NCERT Solutions for Class 12th: Ch 1 Relations and Functions Exercise 1.1

NCERT Solutions for Class 12th: Ch 1 Relations and Functions Exercise 1.1 Maths

Page No: 5

Exercise 1.1

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 an 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) : 3 x – y ≠} = {(x, y) : y = 3 x} = {(1,3), (2, 6), (3, 9), (4, 12)}

(a) R is not reflexive as (x, x) ∉ R    [ ∵ 3 x – x ≠ 0]

(b) R is not symmetric as (x,y) ∈ R does not imply (y, x) ∈ R
[∴ (1, 3) ∈ R does not imply (3. 1) ∈ R]

(c) R is not transitive as (1.3) ∈ R , (3, 9) ∈ R but (1.9) ∉ R

(ii) Relation R in the set N is given by
R = {(x, y) : y = x + 5 and x < 4}
∴ R = {(1,6), (2, 7). (3, 8)}

(a) R is not reflexive as (x, x) ∉ R

(b) R is not symmetric as (x, y) ∈ R ⇏ (v, x) ∈ R

(c) R is not transitive as (x,y) ∈ R, (y, z) ∈ R ⇏ (x, z) ∈ R

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

(a) R is reflexive as (x, x) ∈ R ∀ x ∈ A    [∴ x divides x ∀ x ∈ A]

(b) R is not symmetric as (1, 6) ∈ R but (6, 1) ∉ R.

(c) Let (x, y), (y, z) ∈ A
∴ y is divisible by x and z is divisible by y
∴ z is divisible by x
∴ (r, y) ∈ R (y, z) ∈ R ⇒ (x, z) ∈ R
∴ R is transitive

(iv) Relation R is in the set Z given by R = {(x,y) : x – y is an integer}

(a) R is reflexive as ( x, x) ∈ R [∴ x – x = 0 is an integer]

(b) R is symmetric as (x,y) ∈ R ⇒ (y, x) ∈ A
[∵ x – y is an integer ⇒ y – x is an integer]

(c) R is transitive as (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R
[∵ if x – y, y – z are integers, then (x – y) + (y – z) = x – z is also in integer]

(v)A is the set of human beings in a town at a particular time R is relation in A.

(a) R = {(x, y) : x and y work at the same time}
R is reflexive as (x, x) ∈ R

R is symmetric as ( x, y) ∈ R ∈ (y, x) ∈ R
[∵ x and y work at the same time ⇒ y and x work at the same time]

R is transitive as (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R
[∴ if x and y, y and z work at the same time, then x and z also work at the same time]

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

R is symmetric as ( x, y) ∈ R ⇒ (y, x) ∈ R
[∴ x and y live in the same locality ⇒ y and x live in the same locality]

R is transitive as ( x, y), ( y, z) ∈ R ⇒ (x, z) ∈ R
[∵ if x and y, y and z live in the same locality. then x and z also live in the same locality]

(c) R = {(x,y) : x is exactly 7 cm taller than y}
Since (x, x) ∉ R as x cannot be 7 cm taller than x.
∴ R is not reflexive.

(x, y) ∈ R ⇒ (y.x) ∈ R as if x is taller than y, then y cannot be taller than x.
∴ R is not symmetric.

Again, (x,y), (y,z) ∈ R ⇏ (x, z) ∈ R
[∵ if x is taller than y by 7 cm and y is taller than z by 7 cm, then x is taller than z by 14 cm]
∴ R is not transitive.

(d) R = {(x,y) : x is wife of y}
R is not reflexive as (x,y) ∉ R    [∴ x cannot be wife of x]

Also, (x, y) ∈ R ⇏ (y, x) ∈ R
[∵ if x is wife of y, then y cannot be wife of x]
∴ R is not symmetric.

 Also, this does not imply that x is the wife of z.
∴ (x, z) ∉ R
∴ R is not transitive.

(e) R = {(x,y) : x is father of y}
R is not reflexive as (x, x) ∉ R    [ ∵ x cannot be father of x]

Also, (x,y) ∈ R ⇐ (y, x) ∈ R [ ∵ if x is father of y. then y cannot be father of x]
∴ R is not symmetric.

Let (x, y) ∈ 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.
∴ x is the grandfather of z.
∴ (x, z) ∉ R
∴R is not transitive.

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 ≤  b²}

(i) Since (a, a) ∉ R
[Take a = 1/3 then 1/3 > (1/3)²]
∴ R is not reflexive.

(ii) Also, (a, b) ∈ R ⇏ (b, a) ∈ R
[Take a = 2 ,b = 6, then 2 ≤ 62 but (6)2 < 2 is not true]
∴ R is not symmetric.

(iii) Now (a, b), (b, c) ∈ R ∉ (a, c) ∈ R
[Take a = 1, b = -2, c = -3 ∴ a ≤  b², b ≤ c² but a ≤ c² is not true]
∴ R is not transitive.

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}
R = {(a, b) : b = a + 1} = {(a, a + 1)}
= {(1, 2), (2, 3), (3, 4), (4,5), (5,6)}

(i) R is not reflexive as (a, a) ∉ R ∀ a ∈ A

(ii) (a,b) ∈ R ⇏ (b,a) ∈ R
 [∵ (a, b) ∈ R ⇒ b = a + 1 ⇒ a = b –1]
∴ R is not symmetric.

(iii) (a, b) ∈ R, (b, c) ∈ R ⇏ (a, c) ∈ R
[∵ (a, b), (b, c) ∈ R ⇒ b = a + 1, c = b + 1 ⇒ c = a + 2]
∴ R is not transitive.

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}

(i) Since (a, a) ∈ R ∀ a ∈ R
[∵ a ≤ a ∀ a ∈ R]
∴ R is reflexive.

(ii) (a, b) ∈ R ⇏ (b, a) ∈ R
[∵ if a ≤ b. then b ≤ a is not true]
∴ R is not symmetric.

(iii) Let (a, b), (b, c) ∈ R
∴ a ≤ b, b ≤ c
∴ a ≤ c ⇒ (a, c) ∈ R
∴ (a, b), (b. c) ∈ R ⇒ (a, c) ∈ R
∴ R is transitive.

5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b³ } is reflexive, symmetric or transitive.

Answer

R = {(a, b) : a ≤ b³}

(i) Since (a, a) ∉ R as a ≤ a³ is not always true.
[Take a = 1/3. then a ≤ a³ is not true]
∴ R is not reflexive.

(ii) Also (a, b) ∈ R ⇏ (b, a) ∈ R
[Take a = 1, b = 4, ∴ 1 ≤ 4³ but 4 ≰ (l)³]
∴ R is not symmetric.

(iii) Now (a, b) ∈ R, (b, c) ∈ R ⇏ (a, c) ∈ R
[Take a = 100, b = 5, c = 3, ∴ 100 ≤ 5³, 5 ≤ 3³ but 100 ≥ 3³] R is not symmetric.

Page No: 6

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

A = {1,2, 3}
R = {(1.2), (2, 1)}

Since, (a, a) ∈ R ∀ a ∈ A R is not reflexive.

Now (1, 2) ∈ R ⇒ (2, 1) ∈ R and (2, 1) ∈ R ⇒ (1,2) ∈ R
∴ (a, b) ∈ R ⇒ (b, a) ∈ R ∀ (a, b) ∈ R
∴ R is symmetric

Again, (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R
∴ R is not transitive.

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

A is the set of all books in a library of a college.
R = {(x,y) : x and y have same number of pages}

Since, (x, x) ∈ R as x and x have the same number of pages ∀ x ∈ A.
∴ R is reflexive.

Also (x, y) ∈ R
⇒ x and y have the same number of pages
⇒ y and x have the same number of pages
⇒ (y, x) ∈ R
∴ R is symmetric.

Now, (x, y) ∈ R and (y, z) ∈ R.
⇒ x and y have the same number of pages and y and z have the same number of pages
⇒ x and z have he same number of pages
⇒ (x, z) ∈ R
∴ R is transitive.

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

Answer

A = {1, 2, 3, 4, 5}
R = {(a, b) : | a – b | is even}
Now | a – a | = 0 is an even number,
∴ (a, a) ∈ R ∀ a ∈ A
⇒ R is reflexive.

Again (a, b) ∈ R
⇒ | a – b | is even
⇒ | – (b – a) | is even
⇒ | b – a | is even
⇒ (b, a) ∈ R
∴ R is symmetric.

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 – b) + (b – c) is even
⇒ a – c is even ⇒ | a – c | is even
⇒ (a, c) ∈ R
∴ R is transitive.
∴ R is an equivalence relation.

∵ | 1 – 3 | = 2, | 3 – 5 | = 2 and | 1 – 5 | = 4 are even,
All the elements of {1, 3, 5} are related to each other.
∵ | 2 – 4 | = 2 is even,
All the elements of {2, 4} are related to each other.

Now | 1 – 2 | = 1, | 1 – 4 | = 3, | 3 – 2 | = 1, | 3 – 4 | = 1, | 5 – 2 | = 3 and | 5 – 4 | = 1 are all odd
No element of the set {1, 3, 5} is related to any element of (2, 4}.

9. Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
(i) R = {(a, b) : |a – b| is a multiple of 4}
(ii) R = {(a, b) : a = b} is an equivalence relation.
Find the set of all elements related to 1 in each case.

Answer

A= {x ∈ Z : 0 ≤ x ≤ 12}
= {0, 1, 2, 3, 4, 5, 6, 7, 8,9, 10, 11, 12 }

(i) R = {(a, b) : | a – b | is a multiple of 4|
As | a – a | = 0 is divisible by 4
∴ (a, a) ∈ R ∀ a ∈ A.
∴ R is reflexive.

Next, let (a, b) ∈ R
⇒ | a – b | is divisible by 4
⇒ | – (b – a) | is divisible by 4
⇒ | b – a | is divisible by 4 ⇒ (b, a) ∈ R
∴ R is symmetric.

Again. (a, b) ∈ R and (b, c) ∈ R
⇒ |a – A| is a multiple of 4 and |b – c| is a multiple of 4
⇒ a – b is a multiple of 4 and b – c is a multiple of 4
⇒ (a – b) + (b – c) is a multiple of 4
⇒ a – c is a multiple of 4
⇒ |a – c| is a multiple of 4
⇒ (a, c) ∈ R
∴ R is transitive.
∴ R is an equivalence relation.
Set of elements which are related to | = {a ∈ A : (a, 1) ∈ R}
= {a ∈ A : |a – 1| is a multiple of 4]
= {1, 5, 9}
[ ∵ | 1 – 1 | = 0, | 5 – 1 | = 4 and I 9 – 1 | = 8 are multiples of 4] (ii) R = {(a, b) : a = b} ∴ a = a ∀ a ∈ A,

∴ R is reflexive.
Again, (a, b) ∈ R ⇒ a = b ⇒ b = a ⇒ (b,a) ∈ R ∴ R is symmetric.
Next. (a, b) ∈ R and (b, c) ∈ R
⇒    a = b and b = c
⇒    a = c ⇒ (a, c) ∈ R
∴ R is transitive.
∴ R is an equivalence relation.
Set of elements of A which are related to I = {a ∈ A : (a,1) ∈ R }
= {a ∈ A : a = 1} = {1}

10. 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.

Answer

(i) Let A = {1, 2}.
Then A x A = {(1,1), (1,2), (2,1), (2,2) }.
Let R = {(1,2), (2,1 )} .
Then R ⊆ A x A and hence R is a relation on the set A.
R is symmetric since (a, b) ∈ R ⇒ (b. a) ∈ R.
R is not reflexive since I ∈ A but (1,1) ∉ R.
R is not transitive since (1, 2) ∈ R, (2,1) ∈ R but (1,1) ∉ R.

(ii) Let A = {1,2,3}
Then A x A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}.
Let R = {(1,1), (2,2), (1,2), (2,1), (1,3), (2,3)}.
Then R is transitive since (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R.
R is not reflexive since 3 G A but (3.3) ∉ R.
R is not symmetric since (1,3) ∈R but (3,1) ∉ R.

(iii) Let A = {1,2 3}
Then A x A = {(1, 1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3) }.
Let R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}.
R is a relation on A as R ⊆ A x A.
R is reflexive as (a, a) ∈ R ∀ a ∈ A.
Also. R is symmetric since (a. b) ∈ R implies that (b, a) ∈R.
But R is not transitive since (1,2) ∈R arid (2,3) ∈R but (1,3) ∢ R.

(iv) Let A = {1,2,3}.
Then A x A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}.
Let R = {(1,1), (2,2), (3,3), (1,3)}.
Then R is a relation on A as R ⊆ A x A.
R is reflexive since (a, a) ∈R ∀ a ∈ A.
R is not symmetric as (1,3) ∈R and (3,1) ∉ R. R is transitive since (a, b) ∈R and (b, c) ∈R implies that (a, c) ∈R.

(v) Let A = {1,2,3}
Then A x A = {(1,1), (1,2), (1, 3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
Let R = {(1,1), (1,2), (2,1), (2,2)}.
R is not reflexive as 3∈ A and (3,3) ∉ R.
R is symmetric as (a, b) ∈ R ⇒ (b, a) ∈R.
R is transitive since (a, b) ∈ R and (b, c) ∈R implies that (a, c) ∈ R.

11. Show that the relation R in the set A of points in a plane given by R = {(P, 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 P ≠ (0, 0) is the circle passing through P with origin as centre.

Answer

A is the set of points in a plane.
R = {(P. Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}
= {(P, Q) : | OP | = | OQ | where O is origin}
Since, | OP | = | OP |, (P, P) ∈ R ∀ P ∈ A.
∴ R is reflexive.

Also (P. Q) ∈ R
⇒ | OP | = | OQ |
⇒ | OQ | = | OP |
⇒ (Q.P) ∈ R ⇒ R is symmetric.

Next let (P, Q) ∈ R and (Q, T) ∈ R ⇒     | OP | = | OQ | and | OQ | = | OT |
⇒ | OP | = | OT |
⇒ (P,T) ∈ R
∴ R is transitive

∴ R is an equivalence relation
Set of points related to P ≠ O
= {Q ∈ A : (Q,P) ∈ R} = {Q ∈ A : | OQ | = | OP |}
= {Q ∈ A :Q lies on a circle through P with centre O}