,5\} $$

given by

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

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:

The relation is defined by

$$ (a,b)\in R \iff |a-b| \text{ is even.} $$

An even difference means that both numbers have the same parity, i.e., both are odd or both are even.

The set

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

contains

Odd elements: $$ \{1,3,5\} $$
Even elements: $$ \{2,4\} $$

1. Reflexive

A relation is reflexive if

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

for every

$$ a\in A. $$

Since

$$ |a-a|=0, $$

and

$$ 0 $$

is an even number, every element is related to itself.

Therefore, the relation is reflexive.

2. Symmetric

A relation is symmetric if

$$ (a,b)\in R \Longrightarrow (b,a)\in R. $$

$$ |a-b| $$

is even, then

$$ |b-a|=|a-b|, $$

which is also even.

Hence,

$$ (b,a)\in R. $$

Therefore, the relation is symmetric.

3. Transitive

A relation is transitive if

$$ (a,b)\in R \quad\text{and}\quad (b,c)\in R \Longrightarrow (a,c)\in R. $$

$$ |a-b| $$

and

$$ |b-c| $$

are even, then

$$ a $$

and

$$ b $$

have the same parity, and

$$ b $$

and

$$ c $$

also have the same parity.

Hence,

$$ a $$

and

$$ c $$

have the same parity.

Therefore,

$$ |a-c| $$

is even.

Thus,

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

Hence, the relation is transitive.

Therefore, R is an Equivalence Relation.

Since the relation is

Reflexive
Symmetric
Transitive

it is an equivalence relation.

Elements of the set

$$ \{1,3,5\} $$

All these numbers are odd.

The difference between any two odd numbers is always even.

For example,

$$ |1-3|=2, $$ $$ |1-5|=4, $$ $$ |3-5|=2. $$

All these are even numbers.

Hence, every element of

$$ \{1,3,5\} $$

is related to every other element of the same set.

Elements of the set

$$ \{2,4\} $$

Both numbers are even.

Their difference is

$$ |2-4|=2, $$

which is even.

Therefore, the elements

$$ 2 $$

and

$$ 4 $$

are related to each other.

Relation between

$$ \{1,3,5\} $$

and

$$ \{2,4\} $$

Every element of

$$ \{1,3,5\} $$

is odd, while every element of

$$ \{2,4\} $$

is even.

The difference between an odd number and an even number is always odd.

For example,

$$ |1-2|=1, $$ $$ |3-4|=1, $$ $$ |5-2|=3. $$

All these differences are odd.

Hence, no element of

$$ \{1,3,5\} $$

is related to any element of

$$ \{2,4\}. $$

Final Answer

Property	Result
Reflexive	Yes
Symmetric	Yes
Transitive	Yes

Answer:

$$ \boxed{\text{The relation }R\text{ is an equivalence relation.}} $$

The equivalence classes are

$$ \boxed{\{1,3,5\}\quad\text{and}\quad\{2,4\}.} $$

All elements within each class are related to one another, whereas no element of one class is related to any element of the other class.

Question 9 (i)

Show that the relation

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

on the set

$$ A=\{0,1,2,\ldots,12\} $$

is an equivalence relation. Find the set of all elements related to 1.

Proof

To prove that R is an equivalence relation, we verify the three properties.

Reflexive

For every

$$ a\in A, $$ $$ |a-a|=0, $$

and

$$ 0=4\times0, $$

which is a multiple of 4.

Hence,

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

for every

$$ a\in A. $$

Therefore, the relation is reflexive.

Symmetric

Suppose

$$ (a,b)\in R. $$

Then

$$ |a-b| $$

is a multiple of 4.

Since

$$ |b-a|=|a-b|, $$

it is also a multiple of 4.

Therefore,

$$ (b,a)\in R. $$

Hence, the relation is symmetric.

Transitive

Suppose

$$ (a,b)\in R $$

and

$$ (b,c)\in R. $$

Then there exist integers

$$ m,n $$

such that

$$ a-b=4m $$

and

$$ b-c=4n. $$

Adding the two equations,

$$ a-c=4m+4n $$ $$ =4(m+n). $$

Since

$$ m+n $$

is an integer,

$$ |a-c| $$

is also a multiple of 4.

Hence,

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

Therefore, the relation is transitive.

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

Elements Related to 1

We need all elements

$$ x\in A $$

such that

$$ |x-1| $$

is a multiple of 4.

Checking the elements of

$$ A=\{0,1,2,\ldots,12\}, $$ $$ |1-1|=0, $$ $$ |5-1|=4, $$ $$ |9-1|=8. $$

All are multiples of 4.

Answer:

$$ \boxed{\{1,5,9\}} $$

Question 9 (ii)

Show that the relation

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

on the set

$$ A=\{0,1,2,\ldots,12\} $$

is an equivalence relation. Find the set of all elements related to 1.

Proof

We verify the three properties.

Reflexive

For every

$$ a\in A, $$ $$ a=a. $$

Therefore,

$$ (a,a)\in R. $$

Hence, the relation is reflexive.

Symmetric

Suppose

$$ (a,b)\in R. $$

Then

$$ a=b. $$

Therefore,

$$ b=a, $$

which implies

$$ (b,a)\in R. $$

Hence, the relation is symmetric.

Transitive

Suppose

$$ (a,b)\in R $$

and

$$ (b,c)\in R. $$

Then

$$ a=b $$

and

$$ b=c. $$

Therefore,

$$ a=c, $$

which implies

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

Hence, the relation is transitive.

Show that the relation

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

defined on the set of all points in a plane is an equivalence relation. Further, show that the set of all points related to a point

$$ P\neq(0,0) $$

is the circle passing through

$$ P $$

with origin as centre.

Proof

Let

$$ O=(0,0) $$

be the origin.

For any two points

$$ P,Q, $$

the relation

$$ (P,Q)\in R $$

means

$$ OP=OQ, $$

where

$$ OP $$

and

$$ OQ $$

denote the distances of the points

$$ P $$

and

$$ Q $$

from the origin.

We verify the three properties of an equivalence relation.

Reflexive

For every point

$$ P, $$

its distance from the origin is equal to itself.

$$ OP=OP. $$

Hence,

$$ (P,P)\in R. $$

Therefore, the relation is reflexive.

Symmetric

Suppose

$$ (P,Q)\in R. $$

Then

$$ OP=OQ. $$

Since equality is symmetric,

$$ OQ=OP. $$

Therefore,

$$ (Q,P)\in R. $$

Hence, the relation is symmetric.

Transitive

Suppose

$$ (P,Q)\in R $$

and

$$ (Q,S)\in R. $$

Then

$$ OP=OQ $$

and

$$ OQ=OS. $$

Therefore,

$$ OP=OS. $$

Hence,

$$ (P,S)\in R. $$

Thus, the relation is transitive.

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

Set of all Points Related to a Point

Let

$$ P\neq(0,0) $$

and let

$$ OP=r, $$

where

$$ r>0. $$

A point

$$ Q $$

is related to

$$ P $$

if and only if

$$ OQ=OP=r. $$

Thus, every point related to

$$ P $$

is at a distance

$$ r $$

from the origin.

But the set of all points at a fixed distance

$$ r $$

from the origin is a circle having the origin as its centre and radius

$$ r. $$

Since

$$ OP=r, $$

the point

$$ P $$

also lies on this circle.

Hence, the set of all points related to

$$ P $$

is the circle passing through

$$ P $$

with the origin as its centre.

Answer:

The relation is reflexive, symmetric and transitive.
Therefore, it is an equivalence relation.
The equivalence class of a point

$$ P\neq(0,0) $$

$$ \boxed{\{Q:OQ=OP\}}, $$

which is the circle centred at the origin and passing through

$$ \boxed{P.} $$

Question 12

Show that the relation

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

defined on the set of all triangles is an equivalence relation. Consider three right-angled triangles

$$ T_1=(3,4,5),\qquad T_2=(5,12,13),\qquad T_3=(6,8,10). $$

Which triangles among

$$ T_1,\ T_2\ \text{and}\ T_3 $$

are related?

Proof

We verify the three properties of an equivalence relation.

Reflexive

Every triangle is similar to itself because all its corresponding angles are equal and the corresponding sides are in the ratio

$$ 1:1. $$

Hence,

$$ (T,T)\in R $$

for every triangle

$$ T. $$

Therefore, the relation is reflexive.

Symmetric

Suppose

$$ (T_1,T_2)\in R. $$

Then

$$ T_1 $$

is similar to

$$ T_2. $$

Similarity of triangles is a symmetric property.

Therefore,

$$ T_2 $$

is also similar to

$$ T_1. $$

Hence,

$$ (T_2,T_1)\in R. $$

Therefore, the relation is symmetric.

Transitive

Suppose

$$ (T_1,T_2)\in R $$

and

$$ (T_2,T_3)\in R. $$

Then

$$ T_1 $$

is similar to

$$ T_2, $$

and

$$ T_2 $$

is similar to

$$ T_3. $$

Therefore,

$$ T_1 $$

is similar to

$$ T_3. $$

Hence,

$$ (T_1,T_3)\in R. $$

Therefore, the relation is transitive.

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

Triangles Related Among

$$ T_1,\ T_2\ \text{and}\ T_3 $$

Now compare the ratios of the corresponding sides.

Triangles

$$ T_1=(3,4,5) $$

and

$$ T_3=(6,8,10) $$

have

$$ \frac{6}{3}=2,\qquad \frac{8}{4}=2,\qquad \frac{10}{5}=2. $$

Since all corresponding sides are proportional,

$$ T_1\sim T_3. $$

Hence,

$$ (T_1,T_3)\in R $$

and

$$ (T_3,T_1)\in R. $$

Now compare

$$ T_1=(3,4,5) $$

and

$$ T_2=(5,12,13). $$

The corresponding side ratios are

$$ \frac{5}{3},\qquad \frac{12}{4}=3,\qquad \frac{13}{5}, $$

which are not equal.

Therefore,

$$ T_1 $$

is not similar to

$$ T_2. $$

Now compare

$$ T_2=(5,12,13) $$

and

$$ T_3=(6,8,10). $$

The corresponding side ratios are

$$ \frac{6}{5},\qquad \frac{8}{12},\qquad \frac{10}{13}, $$

which are not equal.

Therefore,

$$ T_2 $$

is not similar to

$$ T_3. $$

Answer:

The relation is reflexive, symmetric and transitive. Hence, it is an equivalence relation.

Among the given triangles, only

$$ \boxed{T_1\text{ and }T_3} $$

are related because

$$ T_1\sim T_3. $$

The triangle

$$ T_2 $$

is not related to either

$$ T_1 $$

$$ T_3. $$

Question 13

Show that the relation

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

defined on the set of all polygons is an equivalence relation. What is the set of all elements in

$$ A $$

related to the right-angled triangle

$$ T $$

with sides

$$ 3,\ 4,\ 5? $$

Proof

We verify the three properties of an equivalence relation.

Reflexive

Every polygon has the same number of sides as itself.

Therefore, for every polygon

$$ P, $$ $$ (P,P)\in R. $$

Hence, the relation is reflexive.

Symmetric

Suppose

$$ (P_1,P_2)\in R. $$

Then

$$ P_1 $$

and

$$ P_2 $$

have the same number of sides.

Therefore,

$$ P_2 $$

and

$$ P_1 $$

also have the same number of sides.

Hence,

$$ (P_2,P_1)\in R. $$

Therefore, the relation is symmetric.

Transitive

Suppose

$$ (P_1,P_2)\in R $$

and

$$ (P_2,P_3)\in R. $$

Then

$$ P_1 $$

and

$$ P_2 $$

have the same number of sides, and

$$ P_2 $$

and

$$ P_3 $$

also have the same number of sides.

Therefore,

$$ P_1 $$

and

$$ P_3 $$

have the same number of sides.

Hence,

$$ (P_1,P_3)\in R. $$

Therefore, the relation is transitive.

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

Set of all Elements Related to the Triangle

The given polygon

$$ T $$

is a triangle.

A polygon is related to

$$ T $$

if and only if it has the same number of sides as

$$ T. $$

Since every triangle has

$$ 3 $$

sides, all triangles are related to

$$ T. $$

Thus, the equivalence class of

$$ T $$

is the set of all triangles.

Answer:

The relation is reflexive, symmetric and transitive.
Therefore, it is an equivalence relation.
The set of all elements related to the triangle

$$ \boxed{T} $$

$$ \boxed{\text{the set of all triangles}.} $$

Question 14

Let

$$ L $$

be the set of all lines in the

$$ XY $$

plane and let the relation

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

be defined on

$$ L. $$

Show that

$$ R $$

is an equivalence relation. Find the set of all lines related to the line

$$ y=2x+4. $$

Proof

We verify the three properties of an equivalence relation.

Reflexive

Every line is parallel to itself.

Hence, for every line

$$ L, $$ $$ (L,L)\in R. $$

Therefore, the relation is reflexive.

Symmetric

Suppose

$$ (L_1,L_2)\in R. $$

Then

$$ L_1 $$

is parallel to

$$ L_2. $$

Since the property of parallelism is symmetric,

$$ L_2 $$

is also parallel to

$$ L_1. $$

Hence,

$$ (L_2,L_1)\in R. $$

Therefore, the relation is symmetric.

Transitive

Suppose

$$ (L_1,L_2)\in R $$

and

$$ (L_2,L_3)\in R. $$

Then

$$ L_1 $$

is parallel to

$$ L_2, $$

and

$$ L_2 $$

is parallel to

$$ L_3. $$

Therefore,

$$ L_1 $$

is parallel to

$$ L_3. $$

Hence,

$$ (L_1,L_3)\in R. $$

Therefore, the relation is transitive.

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

Set of all Lines Related to

$$ y=2x+4 $$

The given line has slope

$$ m=2. $$

Any line parallel to it must have the same slope.

The general equation of every line parallel to

$$ y=2x+4 $$

$$ y=2x+c, $$

where

$$ c $$

is any real number.

Each of these lines has slope

$$ 2, $$

so each is parallel to

$$ y=2x+4. $$

Conversely, every line related to

$$ y=2x+4 $$

must have slope

$$ 2, $$

and therefore can be written in the form

$$ y=2x+c. $$

Answer:

The relation is reflexive, symmetric and transitive.
Therefore, it is an equivalence relation.
The set of all lines related to

$$ y=2x+4 $$

$$ \boxed{\{\,y=2x+c\;:\;c\in\mathbb{R}\,\}.} $$

That is, the equivalence class of

$$ y=2x+4 $$

consists of all lines having slope

$$ 2. $$

Question 15

Let the relation

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

be defined on the set

$$ A=\{1,2,3,4\}. $$

Choose the correct answer.

Solution

Step 1: Check Reflexivity

For a relation to be reflexive, the pairs

$$ (1,1),\ (2,2),\ (3,3)\ \text{and}\ (4,4) $$

must belong to the relation.

Since all these pairs are present in

$$ R, $$

the relation is reflexive.

Step 2: Check Symmetry

If the relation were symmetric, then whenever

$$ (a,b)\in R, $$

the pair

$$ (b,a) $$

should also belong to

$$ R. $$

Here,

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

but

$$ (2,1)\notin R. $$

Therefore, the relation is not symmetric.

Step 3: Check Transitivity

We verify the transitive property.

Since

$$ (1,3)\in R \quad\text{and}\quad (3,2)\in R, $$

transitivity requires

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

which is present.

Also,

$$ (1,1),(2,2),(3,3)\ \text{and}\ (4,4) $$

ensure that all remaining possible combinations satisfy the transitive property.

Hence, the relation is transitive.

Answer:

The relation is reflexive and transitive but not symmetric.

$$ \boxed{\text{Option (B)}} $$

CHAPTER-1(RELATIONS AND FUNCTIONS)

Exercise 1.1

Question 1 (i)

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 1 (ii)

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 1 (iii)

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 1 (iv)

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 1 (v)

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 2

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 3

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 4

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 5

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 6

1. Reflexive

2. Symmetric

3. Transitive

Final Answer

Question 7

1. Reflexive

2. Symmetric

3. Transitive

Final Conclusion

Question 8

1. Reflexive

2. Symmetric

3. Transitive

Therefore, R is an Equivalence Relation.

Elements of the set

Elements of the set

Relation between

Final Answer

Question 9 (i)

Proof

Reflexive

Symmetric

Transitive

Elements Related to 1

Question 9 (ii)

Proof

Reflexive

Symmetric

Transitive

Elements Related to 1

Question 10 (i)

Solution

CHAPTER-1
(RELATIONS AND FUNCTIONS)