Mathematics Part I Chapter 1 Relations And Functions

The given information is

The information is represented in the form of a 3 X 2 matrix as follows :

The entry in the third row and second column represents the number of women workers in factory III.
          

We know that a matrix of order m x n has mn elements. Therefore, for finding all possible orders of a matrix with 24 elements, w e will find all ordered pairs with products of elements as 24.
∴ all possible ordered pairs are
(1, 24), (24. 1), (2, 12), (12. 2). (3, 8), (8, 3), (4. 6), (6, 4)
∴ possible orders are
1 x 24. 24 x 1, 2 x 12, 12 x 2, 3 x 8, 8 x 3, 4 x 6. 6 x 4
If number of elements  =13, then possible orders are 1 x 13, 13 x l.

We know that a matrix of order m x n has mnelements. Therefore, for finding all possible orders of a matrix with 18 elements, we will find all ordered pairs with products of elements as 18
∴ all possible ordered pairs areall possible ordered pairs areall possible ordered pairs areall possible ordered pairs areall possible ordered pairs areall possible ordered pairs are
(I, 18). (18, 1). (2, 9). (9. 2). (3, 6). (0, 3)
∴ possible orders are 1 x 18. 18 x 1. 2 x 9,-9 X 2, 3 x 6. 6 x 3.
If number of elements = 5, then possible orders arc 1 x 5. 5 x I.

Number of elements = 12
∴possible orders of the matrix are
1 x 12, 12 x 1,2 X 6. 6 x 2,3 x 4,4 x 3
If numbers of elements = 7, then possible orders are 1 x 7. 7 x 1

Let A = [a i] be required 3 x3 matrix    
where a,ij , = 2 i – 3 j    
∴ a₁₁, = 2–3 = –1. a₁₂ = 2–6 = 4. a₁₃,= 2–9 = 7
a₂₁ = 4 — 3 = 1. a₂₂ = 4 — 6 = —2 . a₂₃  = 4 –9 = 5
a₃₁ = 6 – 3 = 3, = 6 – 6 = 0 , a₃₃ = 6 – 9 = –3

Since the school is boys school i.e. there is not girl student no student of the school can be sister of any student of the school.
∴ R ⇒ R is the empty relation.
Clearly the difference between heights of a and b is less than 3 metres.
∴ R’ = A x A is the universal relation.

Let A = {1. 2, 3}.
Let R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
Here R is reflexive as (a, a)∈ R ∀ a ∈ A.
R is not symmetric as (1, 2) ∈ R but (2, 1) ∈ R.
R is not transitive as (1.2) ∈ R and (2, 3) ∈ R but(1,3) ∈ R.
∴ R is a relation which is reflexive but neither symmetric not transitive.

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.

(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.

R= {(L₁, L₂) : L₁ is perpendicular to L₂} Since no line can be perpendicular to itself ∴ R is not reflexive.
Let (L₁, L₂) ∈ R
∴ L₁ is perpendicular to L₂ ⇒ L₂ is peipendicular to L₁
⇒ (L₂, L₁) ∈ R
∴ (L₁L₂) ∈ R ⇒ (L₂, L₁) ∈ R
∴ R is symmetric
Again we know that if L₁ is perpendicular to L₂ and L₂ is perpendicular to L₃, then L₁can never be perpendicular to L₃.
∴ (L₁, L₂) ∈ R, (L₂, L₃) ∈ R does not imply (L₁, L₃) ∈ R
∴ R is not transitive.

(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 is in the set N 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
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.
R is not transitive.

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.

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

R = {(a, b) : a ≤ b²}
(i) Since (a, a) ∉ R

∴ 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.

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.

Since R and R’ are relations on a set A.
∴ R ⊆ A x A and R’ ⊆ A x A.
⇒ R ∪ R’ ⊆ A x A and R ∩ R’ ⊆ A x A.
∴ R ∪ R’ and R ∩ R’ are also relations on the set A.
We now show that R ∪ R’ is reflesive relation on A.
Let a ∈ A.
∴ (a, a) ∈ R and (a. a) ∈ R’.    (∵ R and R’ are reflexive on A)
⇒ (a. a) ∈ R ∪ R’ and R ∩ R’ ∀ a ∈ A.
∴ R ∪ R’ and R ∩ R’ are reflexive relations on A.

Let a. b ∈ A such that (a, b) ∈ R ∩ R’.
∴ (a. b) ∈ R and (a. b) ∈ R’.
∴ (b, a) ∈ R and (b, a) ∈ R’.    (∵ R and R’ are symmetric)
⇒ (b.a) ∈ R ∩ R’.
we have proved that (a, b) ∈ R ∩ R’ ⇒ (b, a) ∈ R ∩ R’.
∴ R fl R’ is a symmetric relation.

Let R and R’ be two symmetric relations on a set A.
Let a, b ∈ A such that (a, b) ∈ R ∪ R’
∴ Either (a, b) ∈ R or (a, b) ∈ R’
If (a, b) ∴ R then (b, a) ∴ R    (∵ R is symmetric)
∴ (b, a) ∈ R ∪ R’ (since R ⊆ R⊆ R’)
Similarly we can prove that (a, b) ∈ R’ ∈ (b, a) ∈ R ∪ R’
In both the cases (b, a) ∈ R ∪ R’
∴ R ∪ R’ is a symmetric relation on A.

The smallest relation R₁ containing (1, 2) and (2, 3) which is reflexive and transitive but not symmetric is {(1, 1), (2. 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Now, if we add the pair (2. 1) to R₁ to get R₂, then the relation R₂ will be reflexive, transitive but not symmetric. Similarly, we can obtain R₃ and R₄ by adding (3, 2) and (3, 1) respectively, to R₁ to get the desired relations. However, we can not add any two pairs out of (2, 1), (3, 2) and (3, 1) to R₁ at a time, as by doing so, we will be forced to add the remaining third pair in order to maintain transitivity and in the process, the relation will become symmetric also which is not required. the total number of desired relations is four.

The characteristic of sets {1, 4, 7 }, {2, 5, 8} and {3, 6, 9} is that difference between any two elements of these sets is a multiple of 3.
∴ (x,y) ∈ R₁ ⇒ x – y is a multiple of 3
⇒ {x,y} ⊂ {1,4,7} or {x, y} ⊂ {2, 5, 8} or {x,y} ⊂ {3, 6, 9}
⇒ (x,y) ∈ R₂.
Hence R₁ ⊂ R₂.
Similarly, { x, y} ∈ R₂
⇒ {x, y}  ⊂ {1, 4, 7} or {x,y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} ⇒ x – y is divisible by 3 ⇒ {x ,y} ∈ R₁.
∴ R₂ ⊂ R₁.
Hence, R₁ = R₂.

R = {(a, b) : a = b – 2, b > 6}
∴ (a, b) ∈ R ⇒ a = b – 2 where b > 6 ∴ (6, 8) ∈ R as 6 = 8 – 2 where b = 8 > 6 ∴ (C) is correct answer.

A = {1, 2, 3}
R₁ = {(1,2), (1,3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)} is the only relation on {1, 2, 3} which is reflexive, symmetric but not transitive and is such that (1, 2), (1, 3) ∈ R₁.
∴ (A) is correct 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.

R = {(T₁, T₂) : T₁ is congruent to T₂}
(i) Since every triangle is congruent to itself
∴ R is reflexive.
(ii) Also (T₁, T₂) ∈ R ⇒ T₁ is congruent to T₂ ⇒ T₂ is congruent to T₁ (T₂,T₁) ∈ R
(T₁,T₂) ∈ R ⇒ (T₂,T₁) ∈ R ⇒ R is symmetric.
(iii) Again (T₁, T₂), (T₂, T₃) ∈ R ⇒ T₁ i is congruent to T₂ and T₂ is congruent to T₃ ∴ T₁ is congruent to T₃ ∴ (T₁,T₃) ∈ R (T₁,T₂), (T₂,T₃) ∈ R ⇒ (T₁,T₃) ∈ R ∴ R is transitive.
From (i), (ii), (iii), it is clear that R is reflexive, symmetric and transitive
∴ R is an equivalence relation.

R = {(T₁, T₂) : T₁ is similar to T₂}
Since every triangle is similar to itself
∴ R is reflexive.
Also (T₁ T₂) ∈ R ⇒ T₁ is similar to T₂ ⇒ T₂ is similar to T₁ ∴ (T₂,T₁) ⇒ R
∴ (T₁,T₂) ∈ R ⇒ (T₂,T₁) ∈ R ⇒ R is symmetric.
Again (T₁, T₂), (T₂, T₃) ∈ R
⇒ T₁ is similar to T₂ and T₂ is similar to T₃
∴ T₁ is similar to T₃ ⇒ (T₁,T₃) ∈ R ∴ (T₁, T₂), (T₂,T₃) ∈ R ⇒ (T₁, T₃) ∈ R ∴ R is transitive.
∴ R is reflexive, symmetric and transitive ∴ R is an equivalence relation.
Now T₁, T₂, T₃ are triangles with sides 3, 4, 5 ; 5, 12, 13 and 6, 8, 10.
Since         

∴ T₁ is similar to T₃ i.e. T₃ is similar to T₁.
No two other triangles are similar.

A is the set of all polygons
R = {(P₁, P₂) : P₁ and P₂ have same number of sides }
Since P and P have the same number of sides
∴ (P.P) ∈ R ∀ P ∈ A.
∴ R is reflexive.
Let (P₁, P₂) ∴ R
⇒ P₁ and P₂ have the same number of sides ⇒ P₂ and P₁ have the same number of sides ⇒ (P₂, P₁) ∈ R
∴ (P₁, P₂) ∈ R ⇒ (P₂, P₁) ∈ R ∴ R is symmetric.
Let (P₁, P₂) ∈ R and (P₂, P₃) ∈ R.
⇒ P₁ and P₂ have the same number of sides and P₂ and P₃ have same number of sides
⇒ P₁ and P₃ have the same number of sides
⇒ (P₁, P₃) ∈ R
∴ (P₁, P₂), (P₂, P₃) ∈ R ∈ (P₁, P₃) ∈ R ∴ R is transitive.
∴ R is an equivalence relation.
Now T is a triangle.
Let P be any element of A.
Now P ∈ A is related to T iff P and T have the same number of sides P is a triangle
required set is the set of all triangles in A.

L is the set of all lines in XY plane.
R = {(L₁, L₂) : L₁ is parallel to L₂}
Since every line l ∈ L is parallel to itself,
∴ (l,l) ∴ R ∀ l ∈ L
∴ R is reflexive.
Let (L₁, L₂) ∈ R ∴ L₁ || L₂ ⇒ L₂ || L₁
⇒ (L₂, L₁) ∈ R.
∴ R is symmetric.
Next, let (L₁ L₂) ∈ R and (L₂, L₃) ∈ R ∴ L₁ || L₂ and L₂ || L₃
∴ L₁ || L₃    (L₁ , L₃) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
Let P be the set of all lines related to the line y = 2 x + 4.
∴ P = {l : l is a line related to the line y = 2 x + 4}
= {l : l is a line parallel to the line y = 2 x + 4}
= { l : l is a line with equation y = 2 x + c, where c is an arbitrary constant }

R = {(a, b) : 2 divides a – b}
where R is in the set Z of integers.
(i)    a – a = 0 = 2 .0
∴ 2 divides a – a ⇒ (a, a) ∈ R ⇒ R is reflexive.
(ii) Let (a, a) ∈ R ∴ 2 divides a – b ⇒ a – b = 2 n for some n ∈ Z ⇒ b – a = 2 (–n)
⇒ 2 divides b – a ⇒ (b. a) ∈ R
(a, ft) G R ⇒ (b, a) ∈ R ∴ R is symmetric.
(iii) Let (a, b) and (b, c) ∈ R
2 divides a – b and b – c both ∴ a – b = 2 n₁ and b – c = 2 n₂ for some n₁, n₂ ∈ Z ∴ (a – b) + (b – c)= 2 n₁ + 2 n₂ ⇒ a – c = 2 (n₁ + n₂ )
⇒ 2 divides a – c
⇒ (a, c) ∈ R
∴ (a,b), (b,c) ∈ R ⇒ (a, c) ∈ R
∴ R is transitive
From (i), (ii), (iii) it follows that R is an equivalence relation.

The relation R ⊆ N x N is defined by by (a. b)∈ R if and only if 5 divides b – a.
This means that R is a relation on N defined by , if a. b ∈ N then (a, b) ∈ R if and only if 5 divides b – a.
Let a, b, c belongs to N. Then (i)    a – a = 0 = 5 . 0.
5 divides a – a.
⇒ (a. a) ∈ R .
⇒ R is reflexive.
(ii) Let (a, b) ∈ R.
∴ divides a – b.
⇒ a – b = 5 n for some n ∈ N.
⇒ b – a = 5 (–n).
⇒ 5 divides b – a ⇒ (b, a) ∈ R.
∴ R is symmetric.
(iii) Let (a, b) and (b, c) ∈ R.
5 divides a – b and b – c both
∴ a – 6 = 5 n₁ and b – n = 5 n₂ for some n₁ and n₂ ∈ N ∴ (a – b) + (b – c) = 5 n₁ + 5 n₂⇒ a – c = 5 (n₁ + n₂)
⇒ 5 divides a – c ⇒ (a, c) ∈ R
∴ R is transitive relation in N.

2 x + y = 41 ⇒ y = 41 – 2 x

x = 1 y = 41 – 2 (1) = 41 – 2 = 39 x = 2 ⇒ y = 41 – 2 (2) = 41 – 4 = 37 x = 3 ⇒ y = 41 – 2 (3) = 41 – 6 = 35 x = 4 ⇒ y = 41 – 2 (4) = 41 – 8 = 33

x =19 ⇒ y = 41 – 2 (19) = 41 – 38 = 3 x = 20 ⇒ y = 41 – 2 (20) = 41 – 40 = 1 x = 21 ⇒ y = 41 – 2 (21) = 41 – 42 = –1 ∉ N

∴ R = {(1,39), (2, 37), (3, 35), (4, 33)., (20, 1)}

domain of R = {1,2,3,4,........., 20}

and range of R = {1, 3, 5, 7,......,39}

(i) Now 1∉ N but (1, 1) ∉ R

(ii) (1,39), ∈ R but (39, 1) ∴ R ∴ R is not symmetric (iii) (20,1), (1,30) ∈ R but (20. 39) ∉ R ∴ R is not transitive.

We are given that
R = { (x, y) : x < y, x ∈ N, y ∈ N }
S = { (x, y) : x + y = 10, x ∈ N, y ∈ N }
T = { (x. y) : x = y or x – y = 1, x ∈ N, y ∈ N } (i) R is not reflexive as (x ,x) ∉ R ∀ x ∈ N
R is not symmetric as 1 < 2 ⇒ (1, 2) ∈ R but 2 < 1 ⇒ (2, 1) ∉ R
R is transitive as x and so ( x, y ) ∈ R and ( y, z ) ∈ R ⇒ ( x, z) ∈ R.
(ii) S is not reflexive as (x, x) ∉ R ∀ x ∈ IN

S is symmetric as x + y = 10 ⇒ y + x =10 and so (x, y) ∈ R ⇒ (y, x ) ∈ R ∀ x, y ∈ N S is not transitive as 3 + 7 = 10 ⇒ (3, 7) ∈ R and 7 + 3 = 10 ⇒ (7, 3) ∈ R but 3 + 3 ≠ 10 ⇒ (3,3) ∉ R (iii) T is not reflexive as x = x ∀ x ∈ N ⇒ (x, x) ∈ R

Now 2 – 1 = 1 as (2, 1) ∈ R but 1 ≠ 2  or 1 – 2 ≠ 1 ⇒ (1,2) ∉ R T is not symmetric.

Again 4 – 3=1 ⇒ (4,3) ∈ R and 3 – 2 = 1 ⇒(3,2) ∈ R but 4 ≠ 2 or 4 – 2 ≠ 1 ⇒ (4, 2) ∉ R.

T is not transitive.

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}.

(i) Since (x,y) R (x,y), ∀ (x, y) ∈ A, as x y = y x.
R is reflexive.
(ii) Again (x,y)R (u,v)
⇒ x v = y u ⇒ u y = v x and so (u, v) R (x, y).
∴ R is symmetric.
(iii) Again (x, y) R (u, v) and (u, v) R (a, b)

∴ (x, y)R (a,b).
∴ R is transitive.
From (i), (ii), (iii), it follows that R is an equivalence relation.

P(X) = { A : A is a subset of X }
(i) Since A C A ∀ ∈ P(X)
∴ ARA ∀ A ∈ P(X)
∴ R is reflexive relation (ii) Let A, B, C ∈ P(X) such that ARB and BRC A ⊂ B and B ⊂ C ⇒ A ⊂ C
⇒ A R C
∴ ARB and BRC ⇒ ARC
∴ R is transitive relation.
(iii) Now if A ⊂ B, then B may not be a subset of A i.e. ARB ⇏ BRA
∴ R is not a symmetric relation.
From (i), (ii), (iii), it follows that R is not an equivalence relation on P(X).

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}.

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}.

The smallest equivalence relation R₁ containing (1, 2) and (2, 1) is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}.
Now we are left with only 4 pairs namely (2, 3), (3, 2), (1, 3) and (3, 1).
If we add any one, say (2, 3) to R₁. then for symmetry we must add (3, 2) also and now for transitivity we are forced to add (1,3) and (3, 1). Thus, the only equivalence relation bigger than R₁ is the universal relation. This shows that the total number of equivalence relations containing (1,2) and (2, 1) is two.

Here (a, b) R (c, d) ⇔ a + d = b + c.
(i) Now (a, b) R (a, b) if a + b = b + a, which is true.
∴ relation R is reflexive.
(ii) Now (a, b) R (c, d)
⇒ a + d = b + c ⇒ d + a = c + b
⇒ c + b = d + a ⇒ (c, d) R (a, b)
∴ relation R is symmetric.
(iii) Now (a, b) R (c, d) and (c, d) R (e,f)
⇒ a + d = b + c and c + f = d + e
⇒ (a + d) + (c + f) = (b + c) + (d + e) ⇒ a + f = b + e
⇒ (a , b) R (e, f)
∴ relation R is transitive.
Now R is reflexive, symmetric and transitive
∴ relation R is an equivalence relation.

Here (a, b) R (c, d) ⇔ a d = b c
(i) Now (a, b) R (a, b) if a, b = b a, which is true
∴ relation R is reflexive.
(ii) Now (a, b) R (c, d)
⇒ a d = b c ⇒ d a = c b ⇒ c b = d a ⇒ (c, d) R (a, b)
∴ relation R is symmetric.
(iii) Now (a, b) R (c, d) and (c, d) R (e,f)
⇒ a d = b c and c f = d e ⇒ (a d) (c f) = (b c) (d e)
⇒ a d c f = b e d e ⇒ (a f) (d c) = (b e) (d c)
⇒ a f = b e ⇒ (a, b) R (e, f) ∴ relation R is transitive Now R is reflexive, symmetric and transitive ∴ relation R is an equivalence relation.

(i) Let (a, b) be any element of N x N
Now (a, b) ∈ N x N ⇒ a, b ∈ N ∴ a b (b + a) = b a (a + b)
⇒ (a, b) R (a, b)
But (a, b) is any element of N x N ∴ (a, b) R (a, b) ∀ (a, b) ∈ N x N ∴ R is reflexive on N x N.
(ii)    Let (a, b), (c, d ) ∈ N x N such that (a, b) R (c, d)
Now (a, b) R (c, d) ⇒ a d (b + c) = b c (a + d)
⇒ c b (d + a) = d a (c + b)
⇒ (c, d) R (a, b)
∴ (a, b) R (c, d ) ⇒ (c, d) R (a, b) ∀ (a, b), (c, d) ∈ N x N ∴ R is symmetric on N x N.
(iii)    Let (a, b), (c, d ), (e, f) ∈ N x N such that
(a, b) R (c, d ) and (c, d ) R (e, f)
(a, b) R (c, d) ⇒ a d (b + c) = b c (a + d)
         
Also   (c, d)  R (e, f)    cf(d + c) = d e (c + f)

Adding (1) and (22),  we get

⇒ a f (b + e) = b e (a + f) ⇒ (a, b) R (e,f)

∴ (a, b) R (c, d) and (c, d) R (e.f) ⇒ (a, b) R (e, f) ∀ (a, b), (b, c), (c, d) ∈ N x N ∴ R is transitive on N x N ∴ R is reflexive, symmetric and transitive ∴ R is an equivalence relation on N x N

Q arbitrarily, then

(i) Since      

therefore, a, b are integers.
∴ ab = ba, since multiplication is comutative in Z.

∴ R is reflexive.
(ii)   Let  
∴ a d = b c ⇒ d a = c b ⇒ c b = d a
⇒ R is symmetric.
(iii)   

∴ ad = bc and cf = de ⇒ (a d) (c f) = (b c) (d e)
⇒ (c d) (a f) = (c d) (b e), by using commutative and associative laws of multiplication in Z.
⇒ a f = be

R is transitive.

Thus R is an equivalence relation.

Suppose that R₁ and R₂ are two equivalence relations on a non-empty set X.
First we prove that R₁ ∩ R₂ in an equivalence relation on X.
(i) R₂ ∩ R₂ is reflexive :
Let a ∈ X arbitrarily.
Then (a, a) ∈ R₁ and (a, a) ∈ R₂ , since R₁, R₂ both being equivalence relations are reflexive.
So. (a, a) ∈ R₁ ∩ R₂
⇒ R₁ ∩ R₂ is reflexive.
(ii) R₁ ∩ R₂ is symmetric :
Let a, b ∈ X such that (a, b) ∈ R₁ ∩ R₂ ∴ (a, b) ∈ R₁ and (a, b) ∈ R₂ ⇒ (b, a) ∈ R₁and (b, a) ∈ R₂, since R₁ and R₂ being equivalence relations are also symmetric.
(b, a) ∈ R₁∩ R₂
(a, b) ∈ R₁ ∩ R₂ implies that (b, a) ∈ R₁ ∩ R₂.
∴ R₁ ∩ R₂ is a symmetric relation.
(iii) R₁ ∩ R₂ is transitive :
Let a, b, c ∈ X such that (a, b) ∈ R₁ ∩ R₂ and (b, c) ∈ R₁ ∩ R₂.
(a, b) ∈ R₁ ∩ R₂ ⇒ (a, b) ∈ R₁ and (a, b) ∈ R₂    ...(i)
(b, c) ∈ R₁ ∪ R₂ ⇒ (b, c) ∈ R₁ and (b, c) ∈ R₂    ...(ii)
(i) and (ii) ⇒ (a, b) and (b, c) ∈ R₁
⇒ (a. c) ∈ R₁, since R₁ being an equivalence relation is also transitive.
Similarly, we can prove that (a, c) ∈ R₂ ∴ (a, c) ∈ R₁ ∩ R₂ So, R₁ ∩ R₂ is transitive.
Thus R₁ ∩ R₂ is reflexive, symmetric and also transitive. Thus R₁ ∩ R₂ is an equivalence relation.

Let A = {1, 2, 3, 4}
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1,3), (3, 3), (3, 2)}
R is reflexive as (a, a) ∈ R ∀ a ∈ A R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R R is transitive as (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R ∴ R is reflexive and transitive but not symmetric.

Here 1 is related to 2.
there are two possible cases :
Case I : When 1 is not related to 3, then the relation R₁ = {(1, 1), (1,2), (2, 1), (2, 2), (3, 3)} is only equivalence relation containing (1,2).
Case II : When 1 is related to 3, then
A x A = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)} is the only equivalence relation containing (1, 2). there are two equivalence relations on A containing (1, 2).  (B) is correct answer.

Here A = {1, 2, 3}, B = {4, 5, 6, 7}
and f = {(1, 4), (2, 5), (3, 6)}
∴(1) = 4, f(2) = 5, f(3) = 6 different elements of A have different images in B under f.
∴ is one-to-one.

A is the set of all 50 students of class X in a school.
No two different students of the class can have same roll number.
Therefore, f must be one-one.
We can assume without any loss of generality that roll numbers of students arc from 1 to 50. This implies that 51 in N is not roll number of any student of the class. so that 51 can not be image of any element of X under f. Hence, f is not onto.

f : N → N is given by f (x) = 2x Let ,x₁, x₂ ∈ N such that f (x₁) = f (x₂)
∴ 2 x₁ = 2 x₂ ⇒ x₁ = x₂ ∴ f is one-one.
f is not onto as for 1 ∈ N, there does not exist any x in N such that f (x) = 2 x = 1.

f : N → N is given by f (x) = 5 x
Let x₁, x₂ ∈ N such that f (x₁) = f (x₂)
∴ 5 x₁ = 5 x₂ ⇒ x₁ = x₂ ∴ f is one-one i.e. injective.
f is not onto i.e. surjective as for 1 ∈ N, there docs not exist any in N such that f (x) = 5 x = 1

f : R → R is given by f (x) = 2x
Let x₁, x₂ ∈ R such that f (x₁) = f (x₂)
∴ 2x₁ = 2 x₂ ⇒ x₁ = x₂ ∴ f is one-one.
Also, given any real number y ∈ R, there exists
 
such that 

∴ f is onto.

(i) f : N → N is given by f (x) = x²
Let x₁ x_{2 ∈ N be such that f(x1) ⁼ f(x2)}
∴ x₁² = x ₂² ⇒ x₂ ² x₁ ² = 0
⇒ (x₂ –x₁) (x₂ + x₁) = 0
⇒    x₂ – x₁ = 0    [x₁ + x₂ ≠ 0 as x₁, x₂ ∈ N]
⇒    x₂ = x₁ ⇒ x₁ = x₂
∴ f is one-one, i.e.. f is injective.
Since range of f = { 1² , 2², 3²............}
= {1.4.9......} ≠ N.
∴ f is not surjective.
(ii) f : Z → Z is given by f (x) = x²
Let x₁, x₂ ∈ Z be such that f(x₁)= f (x₂)
∴ x₂² =x₂² ⇒ x₂² = 0
⇒ (x₂ – x₁) (x₂ + x₁) = 0
⇒    x₂ = x₁    or x₂ = – x₁
∴ f (x₁) = f(–x₁) ∀ x₁ ∈ Z
∴ f is not one-one, i.e. f is not injective.
Also range of f = { 0², 1², 2²,.....}
= {0, 1,4, 9,.........}
≠     Z
∴ f is not onto i.e.. f is not surjective (iii) f : R → R is given by f (x₁) = x² Let x₁, x₂ ∈ R be such that f (x₁) = f (x₂)
⇒    x₁² = x₂² ⇒ (x₂ – x₂) (x₂ + x₁) = 0
⇒    x₂ = x₁ or x₂ = – x₁
⇒    f(x₁) = f (–x₁) ∀ x₁ ∈ R
∴ f is not one-one, i.e., f is not injective.
As range of f does not contain any negative real, therefore, range of ≠ R.
Hence. f is not onto, i.e., f is not surjective.
(iv) f : N → N is given by f (x) = x³ Let x₁ ,. x₂ ∈ N be such that f (x₁) = f(x₂)
⇒    x₁³ = x₂³ ⇒ x₁ = x₂
∴ f is one-one, i.e., injective.
Also range of f = {1³, 2³, 3³,.........}
= {1,8,27,.....}
≠ N
∴ f is not onto, i.e.,f is not surjective.
(v) f : Z → Z is given by f (x) = x³ Let x₁, x₂ ∈ Z be such that f (x₁) = f (x₂)
⇒    x₁³ = x₂³ ⇒ x₁ = x₂
Also range of f = {0³ ± 1³, ± 2³, ± 3³,....}
= {0, ± 1, ± 8, ± 27.............}
≠ Z ∴ f is not onto, i.e., f is not surfective.

Here f(x) = x² D_f = R Now 1,–1 ∈ R Also f (1)= 1, f (–1) = 1 Now 1 ≠ –1 but f (1) = f (–1)
∴ f is not one-to-one.
Again, the element – 2 in the co-domain of R is not image of any element x in the domain R.
∴ f is not onto.

f : R → R is given by f (x) = | x |
Different elements in R can have the same image
[∵ f (–1) = |–1| = 1, f(1) = |1| = 1]
∵ f is not one-one.
Also, R_f = set of non-negative reals ≠ R
∵ f is not onto.

f : R → R is given by f (x) = [x]
Different elements in R can have the Same image {∵ for all x ∈ [0, 1), f (x) = 0 } ∵ f is not one-one Also R_f = set of integers ≠ R
∵ f is not onto.

f : A x B → B x A is such that f (a,b) = (b,a).
Let (a₁, b₁) and (a₂, b₂) any two elements of A x B such that
f (a₁, b₁) = f (a₂, b₂)
∴ (b₁, a₁) = (b₂, a₂) ⇒ b₁= b₂ and a₁ = a₂
⇒    = (a₂,b₂)
∴ f (a₁,b₁) = f (a₂,b₂) ⇒ (a₁b₁) = (a₂,b₂) ∀ (a₁, b₁) (a₂, b₂) ∈ A x B
∴ f is one-to-one
Again let (b, a) be any element of B X A ∴ b ∈ B and a ∈ A. So (a, b) ∈ A X B ∴ for all (b, a) ∈ B X A, there exists (a, b) ∈ A x B such that f (a, b) = (b, a)
∴ f : A x B → B x A is an onto function ∴ f is one-to-one and onto

f : N → N is given by f (1) = f (2) = 1 and f (x) = x – 1.
Since    f (1) = f (2)
∴ f is not one-one Given any y ∈ N, y ≠ 1, we can choose x as y + 1 such that f(y +1) = y + 1 – 1 = y Also    f (1) = 1
∴ f is onto.

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

Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the codomain {1, 2, 3} under f.
∴ f has to be onto.

(i) Here f (x) =3 – 4 x ∀ x ∈ R
Let x₁, x₂ ∈ R be such that f (x₁) =  f (x₂) ∴ 3 – 4 x₁ = 3 – 4 x₂
⇒ –4 x₁ = – 4 x₂ ⇒ x₁= x₂
∴ f is one-one.
Let y ∈ R be any real number.
Put f(x) = y , ∴ 3 – 4 x = y, i.e.,

Thus, corresponding to every y ∈ R, there exists

such that 

∴ f is onto.
∴ f is a bijection.
(ii) Let x₁, x₂ ∈ R be such that f (x₁) = f (x₂)
∴ 1 + x₁² = 1 + x₂² ∴ x = ∴ x₁ =  ± x₂ ∴ f(x₁) = f(–x₁)
⇒ f is not one-one.
Also, range of f contains only those reals which are greater than or equal to 1.
[∵ x² ≥ 0 ∀ x ∈ R, ∴ 1 + x² ∁ 1 ∀ x ∈ R]
∴ R_f ≠ R,
⇒ f is not onto.
Thus, f is neither one-one nor onto.

f : N – {1} → N is defined by
f (n) = the highest prime factors of n.
∴ f (6) = the highest prime factor of 6 = 3 f (12) = the highest prime factor of 12 = 3 Now 6 and 12 are associated to the same element.
∴ f is not one-to-one Also range of f consists of prime numbers only ∴ range of f ≠ N ∴ f is not onto function.
Range of f is the set-of all prime numbers.

Let A = {a₁,a₂.....,a_n } where n is finite.
Range of f = {(f (a₁) , f (a₂), ...., f (a_n)}
Since f : A → A is an onto function
∴ range of f = A
⇒ {f (a₁),f(a₂),...f(a_n)} = (a₁,a₂,....,a_n}
But A is a finite set consisting of n elements
∴ f(a_n), f(a₂), ... ,f(a_n) are district elements of A.
∴ f is one-to-one.

Let A = {a₁,a₂, ...a_n} where n is finite.
Since f is one-to-one.
∴ f(a₁), f(a₂)....,f (a_n)    district element of A.
∴ A = {f (a₁), f(a₂).....,f (a_n)}
Let b be any element of A. Then
b = f(a_i) for some i, 1 ≤ i ≤ n. ∴ f is onto.

A = {1, 2, 3}

Let f = A → A be a one-to-one function. Then f (1) has three choices, namely 1, 2 or 3. So f (1) = 1 or f (1) = 2 or f (1) = 3.

Similarly f (2) and f (3) have three choices each.
all the one-to-one functions from A to A are
(i) {(1, 1), (2, 2), (3, 3)}  (ii) (1, 1), (2, 3), (3, 2)}
(iii) {(1.2). (2, 3), (3, 1)} (iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1,3), (2, 2), (3, 1)} (vi) {(1, 3), (2, 1), (3, 2)}
required number of functions = 6

For every a ∈ X. (a. a) ∈ R as f (a) = f (a),
∴ R is reflexive.
Similarly. (a, b) ∈ R ⇒ f(a) = f(b) ⇒ f(b) = f(a) ⇒ (b, a) ∈ R ∴ R is symmetric.
Further, (a, b) ∈ R and (b, c) ∈ R
⇒ f (a) = f (b) and f (b) = f(c) ⇒ f (a) = f (c) ⇒ (a, c) ∈ R ∴ R is transitive.
∴ R is an equivalence relation.

f : Z → Z, is defined by f = {(n, n²) : n ∈ Z}
∴ D_f = Z and f (n) = n² Again g : Z → Z is defined by g = {(n, | n |² ) : n ∈ Z}
∴ D_g = Z and g (n) = | n |² = n² = f (n)
∴ D_f = D_g and f (n) = g (n) ∀ n ∈ D_f or D_g
∴ f = g

f : R → R is given by f (x) = x⁴
Different elements in R can have the same image
[∵ f (–2) = (–2)⁴ = 16, f (2) = (2)⁴ = 16]
∴ f is not one-one.
Also R_f = set of non-negative reals ≠ R ∴ f is not onto.
∴ f is neither one-one nor onto.
∴ (D) is correct answer.

f is one-one, as f (x₁) = f (x₂)
⇒ 3 x₁ = 3 x₂
⇒ x₁ = x₂.
Also, given any real number y in R, there exist

in R such that

Hence. f is onto
∴ (A) is correct answer.

f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} are functions such that
f (2) = 3, f (3) = 4, f (4) = f (5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11
∴ (g o f) (2) = g (f (2)) = g (3) = 7
(g o f) (3) = g (f (3)) = g (4) = 7
(g o f) (4) = g (f(4)) = g (5) = 11
(g o f) (5) = g (f(5)) = g (5) = 11
∴ g o f = {(2, 7), (3, 7), (4, 11), (5, 11)}

f : {1, 3, 4}  → {1, 2, 5} is given by
f = {(1, 2), (3, 5), (4, 1)}
∴ f (1) = 2, f (3) = 5, f(4) = 1
Also g : {1, 2, 5}  → {1, 3} is given by
g = {(1, 3), (2. 3), (5, 1)}
∴ g (1) = 3, g(2) = 3, g(5) = 1
Since co-domain of f is same as the domain of g
∴ g o f exists and (g o f) : {1,3,4 } → {1,3}
Now (g o f) (1) = g (f(1)) = g(2) = 3
(g o f)(3) = g (f(3)) = g(5) = 1
(g o f) (4) = g (f (4)) = g(1) = 3
∴ g o f = {(1, 3), (3, 1), (4, 3)}

f : R → R such that f (x) = cos x and g : R → R such that g (a ) = 3 x².
Now (g o f) (x) = g (f(x)) = g (cos x) = 3 cos² x and (f o g)) (x) = f (g (x)) = f(3 x²) = cos (3 x²)
Now 3 cos² x ≠ cos (3x²) ∴ g o f ≠ f o g.

f (x) = x² , g (x) = x + 1
(g o f) (x) = g (f(x)) = g (x²) = x² + 1
(f o g ) (x) = f (g (x)) = f (x + 1) = (x + 1)² ∴ g o f ≠ f o g

Here f(x) = x + 7, g(x) = x – 7
(f o g) (x) = f(g(x)) = f(x – 7) = (x – 7) + 7 = x – 7 + 7 ∴ (f o g) (x) = x ∴ (f o g)(7) = 7.

f : N → N defined by f(x) = 2 x ∀ n ∈ N
g : N → N defined by g(y) = 3 y + 4 ∀ y ∈ N
h : N → R defined by h(z) = sin z ∀ z ∈ N.
Now (h o (g o f))(x) = (h o (g o f) (x))
= h(g(f(x)) = h(g(2 x)) = h (3(2 x) + 4) = h(6 x + 4) = sin (6 x + 4) ∀ x ∈ N.
A Iso, ((h o g) o f) (x) = ((h o g) (f (x)) = (h o g) (2 x) = h(g (2 x))
= h(3(2 x) + 4) = h(6 x + 4)
= sin (6 x + 4), ∀ x ∈ N.
∴ h o(g o f) =(h o g) o f.

X= {1, 2, 3}, Y = {a, b, c}
f : X → Y is one-one and onto such that
f(1) = a f(2) = b, f(3) = c
Let g : Y ∴ Y X be a function such that g (a) = 1, g (b) = 2, g (c) = 3
Now (g o f) (1) = g (f (1)) = g(a) = 1
(g o f) (2) = g (f (2)) = g (b) = 2
(g o f)(3) =g(f(3)) = g (c) = 3 ∴ g o f = 1_x
Again (f o g) (a) = f(g (a)) = f(1) = a (f o g) (b) = f(g (b)) = f(2) = b
(f o g) (c) = f(g (c)) = f (3) = c
∴ f o g = I_Y.

i) Since f, g and h are functions from R to R.
∴ (f + g) o h : R ∴ R and f o h + g o h : R ∴ R Now, ((f + g) oh) (x) = ((f + g) h(x))
⇒ ((f + g) o h) (x) = f (h (x)) + g(h(x))
⇒ ((f + g) o h) (x) = (f o h) (x) + (g o h) (x) for all x ∈ R ∴ ( f + g) o h = f o h + g o h (ii) Again (f . g) o h : R → R and (f o h) . (g o h) : R → R Now {(f . g) o h } (x) = ( f . g) (h(x))
⇒ { (f . g) o h } (x) = f(h(x)) . g(h(x))
⇒ {(f . g) o h }(x) = (f o h) (x) . (g o h) (x)
⇒ { (f . g) o h }(x) = {(f o h) . (g o h)} (x) for all x ∈ R ∴ (f . g) o h = (f o h) . (g o h).

We have f (x) = x + 2    ....(1)
Also gof = 1_z ⇒ (gof) (x) = x ∀ x ∈ Z ∴ g(f(x)) = x ∀ x ∈ Z ⇒ g (x + 2) = x ∀ x ∈ Z
∴ g (x) = x – 2 is required function.

f(x) = x² – 3 x + 2 Replacing x by f (x). we get.
f (f(x)) = ( f(x))² –3 f(x) + 2
= (x² – 3 x + 2)² –3(x² – 3 x + 2) + 2
= x⁴  + 9 x² + 4 – 6 x³ – 12 x + 4 x² – 3 x² + 9 x – 6 + 2 = x⁴ – 6 x³ +10 x² – 3 x

Here f : A → B and g : B → C are one-to-one functions ∴  g o f is a function from A to C.
Let x₁, x₂ ∈ A
Now (g o f) (x₁) = (g o f) (x₂)
⇒ g (f (x₁)) = g (f(x₂))
⇒    f(x₁) = f(x₂)    [∵ g is one-to-one]
⇒    x₁ = x₂    [∵ f is one-to-one]
⇒ (g o f) (x₁) = (g o f) (x₂) ⇒ x₁ = x₂ , ∀ x₁,x₂ x₁,x₂ ∈ A
∴ gof is one-to-one.

Here f : A → B and g : B → C are onto functions ∴ g o f is defined from A to C. ∴ g is an onto mapping from B → C ∴ to each z ∈ C, there exists y ∈ B such that g (y) = z Again f is an onto mapping from A to B 
∴ to each y ∈ B, there exists y ∈ A such that f(x) = y ∴ to each z ∈ C, there exists x ∈ A such that z = g (y) = g (f(x)) = (g o f) (x) ∴ g o f is onto.

Let f : A → B and g : B → C be two functions such that g o f : A ∴ C is defined.
We are given that g of : A → C is one-one.
We are to prove that f is one-one If possible, suppose that f is not one-one.
there exists x₁, x₂ ∈ A such that x₁ ≠ x₂ but f (x₁) = f (x₂)
But f(x₁) = f (x₂)    ⇒ g (f(x₁)) = g (f(x₂))
⇒ (gof) (x₁) = (gof) (x₂)
∴ x₁, x₂ ∈ A such that x₁ ≠ x₂ but (gof) (x₁) = (gof) (x₂)
∴ gof is not one-one, which is against the given hypothesis that g of is one-one our supposition is wrong.
∴ f is one-one.
Consider f : {1, 2, 3, 4} ≠ {1, 2, 3, 4, 5, 6} defined as f (x) = x, ∀ x and g : {1, 2, 3, 4, 5, 6} → {1, 2, 3, 4, 5, 6} as g. (x) = x, for x = 1, 2, 3, 4 and g (5) = g (6) = 5. Then, g o f (x) = x ∀ x, which shows that g o f is one-one. But g is clearly not one-one.

Let f : A → B and g : B → C be two functions such that g o f : A → C is defined.
We are given that g o f : A → C is onto. We now prove that g is onto.
Let z ∈C.
Since gof : A → C is onto, so there exists x ∈ A such that (g o f) (x) = z
⇒ g (f (x)) = z
⇒ g (y) = z where y = f(x)
Since a ∈ A and f is a map from A to B ∴ f (x) ∈ B ⇒ y ∈
∴ for given z ∈ C, we have determined y ∈ B such that g (y) = z ∴ g : B → C is onto.
Consider f : {1, 2, 3, 4} → {1, 2, 3, 4} and g : {1, 2, 3, 4} → {1, 2, 3} defined as f (1) = 1, f(2) = 2 , f(3) = f(4) = 3, g (1) = 1, g (2) = 2 and g (3) = g (4) = 3. So, g of is onto but f is not onto.

Let    f(x) = x, g(x) | x |
where f : N → Z and g : Z → Z
Now    g(– 1) = | – | | = 1, g(1) = | 1 | = 1
g is not one-one i.e. injective.
Now f : N → Z and g : Z → Z
∴ g o f : N → Z
Let x₁ , x₂ ∈ N such that
(g o f) (x₁) = (g o f)( x₂) ∴ g (f(x₁)) = g(f(x₂))
∴ g(x₁) = g(x₂)
∴ |x₁| = | x₂|
⇒ x₁ = x₂    [∵ x₁ > 0, x₂ > 0 ]
∴ (g o f)(x₁) = (g o f) (x₂) ⇒ x₁ = x₂ ∴ g o f is onto.

Let g₁ and g₂ be two inverses of f for all y ∈ Y, we have, (f o g₁)(y) = y = I_Y (y) and (f o g ₂) (y) = y = I_Y (y)
⇒    (f o g₁) (y) = (f o g₂) (y) for all y ∈ Y
⇒    f(g₁(y)) = f(g₂(y)) for all y ∈ Y
⇒    g₁ (y) = g₂(y)    [∵ f is one-one as f is invertible]
∴ g₁ = g₂ ∴ inverse of f is unique.

Since inverse of a bijection is also a bijection.
∴ f^–1 : Y → X is a bijection and hence invertible.
Since f^–1 : Y → X is a bijection.
∴ (f^–1)^–1 : X → Y is also a bijection.
Let x be an arbitrary element of X such that    f(x) = y. Then.
f^–1 (y) = x    [∵ f^–1 is the inverse of f ]
⇒ (f^–1)^–1 (x) = y    [∵(f^–1)^–1 is the inverse of f^–1]
(f^–1)^–1 (x) = f(x)    [∴ f(x) = y]
Since x is an arbitrary element of X.
∴ (f^–1)^–1 (x) = f(x) for all x ∈ X
∴ (f^–1)^–1 = f .

(i) Given F = {(a, 3), (b, 2), (c, 1)} ∴ range of F = {1, 2, 3} = T ∴ F is onto.
Also, F is one-one as different elements of S have different images.
∴ F^–1 : T → S is defined as
F^–1 (1) = c, F^–1(2) = b, F^–1(3) = a.
(ii) Mere , F = {(a, 2), (b, 1), (c, 1)} is not one-one as F(b) = F(c) = 1.
F is not invertible.

S = {1, 2, 3}
f : S → S is given by
(a) f = {(1, 1), (2, 2), (3, 3)}
Now f is one-one and onto
∴ f^–1 exists and is given by
f^–1 = {(1, 1),(2,2), (3,3)}
(b) f = {(1,2), (2, 1), (3, 3)} Since f (2) = f(3) = 1
∴ f is not one-one, so that f is not invertible.
(c) f = {(1,3), (3, 2), (2, 1)} Now f is one-one and onto
∴ f^–1 exists and is given by
f^–1 = {(3, 1), (2, 3), (1,2)}

(i) f : {(1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
∴ f (1) = f(2) = f(3) = f(4) = 10 ∴ f is not one-one and so f is not invertible.
(ii) g : {5,6,7,8} → {1, 2, 3, 4} with g= {(5, 4), (6, 3), (7, 4), (8, 2)} ∴ g(5) = g(7) = 4
g is not one-one and so f is not invertible.
(iii) h : {2, 3, 4, 5} → {7,9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
∴ h is one-one and onto ∴ h is invertible and
h^–1 = {(7, 2), (9, 3), (11, 4), (13, 5)}

N x N → N, given by (a, b) → a – b. is not binary operation, as the image of (3, 5) under is 3 – 5 = – 2 ∉ N.
Similarly, ÷ : N x N → N, given by (a, b) → a ÷ b is not a binary operation, as the image of (3, 5) under → is

Since V carries each pair (a, b) in R x R to a unique element namely maximum of a and b lying in R, so V is a binary operation.
Similarly A carries each pair (a, b) in R x R to a unique element namely minimum of a and b lying in R, so ∧ is a binary operation.

Since a + b = b + a and a x b = b x a, ∀ a, b ∈ R
However, is not commutative, since 3 – 4 ≠ 4 – 3
Similarly, 3 ÷ 4 * 4 ≠ 3 show that ‘÷’ is not commutative.

Since 3 * 4 = 3 + 8 = 11 and 4 * 3 = 4 + 6 = 10 operation * is not commutative.

Addition and multiplication are associative,
as (a + b) + c = a + (b + c)
and (a x b) x c = a x (b x c) ∀ a, b, c ∈ R.
However, subtraction and division are not associative, as
(8 – 5) – 3 ≠ 8 – (5 – 3) and (8 ÷ 5) ÷ 3 ≠ 8 ÷ (5 ÷ 3).

The operation * is not associative as
(8 * 5) * 3 = (8 + 10)* 3 = (8 + 10) + 6 = 24,
while 8 * (5 * 3) = 8 * (5 + 6) = 8 * 11 = 8 + 22 = 30.

a + 0 = 0 + a = a and a x 1 = a = 1 x a, ∀ a ∈ R implies that 0 and 1 are identity elements for the operations ‘+’ and ‘x’ respectively.
Further, there is no element e in R with a – e = e – a, ∀ a.
Similarly, we can not find any element e in R_* such that
a ÷ e = e ÷ a, ∀ c in R.
∴ ‘–‘ and ‘ ÷’ do not have identity element.

As – a ∉ N, so – a can not be inverse of a for addition operation on N, although – a satisfies a + (– a) = 0 = (– a) + a.
Similarly, for a ≠ 1 in N, 1/a ∉ N, which implies that other than 1 no element of N
has inverse for multiplication operation on N.

We are given that
a * b = 2 a + b – 3 Put a = 3, b = 4
∴ 3 * 4 = 2 (3) + 4 – 3 = 6 + 4 – 3 = 7

Here a * b = (2 a – b)²
∴ 3 * 5 = (6 – 5)² = (1)² = 1
∴ 5 * 3 = (10 – 3)² = (7)² = 49 Now 1 ≠ 49
∴ 3 * 5 ≠ 5 * 3

(i) If a, b ∈ Z₊, then a – b may or may not belong to H ; for example 3–5 = –2 ∉ Z. is not a binary operation on Z⁺.
(ii)    If a, b ∈ Z⁺, then a b also belong to Z⁺ . is a binary operation on Z⁺.
(iii)    If a, b ∈ R, then a b² ∈ R.
(iv) Since | a – b | ≥ 0 , therefore, for all a, b ∈ Z⁺, a * b ∈ Z⁺. is a binary operation on Z⁺.
(v) For all a, b ∈ Z⁺ a * b = a ∈ Z⁺, is a binary operation on Z⁺.

(i) If e is an identity element then
a *e = e * a = a for all a ∈ Q ⇒    a – e = e – a = a for all a ∈ Q
a – e = a and e – a = a ⇒    e = 0 and e = 2 a for all a ∈ Q
which is not possible.    [∵ the above conditions hold only for a = 0]
the identity element does not exist.
(ii) If e is the identity element, then
a * e = e * a = a for all a ∈ Q
⇒    a² + e² = e² + a² = a for all a ∈ Q
⇒    a = a² for all a ∈ Q,
which is not possible.
the identity element does not exist.

Here * is a binary operation on N given by a * b = l.c.m. (a, b), a, b ∈ N

(i) 5 * 7 = 35, 20 * 16 = 80
(ii)    Since l.c.m. (m, n) = l.c.m. (n, m) ∴ m * n = n * m ∀ m, n ∈ N binary operation is commutative
(iii)    Let a, b, c ∈ N
Now a * (b * c) = l.c.m. (a, b * c)
= l.c.m. [a, l.c.m (b. c)]
= l.c.m [l.c.m. (a, b), c]
= l c.m. [(a * b), c]
∴ a * (b * c) = (a * b) * c ∀ a, b, e ∈ N binary operation is associative.
(iv) Let e be identity element. Then
V 
a * e = a = e * a ∀ a ∈ N

⇒ (a * e) = a ∀ a ∈N
⇒ l.c.m. (a, e) = a ∀ a ∈ N
⇒ e = 1
∴ 1 is the identity element in N
(v) Let a be an invertible element in N.
Then there exists    such that
a * b = 1 ⇒ l.c.m. (a, b) = 1 ⇒ a = b = 1
∴ 1 is the invertible element of N.

a * b = l.c.w. of a and b
∴ 2 * 3 = l.c.m. of 2 and 3
∴ 2 * 3 = 6
But 6 ∉ {1, 2, 3, 4, 5}
∴ *    is not a binary operation on {1, 2, 3,4, 5}.

Here a * b = H.C.F. of a and b, a, b ∈ N (i) H.C.F. (a, b) = H.C.F. (b, a)
∴ a * b = b * a
*    is a commutative binary operation.
(ii) Let a, b, c ∈ N
∴ a * (b * c) = (a, H.C.F. (b, c))
= H.C.F. (H.C.F. (a,b),c)
= (a * b) * c ∴ a * (b * c) = (a * b) * c
∴ *    is associative binary operation.
(iii) If e is an identity element, then e * a = a * e = a for a ∈ N.
⇒ H.C.F. of a and e = a ∀ a ∈ N
⇒ a divides e ∀ a ∈ N.
Such a number e, which is divisible by every natural number, does not exist.

A = N x N and (a, b) * (c, d) = (a + c, b + d)
(i) Let (a, b), (c, d), (e, f) be any three elements of A.
∴ (a, b)* {(c, d) * (a, f)} = (a, b) * (a + e, d + f)
= (a + (c + e), b + (d + f))
= ((a + c) + e, (b + d) + f)
= (a + c, b + d) * (e, f)
∴ (a, b) * {(c. d) * (e,f)} = {(a, b) * (c, d) } * (e, f) ∀ (a, b), (c, d), (e, f) ∈ A ∴ (A, *) is associative.
(ii)    Let (a, b), (c, d) be any two elements of A.
∴ (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b)
∴ (a, b) * (c, d) = (c, d) * (a, b) ∀ (a, b), (c, d) ∈ A ∴ (A, *) is commutative
(iii)    if possible , suppose that (x, y) the is identity element in A.
∴ (a, b) * (x + y) = (a, b) ∀ (a, b), ∈ A
⇒ (a + x, b + y) = (a, b) ∀ (a, b)∈ A ⇒ a + x = a, b + y = b ∀ a, b ∈ N ⇒ x = 0, y = 0 ∀ a, b ∈ N This is impossible as 0 ∉ N ∴ our supposition is wrong.
∴ (A, *) does not have any identity element.

Given a ∧ b = min {a, b}
Required composition table for ‘∧’ is as follows :
Table

A = N x N and (a, b) * (c, d) = (ac, bd)
(i) Let (a, b), (c, d), (e, f) be any three elements of A Now (a, b) * {(c, d) * (e, f)} = (a, b) * (ce, d f)
= (a(ce)), b(d f))
= ((a c)e, (b d), f)
= (a c, b d) * (e, f)
∴ (a, b) * {(c. d) * (e,f)} = {(a, b) * (c , d)} * (e, f) ∀ (a b), (c, d), (e, f) ∈ A ∴ (A,*) is associative.

(ii) Let (a, b), (c, d) be any two elements of A Now (a, b) * (c, d) = (a c, b d) = (c a. d b)
∴ (a, b) * (c, d) = (c, d) * (a, b) ∀ (a, b), (c, d) ∈ A ∴ (A, *) is commutative.

A = N x N and (a, b) * (c, d) = (ad + bc, bd)
(i) Let (a, b), (c, d), (e,f) be any three elements of A.
∴ {(a, b) * (c, d)} * (e,f) = (ad + bc, bd) * (e, f)
= ((ad + bc) f + (bd) e, (bd) f)
∴ {(a, b) * (c, d)} * (e,f) = (adf + btf + bde, bdf)    ...(1)
Again (a, b) * {(c, d) * (e,f)} = (a, b) * (c f + de, d f)
= (a(d f) + b (c f + de), b (d f))
∴ (a, b) * {(c, d) * (e,f)} = (adf + bef + bde, bdf)    ...(2)
From (1) and (2), we get,
(a, b) * {(c, d) * (e,f)} = {(a, b) * (c, d) * (e,f)} ∴ (A. *) is associative.
(ii)    If possible, suppose that (x, y) is identity element in A.
∴ (a, b) * (x, y) = (a, b) ∀ (a, b) ∈ A
⇒ (ay + bx by) = (a, b) ∀ (a, b) ∈ A ⇒ ay + bx = a and by = b ∀ a, b ∈ N ⇒ x = 0, y = 1 ∀ a, b ∈ N This is not possible as 0 ∉ N ∴ our supposition is wrong ∴ (A, *) has no identity element.
(iii)    Let (a, b), (c, d) be any two elements of A.
Now (a, b) *(c, d) = (ad + be , bd)
= (bc + ad,db)
= (cb + da , db)
∴ (a, b) * (c, d)) = (c, d) * (a, b) ∀ (a, b), (c, d) ∈A ∴ (A, *) is commutative .

Let    A = {1,2, 3,4, 5}
Multiplication table is given as follows :
Table

(i) (2 *3) *4 = 1 *4 = 1
2* (3 *4) = 2* 1 = 1
(ii) Since the multiplication table is symmetrical about the diagonal starting at the upper left corner and ending at the lower right corner.
* is commutative.
(iii) 2 * 3 = 1, 4 * 5 = 1
∴ (2 * 3) * (4 * 5) = 1 * 1 = 1

a * b = | a – b |
= | – (b – a) = | b – a | = b * a
⇒ a * b = b * a ∀ a, b ∈ R
⇒ ‘*’ is commutative.
Let a, b, c ∈ R,
∴ (a * b) * c = | a – b | * c = || a – b | – c |
and a * (b * c) = a * | b – c | = | a – | b – c||
⇒    (a * b) * c ≠ a * (b * c)
∴ ‘*’ is not associative.
Again    a o b = a and b o a = b
⇒    a o b ≠ b o a
⇒ o is not commutative.
Now, let a, b, c ∈R
∴ (a o b) o c = a o c = a and a o(b o c) = a o b = a ⇒ ‘o’ is associative.
Again a * (b o c) = a * b = | a – b | and (a * b) o (a * c) = | a – b | o | a – c |
= | a – b | ∴ a * (b o c) = (a * b) o (a * c)
Also a o(b * c) = a o | b – c | = a and (a o b) * (a o c) = a * a = | a – a | = 0 ⇒    a o (b * c) ≠ (a o b) *(a o c)
∴ ‘o’ is not distributive over ‘*’.

Let E ∈ P(X) be an identity element, then
A * E = E * A = A for all A ∈ P(X)
⇒    A ∩ E = E ∩ A = A for all A ∈ P(X)
⇒ X ∩ E = X as X ∈ P(X)
⇒    X ⊂ E
Also    E ⊂ X as E ∈ P(X)
∴ E = X
∴ X is the identity element.
Let A ∈ P(X) be invertible, then there exists B ∈ P(X) such that A * B = B * A = X, the identity element.
⇒    A ∩ B =B ∩ A = X
⇒    X ⊂ A and also X ⊂ B
Also. A, B C X as A, B ∈ P(X)
∴ A = X = B
∴ X is the only invertible element and X^–1 = B = X.