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QUESTION: 1

Let R = {(P, Q) : OP = OQ , O being the origin} be an equivalence relation on A. The equivalence class [(1, 2)] is

Solution:

**Correct Answer : a**

**Explanation : A = {(x,y) : x ^{2} + y^{2} = 5}**

**=> (1,2) {(1) ^{2} + (4)^{2} = 5}**

**=> {1 + 4 = 5}**

**=> {5 = 5}**

QUESTION: 2

Let a relation T on the set R of real numbers be T = {(a, b) : 1 + ab < 0, a, ∈ R}. Then from among the ordered pairs (1, 1), (1, 2), (1, -2), (2, 2), the only pair that belongs to T is________.

Solution:

Since T is a set of real number for it's given ordered pairs we have a condition provided that is 1 + ab is less than zero. so in the given option c if we put order of a and b then it satisfy our given condition

QUESTION: 3

For real number x and y, we write _{x}R_{y} ⇔ x-y + √2 is an irrational number. Then the relation R is:

Solution:

_{x}R_{y} => x - y + √2 is an irrational number.

Let R is a binary relation on real numbers x and y.

Clearly, R is reflexive relation

As _{x}R_{x} iff x – x +√2 = √2 ,which is an irrational number.

Here R is not symmteric if we take x =√2 and y =1 then x – y + √2 is an irrational

number but y – x + √2 = 1, which is not irrational number

Now, R is transitive iff for all (x, y) ∈ R and (y, z) ∈ R implies (x, z) ∈ R

But here R is not transitive as we take x = 1, y = 2√2, z=√2

Given, _{x}R_{y} => x - y + √2 is irrational ............1

and _{y}R_{z} => y - z + √2 is irrational ............2

Add equation 1 and 2, we get

(x - y + √2) + (y - z + √2)

= x - z + √2 = 1, which is not an irrational

QUESTION: 4

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is

Solution:

Let R={(1,3),(4,2),(2,4),(2,3),(3,1)} be a relation on the set A={1,2,3,4}, then

(a) Since (2,3) ∈ R but (3,2) ∈/ ′R, so R is not symmetric.

(b) Since (1,3) ∈ R and (3,1) ∈/ R but (1,1) ∈/ R, so R is not transitive:

(c) Since (1,1) ∈/ R, so R is not reflexive.

(d) Since (2,4) ∈ R and (2,3) ∈ R, so R is not a function.

QUESTION: 5

If A = {1, 3, 5, 7} and we define a relation R = {(a, b), a, b ∈ A: |a - b| = 8}. Then the number of elements in the relation R is

Solution:

Clearly there is no pair in set A whose difference is 8 or -8.

so D is the correct option.

QUESTION: 6

If A = {1, 3, 5, 7} and define a relation, such that R = {(a, b) a, b ∈ A : |a + b| = 8}. Then how many elements are there in the relation R

Solution:

Number of relations would be 4 as.. 1 + 7,7 + 1, 3 + 5, 5 + 3 all are equal to 8

QUESTION: 7

In the set N x N, the relation R is defined by (a, b) R (c, d) ⇔ a + d = b + c. Then R is

Solution:

(a, b) R (c, d) <=> a + d = b + c

Reflexive:

(a, b) R (a, b) <=> a + b = b + a

This is true for all (a, b) € N x N

Hence, it is reflexive.

Symmetric:

Let (a, b) R (c, d) <=> a + d = b + c

a + d = b + c

c + b = d + a (same)

By the above equation,

(c, d) R (a, b)

Hence,

(c, d) R (a, b) <=> c + b = d + a

Hence, it is symmetric

Transitive:

Let (a, b) R (c, d) <=> a + d = b + c, --- eqn 1

(c, d) R (e, f) <=> c + f = d + e --- eqn 2

for all a, b, c, d, e, f € N

eqn 1 : a + d = b + c

⇒ a - b = c - d

eqn 2 : c + f = d + e

⇒ c - d = e - f

So, a-b = e-f

⇒ a + f = b + e

⇒ (a, b) R (e, f)

Hence, it is transitive.

This is an equivalence relation

QUESTION: 8

If A = {1, 2, 3, 4} and B = {1, 3, 5} and R is a relation from A to B defined by(a, b) ∈ element of R ⇔ a < b. Then, R = ?

Solution:

A = {1, 2, 3, 4}

B = {1, 3, 5}

(a, b) ∈ element of R ⇔ a < b for all a ∈ A, b ∈ B

(a, b) pairs satisfying the condition of R are:

(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)

So,

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

QUESTION: 9

Let R be the relation on the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3,3), (3,2)}. then R is

Solution:

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)}

it is seen that (a, a) ∈ R for every a ∈ {1, 2, 3, 4}

so,R is reflexive.

it is seen that (a, b) = (b, a) ∈ R

because, (1, 2)∈ R but (2, 1) ∉ R

so, R is not symmetric.

it is seen that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.

so, R is transitive.

Hence, R is reflexive and transitive but not symmetric.

QUESTION: 10

Let A = {1, 2, 3, 4, 5, 6, 7}. P = {1, 2}, Q = {3, 7}. Write the elements of the set R so that P, Q and R form a partition that results in equivalence relation.

Solution:

P{1, 2} union Q {3, 7} union R{} = A {1, 2, 3, 4, 5, 6, 7}

therefore R = {4, 5, 6}

QUESTION: 11

Let R be a relation on set A of triangles in a plane.

R = {(T_{1}, T_{2}) : T_{1}, T_{2} element of A and T_{1} is congruent to T_{2}} Then the relation R is______

Solution:

It is equivalence relation...

as a triangle is congruent to itself that means (T_{1}, T_{1}) exist in relation which implies it is reflexive.

And also if T_{1} is congruent to T_{2} then T_{2} is also congruent to T_{1} as simple that means (T_{1},T_{2}) and (T_{2},T_{1}) both belongs to relation ...which implies it is symmetric.

And is T1 is congruent to T_{2} and T_{2} is congruent to T_{3}, then T_{1} is also congruent to T_{3}, congruency rule that means (T_{1},T_{2}), (T_{2},T_{3}) and (T_{1},T_{3}) belongs to relation which implies it transitive.

this relation is reflexive, symmetric, and transitive, hence it is equivalence relation.

QUESTION: 12

Which one of the following relations on set of real numbers is an equivalence relation?

Solution:

If |a| = |b | then |b| = |a| so this is symmetric as well as reflexive and if |a| = |b| and |b| = |c| then |c| = |a| then it is transitive as well so it is an equivalence relation.

QUESTION: 13

Let C = {(a, b): a^{2} + b^{2} = 1; a, b ∈ R} a relation on R, set of real numbers. Then C is

Solution:

**Correct Answer :- D**

**Explanation:- Check for reflexive **

**Consider (a,a)**

**∴ a ^{2}+a^{2}**

** =1 which is not always true.**

**If a=2**

**∴ 2 ^{2}+2^{2}**

** =1⇒4+4=1 which is false.**

**∴ R is not reflexive ---- ( 1 )**

**Check for symmetric**

**aRb⇒a ^{2}+b^{2}=1**

**bRa⇒b ^{2}+a^{2} =1**

**Both the equation are the same and therefore will always be true.**

**∴ R is symmetric ---- ( 2 )**

**Check for transitive**

**aRb⇒a ^{2}+b^{2}=1**

**bRc⇒b ^{2}+c^{2}=1**

**∴ a ^{2}+c^{2}=1 will not always be true.**

**Let a=−1,b=0 and c=1**

**∴ (−1) ^{2}+0^{2}=1, **

**= 0 ^{2}+1^{2}**

** =1 are true.**

**But (−1) ^{2}+1^{2}**

** =1 is false.**

**∴ R is not transitive ---- ( 3 )**

QUESTION: 14

Let A = {1, 2, 3, 4} and B = {x, y, z}. Then R = {(1, x), (2, z), (1, y), (3, x)} is

Solution:

Let a set of A = (1, 2, 3, 4) and set B (x, y, z) so. set A of all elements in set B then the relation of A to B

QUESTION: 15

Let R be a relation on N, set of natural numbers such that m R n ⇔ m divides n. Then R is

Solution:

Let there be a natural number n,

We know that n divides n, which implies nRn.

So, Every natural number is related to itself in relation R.

Thus relation R is reflexive .

Let there be three natural numbers a,b,c and let aRb,bRc

aRb implies a divides b and bRc implies b divides c, which combinedly implies that a divides c i.e. aRc.

So, Relation R is also transitive .

Let there be two natural numbers a,b and let aRb,

aRb implies a divides b but it can't be assured that b necessarily divides a.

For ex, 2R4 as 2 divides 4 but 4 does not divide 2 .

Thus Relation R is not symmetric .

QUESTION: 16

If R be a relation “less than” from set A = {1, 2, 3, 4} to B = {1, 3, 5}, i.e. (a, b) ∈ R if a < b, if (b, a) ∈ R^{-1}elements in R^{-1} are

Solution:

A = {1, 2, 3, 4}

B = {1, 3, 5}

(a, b) ∈ R if a < b

(b, a) ∈ R^{ - 1}

R ^{- 1} will have all (b, a) pairs where b < a, for all b ∈ B, a ∈ A

R ^{- 1} = {(3, 1), (3, 2), (5, 1), (5, 2), (5, 3), (5, 4)}

QUESTION: 17

Let R be a relation on a finite set A having n elements. Then, the number of relations on A is

Solution:

Number of relations on A =

Step-by-step explanation:

If there are n elements in set A then the total number of ordered pairs in the set A × A = n²

In other words A × A will have n² elements.

We also know that if a set has N elements then the number of subsets of A are 2^{n}

Therefore, for A × A there can as many relations as the number of subsets of A × A

The number of subsets of A × A =

Therefore the number of relations =

QUESTION: 18

Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q) mq(n + p) = np(m + q). Then, R is

Solution:

(m, n) R (p, q) <=> mq(n + p) = np(m + q)

For all m,n,p,q € N

Reflexive:

(m, n) R (m, n) <=> mn(n + m) = nm(m + n)

⇒ mn^{2} + m^{2}n = nm^{2} + n^{2}m

⇒ mn^{2} + m^{2}n = mn^{2} + m^{2}n

⇒ LHS = RHS

So, (m, n) R (m, n) exists.

Hence, it is Reflexive

Symmetric:

Let (m, n) R (p, q) exists

mq(n + p) = np(m + q) --- (eqn1)

(p, q) R (m, n) <=> pn(q + m) = qm (p + n)

⇒ np(m + q) = mq(n + p)

⇒ mq(n + p) = np(m + q)

This equation is true by (eqn1).

So, (p, q) R (m, n) exists

Hence, it is symmetric.

Transitive:

Let (m, n) R (p, q) and (p, q) R (r, s) exists.

Therefore,

mq(n + p) = np(m + q) --- (eqn1)

ps(q + r) = qr (p + s) --- (eqn2)

We cannot obtain ms(n+r) = nr(m+s) using eqn1 and eqn2.

So, ms(n + r) ≠ nr(m + s)

Therefore, (m, n) R (r, s) doesn’t exist.

Hence, it is not transitive.

QUESTION: 19

A situation in which significant power is distributed among three or more states is known as what?

Solution:

Definition of Multipolar system: A multipolar system is a system in which power is distributed at least among 3 significant poles concentrating wealth and/or military capabilities and able to block or disrupt major political arrangements threatening their major interests.

QUESTION: 20

Let R be an equivalence relation on Z, the set of integers.

R = {(a, b): a,b ∈ Z and a – b is a multiple of 3} The Equivalence class of [1] is

Solution:

**Correct Answer :- d**

**Explanation :** R = (a,b) : 3 divides (a-b)

⇒(a−b) is a multiple of 3.

To find equivalence class 1, put b=1

So, (a−0) is a multiple of 3

⇒ a is a multiple of 3

So, In set z of integers, all the multiple

of 3 will come in equivalence

class {1}

Hence, equivalence class {1} = {3x+1}

{-5,-2,1,4,7}

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