Exercise 1.1 : Solutions of Questions on Page Number : 5
Q1 : 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) ∈R, 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 ≠ 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,
(6, 1) ∉ 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) ∈R 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.
Q3 : 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)}
We can find (a, a) ∉ 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) ∈ 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.
Q4 : 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) ∈ 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.
Q6 : 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) ∉R.
∴ 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.
Q7 : 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
∴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
∴R is transitive.
Hence, R is an equivalence relation.
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
Now, all elements of the set {1, 3, 5} 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.
Q10 : Given 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 = {5, 6, 7}.
Define a relation R on A as R = {(5, 6), (6, 5)}.
Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.
Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.
(5, 6), (6, 5) ∈ R, but (5, 5) ∉ R
∴R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.
(ii)Consider a relation R in R defined as:
R= {(a, b): a < b}
For any a ∈ R, we have (a, a) ∉ R since a cannot be strictly less than a itself. In fact, a = a.
∴ R is not reflexive.
Now,
(1, 2) ∈ R (as 1 < 2)
But, 2 is not less than 1.
∴ (2, 1) ∉ R
∴ R is not symmetric.
Now, let (a, b), (b, c) ∈ R.
⇒ a < b and b < c
⇒ a < c
⇒ (a, c) ∈ R
∴R is transitive.
Hence, relation R is transitive but not reflexive and symmetric.
(iii)Let A = {4, 6, 8}.
Define a relation R on A as:
A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}
Relation R is reflexive since for every a ∈ A, (a, a) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.
Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R.
Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.
Hence, relation R is reflexive and symmetric but not transitive.
(iv) Define a relation R in R as:
R = {a, b): a3 ≥ b3}
Clearly (a, a) ∈ R as a3 = a3.
∴R is reflexive.
Now,
(2, 1) ∈ R (as 23 ≥ 13)
But,
(1, 2) ∉ R (as 13 < 23)
∴ R is not symmetric.
Now,
Let (a, b), (b, c) ∈ R.
⇒ a3 ≥ b3 and b3 ≥ c3
⇒ a3 ≥ c3
⇒ (a, c) ∈ R
∴R is transitive.
Hence, relation R is reflexive and transitive but not symmetric.
(v) Let A = { – 5, – 6}.
Define a relation R on A as:
R = {( – 5, – 6), ( – 6, – 5), ( – 5, – 5)}
Relation R is not reflexive as ( – 6, – 6) ∉ R.
Relation R is symmetric as ( – 5, – 6) ∈ R and ( – 6, – 5}∈R.
It is seen that ( – 5, – 6), ( – 6, – 5) ∈ R. Also, ( – 5, – 5) ∈ R.
∴The relation R is transitive.
Hence, relation R is symmetric and transitive but not reflexive.
Q11 : 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 point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Answer :
R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}
Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
∴R is reflexive.
Now,
Let (P, Q) ∈ R.
⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.
⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.
⇒ (Q, P) ∈ R
∴R is symmetric.
Now,
Let (P, Q), (Q, S) ∈ R.
⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.
⇒ The distance of points P and S from the origin is the same.
⇒ (P, S) ∈ R
∴R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.
Then, triangle T1 is similar to triangle T3.
Hence, T1 is related to T3.
Q13 : Show that the relation R defined in the set A of all polygons as R = {(P1, P2): P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
Answer :
R = {(P1, P2): P1 and P2 have same the number of sides}
R is reflexive since (P1, P1) ∈ R as the same polygon has the same number of sides with itself.
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
∴R is symmetric.
Now,
Let (P1, P2), (P2, P3) ∈ R.
⇒ P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides.
⇒ P1 and P3 have the same number of sides.
⇒ (P1, P3) ∈ R
∴R is transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.
Q14 : Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
Answer :
R = {(L1, L2): L1 is parallel to L2}
R is reflexive as any line L1 is parallel to itself i.e., (L1, L1) ∈ R.
Now,
Let (L1, L2) ∈ R.
⇒ L1 is parallel to L2.
⇒ L2 is parallel to L1.
⇒ (L2, L1) ∈ R
∴ R is symmetric.
Now,
Let (L1, L2), (L2, L3) ∈R.
⇒ L1 is parallel to L2. Also, L2 is parallel to L3.
⇒ L1 is parallel to L3.
∴R is transitive.
Hence, R is an equivalence relation.
The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y = 2x + 4.
Slope of line y = 2x + 4 is m = 2
It is known that parallel lines have the same slopes.
The line parallel to the given line is of the form y = 2x + c, where c ∈R.
Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.
Q15 : Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Answer :
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
It is seen that (a, a) ∈ R, for every a ∈{1, 2, 3, 4}.
∴ R is reflexive.
It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.
∴R is not symmetric.
Also, it is observed that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.
The correct answer is B.
Q16 : Let R be the relation in the set N given by R = {(a, b): a = b – 2, b > 6}. Choose the correct answer.
(A) (2, 4) ∈ R (B) (3, 8) ∈R (C) (6, 8) ∈R (D) (8, 7) ∈ R
Answer :
R = {(a, b): a = b – 2, b > 6}
Now, since b > 6, (2, 4) ∉ R
Also, as 3 ≠ 8 – 2, (3, 8) ∉ R
And, as 8 ≠ 7 – 2
(8, 7) ∉ R
Now, consider (6, 8).
We have 8 > 6 and also, 6 = 8 – 2.
∴(6, 8) ∈ R
The correct answer is C.
Exercise 1.2 : Solutions of Questions on Page Number : 10
Q2 : Check the injectivity and surjectivity of the following functions:
(i) f: N → N given by f(x) = x2
(ii) f: Z → Z given by f(x) = x2
(iii) f: R → R given by f(x) = x2
(iv) f: N → N given by f(x) = x3
(v) f: Z → Z given by f(x) = x3
Answer :
(i) f: N → N is given by,
f(x) = x2
It is seen that for x, y ∈N, f(x) = f(y) ⇒ x2 = y2 ⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
(ii) f: Z → Z is given by,
f(x) = x2
It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.
∴ f is not injective.
Now,-2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x2 = -2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iii) f: R → R is given by,
f(x) = x2
It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.
∴ f is not injective.
Now,-2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x2 = -2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iv) f: N → N given by,
f(x) = x3
It is seen that for x, y ∈N, f(x) = f(y) ⇒ x3 = y3 ⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2.
∴ f is not surjective
Hence, function f is injective but not surjective.
(v) f: Z → Z is given by,
f(x) = x3
It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x3 = y3 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
Q3 : Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-once nor onto, where [x] denotes the greatest integer less than or equal to x.
Answer :
f: R → R is given by,
f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one-one.
Now, consider 0.7 ∈ R.
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.
Q6 : Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Answer :
It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f (1) = 4, f (2) = 5, f (3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.
Exercise 1.3 : Solutions of Questions on Page Number : 18
Q5 : State with reason whether following functions have inverse
(i) f: {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Answer :
(i) f: {1, 2, 3, 4} → {10}defined as:
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10
∴f is not one-one.
Hence, function f does not have an inverse.
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4.
∴g is not one-one,
Hence, function g does not have an inverse.
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.
∴Function h is one-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y.
Thus, h is a one-one and onto function. Hence, h has an inverse.
Q12 : Let f: X → Y be an invertible function. Show that the inverse of f-1 is f, i.e.,
(f-1)-1 = f.
Answer :
Let f: X → Y be an invertible function.
Then, there exists a function g: Y → X such that gof = IXand fog = IY.
Here, f-1 = g.
Now, gof = IXand fog = IY
⇒ f-1of = IXand fof-1= IY
Hence,f-1: Y → X is invertible and f is the inverse of f-1
i.e., (f-1)-1 = f.
Exercise 1.4 : Solutions of Questions on Page Number : 24
Q1 : Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.
(i) On Z+, define * by a * b = a – b
(ii) On Z+, define * by a * b = ab
(iii) On R, define * by a * b = ab2
(iv) On Z+, define * by a * b = |a – b|
(v) On Z+, define * by a * b = a
Answer :
(i) On Z+, * is defined by a *b = a – b.
It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 – 2
= -1 ∉ Z+.
(ii) On Z+, * is defined by a *b = ab.
It is seen that for each a, b ∈ Z+, there is a unique element ab in Z+.
This means that * carries each pair (a, b) to a unique element a *b = ab in Z+.
Therefore, * is a binary operation.
(iii) On R, * is defined by a *b = ab2.
It is seen that for each a, b ∈ R, there is a unique element ab2 in R.
This means that * carries each pair (a, b) to a unique element a *b = ab2 in R.
Therefore, * is a binary operation.
(iv) On Z+, * is defined by a *b = |a – b|.
It is seen that for each a, b ∈ Z+, there is a unique element |a – b| in Z+.
This means that * carries each pair (a, b) to a unique element a *b =
|a – b| in Z+.
Therefore, * is a binary operation.
(v) On Z+, * is defined by a *b = a.
* carries each pair (a, b) to a unique element a *b = a in Z+.
Therefore, * is a binary operation.
Q3 : Consider the binary operation ∨ on the set {1, 2, 3, 4, 5} defined by a ∨b = min {a, b}. Write the operation table of the operation∨.
Answer :
The binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as a ∨b = min {a, b}” a, b ∈ {1, 2, 3, 4, 5}.
Thus, the operation table for the given operation ∨ can be given as:
| v | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 2 | 2 | 2 |
| 3 | 1 | 2 | 3 | 3 | 3 |
| 4 | 1 | 2 | 3 | 4 | 4 |
| 5 | 1 | 2 | 3 | 4 | 5 |
Q4 : Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.
(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Is * commutative?
(iii) Compute (2 * 3) * (4 * 5).
(Hint: use the following table)
| * | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 1 | 2 | 1 |
| 3 | 1 | 1 | 3 | 1 | 1 |
| 4 | 1 | 2 | 1 | 4 | 1 |
| 5 | 1 | 1 | 1 | 1 | 5 |
Answer :
(i) (2 * 3) * 4 = 1 * 4 = 1
2 * (3 * 4) = 2 * 1 = 1
(ii) For every a, b ∈{1, 2, 3, 4, 5}, we have a *b = b *a. Therefore, the operation * is commutative.
(iii) (2 * 3) = 1 and (4 * 5) = 1
∴(2 * 3) * (4 * 5) = 1 * 1 = 1
Q5 : Let*”² be the binary operation on the set {1, 2, 3, 4, 5} defined by a *”² b = H.C.F. of a and b. Is the operation *”² same as the operation * defined in Exercise 4 above? Justify your answer.
Answer :
The binary operation *“² on the set {1, 2, 3 4, 5} is defined as a *“² b = H.C.F of a and b.The operation table for the operation *“² can be given as:
| *“² | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 1 | 2 | 1 |
| 3 | 1 | 1 | 3 | 1 | 1 |
| 4 | 1 | 2 | 1 | 4 | 1 |
| 5 | 1 | 1 | 1 | 1 | 5 |
We observe that the operation tables for the operations * and *“² are the same.
Thus, the operation *“² is same as the operation*.
Q6 : Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find
(i) 5 * 7, 20 * 16 (ii) Is * commutative?
(iii) Is * associative? (iv) Find the identity of * in N
(v) Which elements of N are invertible for the operation *?
Answer :
The binary operation * on N is defined as a *b = L.C.M. of a and b.
(i) 5 * 7 = L.C.M. of 5 and 7 = 35
20 * 16 = L.C.M of 20 and 16 = 80
(ii) It is known that:
L.C.M of a and b = L.C.M of b and a ” a, b ∈ N.
∴a * b = b *a
Thus, the operation * is commutative.
(iii) For a, b, c ∈ N, we have:
(a *b) *c = (L.C.M of a and b) * c = LCM of a, b, and c
a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c
∴(a *b) * c = a * (b *c)
Thus, the operation * is associative.
(iv) It is known that:
L.C.M. of a and 1 = a = L.C.M. 1 and a ” a ∈ N
⇒ a * 1 = a = 1 * a ” a ∈ N
Thus, 1 is the identity of * in N.
(v) An element a in N is invertible with respect to the operation * if there exists an element b in N, such that a *b = e = b *a.
Here, e = 1
This means that:
L.C.M of a and b = 1 = L.C.M of b and a
This case is possible only when a and b are equal to 1.
Thus, 1 is the only invertible element of N with respect to the operation *.
Q7 : Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.
Answer :
The operation * on the set A = {1, 2, 3, 4, 5} is defined as
a * b = L.C.M. of a and b.
Then, the operation table for the given operation * can be given as:
| * | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 2 | 3 | 4 | 5 |
| 2 | 2 | 2 | 6 | 4 | 10 |
| 3 | 3 | 6 | 3 | 12 | 15 |
| 4 | 4 | 4 | 12 | 4 | 20 |
| 5 | 5 | 10 | 15 | 20 | 5 |
It can be observed from the obtained table that:
3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A
3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A
Hence, the given operation * is not a binary operation.
Q8 : Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
Answer :
The binary operation * on N is defined as:
a *b = H.C.F. of a and b
It is known that:
H.C.F. of a and b = H.C.F. of b and a ” a, b ∈ N.
∴a * b = b *a
Thus, the operation * is commutative.
For a, b, c ∈ N, we have:
(a *b)*c = (H.C.F. of a and b) * c = H.C.F. of a, b, and c
a *(b *c)= a *(H.C.F. of b and c) = H.C.F. of a, b, and c
∴(a *b) * c = a * (b *c)
Thus, the operation * is associative.
Now, an element e ∈ N will be the identity for the operation * if a *e = a = e*a a ∈ N.
But this relation is not true for any a ∈ N.
Thus, the operation * does not have any identity in N.
Q10 : Find which of the operations given above has identity.
Answer :
An element e ∈ Q will be the identity element for the operation * if
a * e = a = e * a, a ∈ Q.
We are given
Q11 : Let A = N x N and * be the binary operation on A defined by
(a, b) * (c, d) = (a + c, b + d)
Show that * is commutative and associative. Find the identity element for * on A, if any.
Answer :
A = N × N
* is a binary operation on A and is defined by:
(a, b) * (c, d) = (a + c, b + d)
Let (a, b), (c, d) ∈ A
Then, a, b, c, d ∈ N
We have:
(a, b) * (c, d) = (a + c, b + d)
(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
[Addition is commutative in the set of natural numbers]
∴(a, b) * (c, d) = (c, d) * (a, b)
Therefore, the operation * is commutative.
Now, let (a, b), (c, d), (e, f) ∈A
Then, a, b, c, d, e, f ∈ N
We have:
Q12 : State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation * on a set N, a * a = a a * N.
(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a
Answer :
(i) Define an operation * on N as:
a *b = a + b a, b ∈ N
Then, in particular, for b = a = 3, we have:
3 * 3 = 3 + 3 = 6 ≠ 3
Therefore, statement (i) is false.
(ii) R.H.S. = (c *b) * a
= (b *c) * a [* is commutative]
= a * (b *c) [Again, as * is commutative]
= L.H.S.
∴ a * (b *c) = (c *b) * a
Therefore, statement (ii) is true.
Q13 : Consider a binary operation * on N defined as a * b = a3 + b3. Choose the correct answer.
(A) Is * both associative and commutative?
(B) Is * commutative but not associative?
(C) Is * associative but not commutative?
(D) Is * neither commutative nor associative?
Answer :
On N, the operation * is defined as a *b = a3 + b3.
For, a, b, ∈ N, we have:
a *b = a3 + b3 = b3 + a3 = b *a [Addition is commutative in N]
Therefore, the operation * is commutative.
It can be observed that:
Exercise Miscellaneous : Solutions of Questions on Page Number : 29
Q1 : Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that g o f = f o g = 1R.
Answer :
It is given that f: R → R is defined as f(x) = 10x + 7.
One-one:
Let f(x) = f(y), where x, y ∈R.
⇒ 10x + 7 = 10y + 7
⇒ x = y
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 10x + 7.
Therefore, for any y ∈ R, there exists such that
∴ f is onto.
Therefore, f is one-one and onto.
Thus, f is an invertible function.
Let us define g: R → R as
Now, we have:
Hence, the required function g: R→ R is defined as.
Q2: Let f: W → W be defined as f(n) = n – 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.
Answer :
It is given that:
f: W→W is defined as
One-one:
Let f(n) = f(m).
It can be observed that if n is odd and m is even, then we will have n – 1 = m + 1.
⇒ n – m = 2
However, this is impossible.
Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.
∴Both n and m must be either odd or even.
Now, if both n and m are odd, then we have:
f(n) = f(m) ⇒ n – 1 = m – 1 ⇒ n = m
Again, if both n and m are even, then we have:
f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m
∴f is one-one.
It is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N and any even number 2r in co-domain N is the image of 2r + 1 in domain N.
∴f is onto.
Q3: If f: R → R is defined by f(x) = x2 – 3x + 2, find f(f(x)).
Answer :
It is given that f: R → R is defined as f(x) = x2 – 3x + 2.
Q4 : Show that function f: R →{x ∈ R: – 1 < x < 1} defined by f(x) =
, x ∈R is one-one and onto function.
Answer :
It is given that f: R →{x ∈ R: – 1 < x < 1} is defined as f(x) =, x ∈R.
Suppose f(x) = f(y), where x, y ∈ R.
It can be observed that if x is positive and y is negative, then we have:
Since x is positive and y is negative:
x > y ⇒ x – y > 0
But, 2xy is negative.
Then,
.
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out
x and y have to be either positive or negative.
When x and y are both positive, we have:
When x and y are both negative, we have:
∴ f is one-one.
Now, let y ∈ R such that – 1 < y < 1.
If x is negative, then there exists 
such that
If x is positive, then there exists
such that
∴ f is onto.
Hence, f is one-one and onto.
Q5 : Show that the function f: R → R given by f(x) = x3 is injective.
Answer :
f: R→ R is given as f(x) = x3.
Suppose f(x) = f(y), where x, y ∈ R.
⇒ x3 = y3 … (1)
Now, we need to show that x = y.
Suppose x ≠ y, their cubes will also not be equal.
x3 ≠ y3
However, this will be a contradiction to (1).
∴ x = y
Hence, f is injective.
Q6 : Give examples of two functions f: N → Z and g: Z→Z such that g o f is injective but g is not injective.
(Hint: Consider f(x) = x and g(x) =|x|)
Answer :
Define f: N→Z as f(x) = x and g: Z→Z as g(x) =|x|.
We first show that g is not injective.
It can be observed that:
g( – 1) =|-1|=1
g(1) =|1|=
∴ g( – 1) = g(1), but – 1 ≠ 1.
∴ g is not injective.
Now, gof: N → Z is defined as gof(x)=g(f(x))=g(x)=|x|.
Let x, y ∈ N such that gof(x) = gof(y).
⇒ |x|=|y|
Since x and y ∈ N, both are positive.
∴|x|=|y|⇒x=y
Hence, gof is injective
Q7 : Given examples of two functions f: N→N and g: N→N such that gof is onto but f is not onto.
(Hint: Consider f(x) = x + 1 and 
Answer :
Define f: N→N by,
f(x) = x + 1
And, g: N→N by,
We first show that g is not onto.
For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N.
∴ f is not onto.
Now, gof: N→N is defined by,
Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y.
Hence, gof is onto.
Q8 : Given a non empty set X, consider P(X) which is the set of all subsets of X.
Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify you answer:
Answer :
Since every set is a subset of itself, ARA for all A ∈ P(X).
∴R is reflexive.
Let ARB ⇒ A ⊂ B.
This cannot be implied to B ⊂ A.
For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.
∴ R is not symmetric.
Further, if ARB and BRC, then A ⊂ B and B ⊂ C.
⇒ A ⊂ C
⇒ ARC
∴ R is transitive.
Hence, R is not an equivalence relation since it is not symmetric.
Q9 : Given a non-empty set X, consider the binary operation *: P(X) x P(X) → P(X) given by A * B = A ∩ B ” A, B in P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation*.
Answer :
It is given that
.
We know that
.
Thus, X is the identity element for the given binary operation *.
Now, an elementis A ∈ P(X) invertible if there exists
such that
This case is possible only when A = X = B.
Thus, X is the only invertible element in P(X) with respect to the given operation*.
Hence, the given result is proved.
Q10 : Find the number of all onto functions from the set {1, 2, 3, … , n) to itself.
Answer :
Onto functions from the set {1, 2, 3, … ,n} to itself is simply a permutation on n symbols 1, 2, …, n.
Thus, the total number of onto maps from {1, 2, … , n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n!.
Q11 : Let S = {a, b, c} and T = {1, 2, 3}. Find F-1 of the following functions F from S to T, if it exists.
(i) F= {(a, 3), (b, 2), (c, 1)} (ii) F= {(a, 2), (b, 1), (c, 1)}
Answer :
S = {a, b, c}, T = {1, 2, 3}
(i) F: S → T is defined as:
F = {(a, 3), (b, 2), (c, 1)}
⇒ F (a) = 3, F (b) = 2, F(c) = 1
Therefore, F-1: T → S is given by
F-1 = {(3, a), (2, b), (1, c)}.
(ii) F: S → T is defined as:
F = {(a, 2), (b, 1), (c, 1)}
Since F (b) = F (c) = 1, F is not one-one.
Hence, F is not invertible i.e., F-1 does not exist.
Q12 : Consider the binary operations*: R ×R → and o: R × R → R defined as a*b=|a-b| and a o b = a, “a, b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that “a, b, c ∈ R, a*(b o c) = (a * b) o (a * c). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
Answer :
It is given that *: R ×R → and o: R × R → R is defined as
a*b=|a-b| and a o b = a, “a, b ∈ R.
For a, b ∈ R, we have:
a*b = |a-b|
b*a = |b-a|= |-(a-b)|=|a-b|
∴ a * b = b * a
∴ The operation * is commutative.
It can be observed that,
∴The operation * is not associative.
Now, consider the operation o:
It can be observed that 1 o 2 = 1 and 2 o 1 = 2.
∴1 o 2 ≠ 2 o 1 (where 1, 2 ∈ R)
∴The operation o is not commutative.
Let a, b, c ∈ R. Then, we have:
(a o b) o c = a o c = a
a o (b o c) = a o b = a
⇒ a o b) o c = a o (b o c)
∴ The operation o is associative.
Now, let a, b, c ∈ R, then we have:
a * (b o c) = a * b =|a-b|
(a * b) o (a * c) =
Hence, a * (b o c) = (a * b) o (a * c).
Now,
1 o (2 * 3) =
(1 o 2) * (1 o 3) = 1 * 1 =|1-1|=0
∴1 o (2 * 3) ≠ (1 o 2) * (1 o 3) (where 1, 2, 3 ∈ R)
The operation o does not distribute over *.
Q13 : Given a non-empty set X, let *: P(X) x P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ” A, B ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A-1 = A. (Hint: (A – Φ) ∪ (Φ – A) = A and (A – A) ∪ (A – A) = A * A = Φ).
Answer :
It is given that *: P(X) × P(X)→ P(X) is defined as
A * B = (A – B) ∪ (B – A) ” A, B ∈ P(X).
Let A ∈ P(X). Then, we have:
A * Φ = (A – Φ) ∪ (Φ – A) = A ∪ Φ = A
Φ * A = (Φ – A) ∪ (A – Φ) = Φ ∪ A = A
∴A * Φ = A = Φ * A. ” A ∈ P(X)
Thus, Φ is the identity element for the given operation*.
Now, an element A ∈ P(X) will be invertible if there exists B ∈ P(X) such that
A * B = Φ = B * A. (As Φ is the identity element)
Now, we observed that.
Hence, all the elements A of P(X) are invertible with A – 1 = A.
Q14 : Define a binary operation *on the set {0, 1, 2, 3, 4, 5} as
Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.
Answer :
Let X = {0, 1, 2, 3, 4, 5}.
The operation * on X is defined as:
An element e ∈ X is the identity element for the operation *, if
Thus, 0 is the identity element for the given operation *.
An element a ∈ X is invertible if there exists b∈ X such that a * b = 0 = b * a.
i.e.
,a = – b or b = 6 – a
But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ – b.
∴b = 6 – a is the inverse of a ” a ∈ X.
Hence, the inverse of an element a ∈X, a ≠ 0 is 6 – a i.e., a – 1 = 6 – a.
Q15 : Let A = { – 1, 0, 1, 2}, B = { – 4, – 2, 0, 2} and f, g: A → B be functions defined by f(x) = x2 – x, x ∈ A and
. Are f and g equal?
Justify your answer. (Hint: One may note that two function f: A →B and g: A→B such that f(a) = g(a) “a ∈A, are called equal functions).
Answer :
It is given that A = { – 1, 0, 1, 2}, B = { – 4, – 2, 0, 2}.
Also, it is given that f, g: A→B are defined by f(x) = x2 – x, x ∈ A and
.
It is observed that:
Hence, the functions f and g are equal.
Q16 : Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(A) 1 (B) 2 (C) 3 (D) 4
Answer :
The given set is A = {1, 2, 3}.
The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transitive is given by:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}
This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.
Relation R is symmetric since (1, 2), (2, 1) ∈R and (1, 3), (3, 1) ∈R.
But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.
Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.
Hence, the total number of desired relations is one.
The correct answer is A.
Q17 : Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(A) 1 (B) 2 (C) 3 (D) 4
Answer :
It is given that A = {1, 2, 3}.
The smallest equivalence relation containing (1, 2) is given by,
R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and (3, 1).
If we odd any one pair [say (2, 3)] to R1, then for symmetry we must add (3, 2). Also, for transitivity we are required to add (1, 3) and (3, 1).
Hence, the only equivalence relation (bigger than R1) is the universal relation.
This shows that the total number of equivalence relations containing (1, 2) is two.
The correct answer is B.
Q18 : Let f: R → R be the Signum Function defined as
and g: R→R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1]?
Answer :
It is given that,
f: R → R is defined as
Also,g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.
Now, let x ∈ (0, 1].
Then, we have:
[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1.
Thus, when x ∈ (0, 1), we have fog(x) = 0 and gof (x) = 1.
Hence, fog and gof do not coincide in (0, 1].
Q19 : Number of binary operations on the set {a, b} are
(A) 10 (B) 16 (C) 20 (D) 8
Answer :
A binary operation * on {a, b} is a function from {a, b} x {a, b} → {a, b}
i.e., * is a function from {(a, a), (a, b), (b, a), (b, b)} → {a, b}.
Hence, the total number of binary operations on the set {a, b} is 24 i.e., 16.
The correct answer is B.