Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(a) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(a)

Question 1.

If A = {a,b,c,d} mention the type of relations on A given below, which of them are equivalence relations?

(i) {(a, a), (b, b)}

(ii) {(a, a), (b, b), (c, c), (d, d)}

(iii) {(a, b), (b, a), (b, d), (d, b)}

(iv) {(b, c), (b, d), (c, d)}

(v) {(a, a), (b, b), (c, c), (d, d), (a, d), (a, c), (d, a), (c, a), (c, d), (d, c)}

Solution:

(i) Symmetric and transitive but not reflexive.

(ii) Reflexive, symmetric as well as transitive. Hence it is an equivalence relation.

(iii) Only symmetric

(iv) Only transitive

(v) Reflexive, symmetric and transitive. Hence it is an equivalence relation.

Question 2.

Write the following relations in tabular form and determine their type.

(i) R = {(x, y) : 2x – y = 0] on A = {1,2,3,…, 13}

(ii) R = {(x, y) : x divides y} on A = {1,2,3,4,5,6}

(iii) R = {(x, y) : x divides 2 – y} on A = {1,2,3,4,5}

(iv) R = {(x, y) : y ≤, x ≤, 4} on A = {1,2,3,4,5}.

Solution:

(i) R = {(x, y) : 2x- y = 0} on A

= {(x, y) : y = 2x} on A

= {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10), (6, 12)}

R is neither reflexive nor symmetric nor transitive.

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

R is reflexive transitive but not symmetric.

(iii) R = {(x, y) : x divides 2 – y} on A

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

= {(x, y) : 2-y is a multiple of x}

= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 4), (3, 2), (3, 5), (4, 2), (5, 2)}

R is neither reflexive nor symmetric nor transitive.

(iv) R = {(x, y) : y ≤ x ≤ 4} on A

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

= {(1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (4, 4)}

R is neither reflexive nor symmetric but transitive.

Question 3.

Test whether the following relations are reflexive, symmetric or transitive on the sets specified.

(i) R = {(m,n) : m-n ≥ 7} on Z.

(ii) R = {(m,n) : 2|(m+n)} on Z.

(iii) R = {(m,n) : m+n is not divisible by 3} Z.

(iv) R = {(m,n) : is a power of 5} on Z – {0}.

(v) R = {(m,n) : mn is divisible by 2} on Z.

(vi) R = {(m,n) : 3 divides m-n} on {1,2,3…,10}.

Solution:

(i) R = {{m, n) : m- n ≥ 7} on Z

Reflexive:

∀ m ∈ Z, m – m = 0 < 7

⇒ (m, m) ∉ R

Thus, R is not reflexive.

Symmetry:

Let (m, n) ∈ R

⇒ m – n ≥ 7

⇒ n – m < 7

∴ (n, m) ∉ R

⇒ R is not symmetric.

Transitive:

Let (m, n), (n, p) ∈ R

m – n ≥ 7

and n – p > 7

⇒ m – p ≥ 7

⇒ (m, p) ∈ R

⇒ R is transitive.

(ii) R = {(m, n) : 2 | (m + n)} on Z

Reflexive:

∀ m ∈ Z, m + m = 2m

which is divisible by 2.

⇒ 2 | (m + m)

⇒ (m, m) ∈ R

⇒ R is reflexive.

Symmetry:

Let (m, n) ∈ R

⇒ 2 | (m + n)

⇒ 2 | (n + m)

(n, m) ∈ R

⇒ R is symmetric.

Transitive:

Let (m, n), (n, p), ∈ R

⇒ 2 | (m + n) and 2 | (n + p)

⇒ m + n = 2k_{1}

⇒ n + p = 2k_{2}

⇒ m + 2n + p = 2k_{1} + 2k_{2}

⇒ m + p = 2(k_{1} + k_{2 }– 1)

⇒ 2 | (m + p)

⇒ (m, p) ∈ R

⇒ R is transitive.

Thus, R is an equivalence relative.

(iii) R = {(m, n) : m + n is not divisible by 3} on Z

Reflexive:

As 3 + 3 is divisible by 3

we have (3, 3) ∉ R

⇒ R is not reflexive.

Symmetric:

Let (m, n) ∈ R

⇒ m + n is not divisible by 3

⇒ n + m is not divisible by 3

⇒ (n, m) ∈ R

⇒ R is symmetric.

Transitive:

(3, 1), (1, 6) ∈ R

But (3, 6) ∉ R

⇒ R is not transitive.

(iv) R = {(m, n) : \(\frac{m}{n}\) is a power of 5} on Z – {0}

Reflexive:

∀ m ∈ Z – {0}

\(\frac{m}{m}\) = 1 = 5°

⇒ (m, m) ∈ R

⇒ R is reflexive.

Symmetric:

Let (m, n) ∈ R

\(\frac{m}{n}\) = 5^{k}

\(\frac{n}{m}\) = 5^{-k}

⇒ (n, m) ∈ Z

⇒ R is symmetric.

Transitive:

Let (m, n), (n, p) ∈ R

⇒ \(\frac{m}{n}\) = 5^{k1} , \(\frac{n}{p}\) = 5^{k2}

⇒ \(\frac{m}{n}\) . \(\frac{n}{p}\) = 5^{k1} . 5^{k2}

⇒ \(\frac{m}{p}\) = 5 ^{k1+k2}

⇒ (m, p) ∈ R

⇒ R is transitive.

Thus R is an equivalence relation.

(v) R = {(m, n) : mn is divisible by 2} on Z

Reflexive:

3 ∈ Z

3 x 3 = 9

which is not divisible by 2.

∴ (3, 3) ∉ R

⇒ R is not reflexive.

Symmetric:

Let (m, n) ∈ R

⇒ mn is divisible by 2

⇒ nm is divisible by 2

⇒ (n, m) ∈ R

⇒ R is symmetric.

Transitive:

⇒ (3, 2), (2, 5) ∈ R

⇒ But 3 x 5 = 15,

⇒ which is not divisible by 2.

⇒ (3, 5) ∉ R

R is not transitive.

(vi) R = {(m, n) : 3 divides m-n} on A = {1, 2, 3……,10}

Reflexive:

Clearly ∀ m ∈ A, m – m = 0

which is divisible by 3

⇒ (m, m) ∈ R

⇒ R is reflexive

Symmetric:

Let (m, n) ∈ R

⇒ m – n is divisible by 3

⇒ n – m is also divisible by 3

⇒ (n, m) ∈ R

⇒ R is symmetric

Transitive:

Let (m, n), (n, p) ∈ R

⇒ m – n and n – p are divisible by 3

⇒ m – n + n – p is also divisible by p.

⇒ m – p is divisible by p.

⇒ (m, p) ∈ R

⇒ R is transitive.

Thus R is an equivalence relation.

Question 4.

List the members of the equivalence relation defined by the following partitions on X= {1,2,3,4}. Also find the equivalence classes of 1,2,3 and 4.

(i) {{1},{2},{3, 4}}

(ii) {{1, 2, 3},{4}}

(iii) {{1,2, 3, 4}}

Solution:

(i) The equivalence relation is

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

[1] = {1}, [2] = {2}, [3] = {3, 4} and [4] = {3, 4}

(ii) The equivalence relation is

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

[1] = [2] = [3] = {1, 2, 3}

[4] = {4}

(iii) The equivalence relation is

R = A x A, [1] = [2] = [3] = [4] = A

Question 5.

Show that if R is an equivalence relation on X then dom R = rng R = X.

Solution:

Let R is an equivalence relation on X.

⇒ R is reflexive

⇒ (x, x) ∈ R ∀ x ∈ X

⇒ Dom R = Rng R = X

Question 6.

Give an example of a relation which is

(i) reflexive, symmetric but not transitive.

(ii) reflexive, transitive but not symmetric.

(iii) symmetric, transitive but not reflexive.

(iv) reflexive but neither symmetric nor transitive.

(v) transitive but neither reflexive nor symmetric.

(vi) an empty relation.

(vii) a universal relation.

Solution:

(i) The relation R = {(a, b), (b, a), (a, c), (c, a), (a, a), (b, b), (c, c)} defined on the set {a, b, c} is reflexive, symmetric but not transitive.

(ii) “The relation x ≤ y on z” is reflexive, transitive but not symmetric.

(iii) The relation R = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)} defined on the set {a, b, c, d} is symmetric, transitive but not reflexive.

(iv) The relation R = {(a, a), (b, b), (c, c), (a, b), (b, c)} defined on the set A = {a, b, c} is reflexive but neither symmetric nor transitive.

(v) R = {(a, b), (b, c), (a, c)} on A = {a, b, c} is transitive but neither reflexive nor symmetric.

(vi) On N the relation R= {(x, y) : x + y = – 5} is an empty relation.

(vii) On N the relation R = {(x, y) : x + y > 0} is an universal relation.

Question 7.

Let R be a relation on X, If R is symmetric then xRy ⇒ yRx. If it is also transitive then xRy and yRx ⇒ xRx. So whenever a relation is symmetric and transitive then it is also reflexive. What is wrong in this argument?

Solution:

Let R is a relation on X.

If R is symmetric then xRy ⇒ yRx

If R is also transitive then xRy and yRx ⇒ xRx

⇒ Whenever a relation is symmetric and transitive, then it is reflexive. This argument is wrong because the symmetry of R does not imply dom R = X and for reflexive xRx ∀ x ∈ X.

Question 8.

Suppose a box contains a set of n balls (n ≥ 4) (denoted by B) of four different colours (may have different sizes), viz. red, blue, green and yellow. Show that a relation R defined on B as R={(b1, b2): balls b1 and b2 have the same colour} is an equivalence relation on B. How many equivalence classes can you find with respect to R?

[Note: On any set X a relation R={(x, y): x and y satisfy the same property P} is an equivalence relation. As far as the property P is concerned, elements x and y are deemed equivalent. For different P we get different equivalence relations on X]

Solution:

On B, R = {(b1, b2) : balls b1 and b2 have the same colour}

Reflexive:

∀ b ∈ B, b and b are of same colour

⇒ (b, b) ∈ R

⇒ R is reflexive.

Symmetric:

Let (b1, b2) ∈ R

⇒ b1 and b2 are of same colour

⇒ b2 and b1 are of same colour

⇒ (b2, b1) ∈ R

⇒ R is symmetric.

Transitive :

Let (b1, b2) and (b2, b3) ∈ R

⇒ b1 and b2 are of same colour

b2 and b3 are of same colour

⇒ b1, b3 are of same colour

⇒ (b1, b3) ∈ R

⇒ R is transitive

∴ R is an equivalence relation.

As there are 4 types of balls there are 4 equivalence relations with respect to R.

Question 9.

Find the number of equivalence relations on X={1,2,3}. [Hints: Each partition of a set gives an equivalence relation.]

Solution:

Method – 1: Number of equivalence relations on a set A with | A | = n.

= The number of distinct partitions of A

= B_{n}

where B_{n+1} = \(\sum_{k=0}^n \frac{n !}{k !(n-k) !} \mathrm{B}_k\)

with B_{0 }= 1

Here n = 3

B_{1} = 1

B_{2 = }\(\frac{1 !}{0 ! 1 !}\) B_{0 }+ \(\frac{1 !}{1 ! 1 !}\) B_{1}

= 1 + 1 = 2

B_{3} = \(\frac{2 !}{0 ! 2 !}\) B_{0} + \(\frac{2 !}{1 ! 1 !}\) B_{1} + \(\frac{2 !}{2 ! 0 !}\) B_{2}

= 1 + 2 + 2 = 5

Thus there are 5 equivalence relations.

Method – 2:

X= {1, 2, 3}

Number of equivalence relations = number of distinct partitions.

Different partitions of X are

{{1} {2}, {3}}

{{1}, {2, 3}}, {{2}, {1,3}},

{{3}, {1,2}} and {{1, 2,3}}

Thus number of equivalence relations = 5.

Question 10.

Let R be the relation on the set R of real numbers such that aRb iff a-b is an integer. Test whether R is an equivalence relation. If so find the equivalence class of 1 and ½ w.r.t. this equivalence relation.

Solution:

The relation R on the set of real numbers is defined as

R = {(a, b) : a – b ∈ Z}

Reflexive:

∀ a ∈ R (set of real numbers)

a – a = 0 ∈ Z

⇒ (a, a) ∈ R

⇒ R is reflexive.

Symmetric:

Let (a, b) ∈ R

⇒ a – b ∈ Z

⇒ b – a ∈ Z

⇒ (b, a) ∈ R

⇒ R is symmetric.

Transitive:

Let (a, b), (b, c) ∈ R

⇒ a – b and b – c ∈ Z

⇒ a – b + b – c ∈ Z

⇒ a – c ∈ Z

⇒ (a, c) ∈ R

⇒ R is transitive.

Thus R is an equivalence relation.

[1] = {x ∈ R : x -1 ∈ Z} = Z

\(\begin{aligned}

{\left[\frac{1}{2}\right] } &=\left\{x \in \mathrm{R}: x-\frac{1}{2} \in \mathrm{Z}\right\} \\

&=\left\{x \in \mathrm{R}: x=\frac{2 k+1}{2}, k \in \mathrm{Z}\right\}

\end{aligned}\)

Question 11.

Find the least positive integer r such that

(i) 185 ∈ [r]_{7}

(ii) – 375 ∈ [r]_{11}

(iii) -12 ∈ [r]_{13}

Solution:

(i) 185 ∈ [r]_{7}

⇒ 185 – r = 7k, k ∈ z and r < 7

⇒ r = 3

(ii) – 375 ∈ [r]_{7}

⇒ – 375 – r = 11k, k ∈ z and r < 11

⇒ r = 10

(iii) – 12 ∈ [r]_{13}

⇒ – 12 – r = 13k, k ∈ z and r < 13

⇒ r= 1

Question 12.

Find least non negative integer r such that

(i) 7 x 13 x 23 x 413 r (mod 11)

(ii) 6 x 18 x 27 x (- 225) = r (mod 8)

(iii) 1237(mod 4) + 985 (mod 4) = r (mod 4)

(iv) 1936 x 8789 = r (mod 4)

Solution:

(i) 7 x 13 x 23 x 413 ≡ r (mod 11)

Now 7 x 13 ≡ 3 mod 11

23 ≡ 1 mod 11

413 ≡ 6 mod 11

∴ 7 x 13 x 23 x 413 ≡ 3 x 1 x 6 mod 11

≡ 18 mod 11

≡ 7 mod 11

∴ r = 7

(ii) 6 x 18 x 27 x – 225 ≡ r (mod 8)

Now 6 x 18 ≡ 108 = 4 mod 8

27 ≡ 3 mod 8

– 225 ≡ 7 mod 8

⇒ 6 x 18 x 27 x – 225 ≡ 4 x 3 x 7 mod 8

≡ 84 mod 8

≡ 4 mod 8

∴ r = 4

(iii) 1237 (mod 4) + 985 (mod 4) r (mod 4)

Now 1237 ≡ 1 mod 4

985 ≡ 1 mod 4

⇒ 1237 (mod 4) + 985 (mod 4)

≡ (1 + 1) mod 4

≡ 2 mod 4

⇒ r = 2

(iv) 1936 x 8789 ≡ r (mod 4)

1936 x 8789 ≡ 0 mod 4

∴ r = 0

Question 13.

Find least positive integer x satisfying 276x + 128 ≡ (mod 7)

[Hint: 276 ≡ 3, 128 ≡ 2 (mod 7)]

Solution:

Now 128 ≡ 2 mod 7

Now 176 x + 128 ≡ 4 mod 7

⇒ 176 x ≡ (4 – 2) mod 7

⇒ 176 x ≡ 2 mod 7

176 x x ≡ 2 mod 7,

But 276 ≡ 3 mod 7

Thus x = 3.

Question 14.

Find three positive integers x_{i}, i =1, 2, 3 satisfying 3x ≡ 2 (mod 7)

[Hint: If X_{1 }is a solution then any member of [X_{1}] is also a solution]

Solution:

3x ≡ 2 mod 7

Least positive value of x ≡ 3

Each member of [3] is a solution

∴ x = 3, 10, 17 …..