## NCERT Solutions for Class 12th Maths Chapter 1 – Relations and Functions

**Question 1:**

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

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

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

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

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

(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 live in the same locality}

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

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

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

**Answer**

(i) A = {1, 2,
3 ... 13, 14}

R = {(x, y): 3x
- y = 0}∴R
= {(1, 3), (2, 6), (3, 9), (4, 12)}

R is not reflexive since (1, 1), (2, 2) ...
(14, 14) ∉ R.

Also, R is not symmetric as (1, 3) ER, but (3,
1) ∉
R. [3(3) - 1. = 0]

Also, R is not
transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉
R. [3(1) - 9 2 0]

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

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

It is seen that (1, 1) ∉ R.

∴
R is not reflexive.

(1, 6) R

But,

(1, 6) ∉
R.

∴
R is not symmetric.

Now, since
there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong
to R.

R is not
transitive.

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

(iii) A = {1, 2, 3, 4, 5, 6}

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

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

(x,x)∈R

∴R
is reflexive.

Now,

(2, 4) ∈R [as 4 is divisible by 2]

But,

(4, 2) ∉
R. [as 2 is not divisible by 4]

∴R is not symmetric.

Let (x, y), (y,
z) ∈ R. Then, y is divisible by x and z is
divisible by y.

∴z
is divisible by x.

(x, z) ∈R

∴R
is transitive.

Hence, R is
reflexive and transitive but not symmetric.

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

Now, for every
x ∈Z,
(x, x) ER as x - x = 0 is an integer.

∴R is reflexive.

Now, for every
x, y ∈
Z if (x, y) ∈ R, then x - y is an integer.

-(x - y) is
also an integer.

(y - x) is an
integer.

∴
(y, x) ∈ R

∴R
is symmetric.

Now,

Let (x, y) and
(y, z) R, where x, y, z ∈ Z.

(x - y) and (y - z) are integers.

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

∴
(x, z) ∈R

∴R
is transitive.

Hence, R is reflexive, symmetric, and
transitive.

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

(X, X) ∈
R

∴
R is reflexive.

If (x, y) ∈ R, then x and y work at the same
place.

y and x work at
the same place.

(y, x) ∈R.

∴ R is symmetric.

Now, let (x, y), (y, z) ∈R

x and y work at the same place and y and z
work at the same place.

x and z work at
the same place.

(x, z) ∈R

∴
R is transitive.

Hence, R is
reflexive, symmetric, and transitive.

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

Clearly (x, x)
∈R as x and x is the same human being.

∴ R is reflexive.If (x, y) ∈R, then x and y live in the same locality.

y and x live in the same locality.

(y, x) ∈ R

∴R is symmetric.

Now, let (x, y)
∈
R and (y, z) ∈R.

x and y live in the same locality and y and z
live in the same locality.

x and z live in the same locality.

(x, z) ∈
R

∴
R is transitive.

Hence, R is reflexive, symmetric, and
transitive.

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

Now,

(x, x) ∉
R

Since human
being x cannot be taller than himself.

∴R is not reflexive.

Now, let (x, y) ∈R.

x is exactly 7 cm taller than y.

Then, y is not taller than x.

∴ (y, x) ∉R

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

∴R is not symmetric.

Now, Let (x,
y), (y, z) ∈ R.

x is exactly 7
cm taller than y and y is exactly 7 cm taller than z.

x is exactly 14
cm taller than z .

∴ (x, z) ∉R

∴
R is not transitive.

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

(d) R = {(x,
y): x is the wife of y}

Now,

(x, x) ∉ R

Since x cannot
be the wife of herself.

∴R is not reflexive.

Now, let (x, y)
∈
R

x is the wife
of y.

Clearly y is
not the wife of x.

∴ (y, x) ∉ R

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

∴
R is not transitive.

Let (x, y), (y,
z) ∈
R

x is the wife
of y and y is the wife of z.

This case is not possible. Also, this does not
imply that x is the wife of z.

∴
(x, z) ∉ R

∴R is not transitive.

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

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

(x, x) ∉
R As x cannot be the father of himself.

∴R
is not reflexive.

Now, let (x, y)
∈
R.

x is the father of y.

y cannot be the
father of y.

Indeed, y is
the son or the daughter of y.

∴ (y, x) ∉ R

∴
R is not symmetric.

Now, let (x, y) E R and (y, z) ∈
R.

x is the father of y and y is the father of z.

x is not the father of z.

Indeed x is the grandfather of z.

∴
(x, z) ∉ R

∴R is not transitive.

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

**Question 2:**

Show that the
relation R in the set R of real numbers, defined as

R = {(a, b): a b2} is neither reflexive nor
symmetric nor transitive.

Answer

R = {(a, b): a=b

^{2}}
It can be
observed that ( ½ , ½ ) e R, since ½ > ½

^{2}= ¼ .
∴R
is not reflexive.

Now, (1, 4) E R
as 1 < 4

^{2}
But, 4 is not
less than 1

^{2}.
∴
(4, 1) ∉ R

∴R
is not symmetric.

Now,(3, 2), (2,
1.5) ∈ R

(as 3 < 22 =
4 and 2 < (1.5)2 = 2.25)

But, 3 > (1.5)2 = 2.25

∴ (3, 1.5) ∉R

∴
R is not transitive.

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

**Question 3:**

Check whether
the relation R defined in the set {1, 2, 3, 4, 5, 6} as

R = {(a, b): b = a + 1} is reflexive,
symmetric or transitive.

Answer

Let A = {1, 2,
3, 4, 5, 6}.

A relation R is
defined on set A as:

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

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

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

We can find (a, a) e R, where a ∈ A.

For instance,

(1, 1), (2, 2),
(3, 3), (4, 4), (5, 5), (6, 6) ∉ R

∴R
is not reflexive.

It can be
observed that (1, 2) E R, but (2, 1) ∉ R.

∴R is not symmetric.

Now, (1, 2),
(2, 3) ∈R

But,

(1, 3) ∉ R

∴R is not transitive

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

**Question 4:**

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

R = {(a, b); a
b}

Clearly (a, a) ∈ R as a = a.

∴R is reflexive.

Now,

(2, 4) E R (as
2 < 4)

But, (4, 2) ∉ R as 4 is greater than 2.

∴R
is not symmetric.

Now,

let (a, b), (b,
c) ∈ R.

Then,

a∈ b and b∈ c

a ∈c

(a, c) ∈ R

∴R
is transitive.∈

Hence,R is reflexive and transitive but not
symmetric.

**Question 5:**

Check whether the relation R in R defined as R = {(a, b): a
< b3} is reflexive, symmetric or transitive. Answer

R = {(a, b): a b

^{3}}
It is observed that ( ½ , ½ ) ∉ R as 3< ½ > ( ½ )

^{3}= 1/8
∴R
is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2

^{3}= 8)
But,

(2, 1) e R (as 23 > 1)

R is not symmetric.

We have (3, 3/2 ), (3/2 , 6/5) ∈ R as 3 < (3/2)

^{3}and 3/2 < (6/5)^{3}.
But (3, 6/5) ∉ R as 3 > (6/5)

^{3}^{}
∴R
is not transitive.

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

**Question 6:**

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

Answer

Let A = {1, 2, 3}.

A relation R on A is defined as R = {(1, 2), (2, 1)}.

It is seen that (1, 1), (2, 2), (3, 3) OR. R is not
reflexive.

Now,

as (1, 2) ∈
R and (2, 1) ∈ R, then R is symmetric.

Now, (1, 2) and (2, 1) ∈ R However, (1, 1) ∉
R

∴R
is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

**Question 7:**

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 and y have same number of pages}
is an equivalence relation.

Answer

Set A is the set of all books in the library of a college.

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

Now, R is reflexive since (x, x) ∈ R as x and
x has the same number of pages.

Let (x, y) ∈
R = x and y have the same number of pages.

y and x have the same number of pages.

(y, x) ∈
R

∴is
symmetric.

Now,

let (x, y) ∈
R and (y, z) ∈ R.

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

x and z have the same
number of pages.

(x, z) ∈
R

∴ is transitive.

Hence, R is an equivalence relation.

**Question 8:**

Show that the relation R in the set A = {1, 2, 3, 4, 5}
given by R = {(a, b):|a -6| 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}.

Answer

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

R = {(a, b):l a – b l is even}

It is clear that for any element a ∈ A, we have
|a-a l= 0(which is even).

∴ is reflexive. Let (a, b) ∈
R.

|a-b| is even.

|-(a - b)= | b – a | is also even.

(b, a) ∈ R

∴R
is symmetric.

Now,

let (a, b) ∈
R and (b, c) ∈ R.

|a-b|is even and |b-c
| is even.

(a - b) is even and (b - c) is even.

(a -c)= (a - b)+(b -c) is even. [Sum of two even integers is
even]

|a – c | is even.

(a, c) ∈ R

∴R is transitive.

Hence, R is an
equivalence relation. Now, all elements of the set {1, 2, 3} are related to
each other as all the elements of this subset are odd. Thus, the modulus of the
difference between any two elements will be even. Similarly, all elements of
the set {2, 4} are related to each other as all the elements of this subset are
even. Also, no element of the subset {1, 3, 5} can be related to any element of
{2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are
even. Thus, the modulus of the difference between the two elements (from each
of these two subsets) will not be even.