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i. Since, 

5 – 3 = 6 – 4 

= 7 – 5 = 2 

∴ x – y = 2 

Where xP and yQ 

So, 

R = {(x, y): x – y = 2, xP, y Q} 

ii. Now, 

R = {(5, 3), (6, 4), (7, 5)} 

iii. Domain of relation R = {5, 6, 7} 

Range of relation R = {3, 4, 5}

Given, 

R= {(a, b): a, b A, b is exactly divisible by a} 

A= {1, 2, 3, 4, 5, 6} 

Here, 

6 is exactly divisible by 1, 2, 3 and 6 

5 is exactly divisible by 1 and 5 

4 is exactly divisible by 1, 2 and 4 

3 is exactly divisible by 1 and 3 

2 is exactly divisible by 1 and 2 

1 is exactly divisible by 1 

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

ii. Domain of relation R = {1, 2, 3, 4, 5, 6} 

iii. Range of relation R = {1, 2 , 3, 4, 5, 6}

We have,

R = {(a, b) : a – b is divisible by 2; a, b ∈ Z}

To prove : R is an equivalence relation

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive.

Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a ∈ Z

⇒ a – a = 0

⇒ a – a is divisible by 2

⇒ (a, a) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that-

If (a, b) ∈ R, then (b, a) ∈ R

Let a, b ∈ Z and (a, b) ∈ R

⇒ a – b is divisible by 2

⇒ a – b = 2p For some p ∈ Z

⇒ b – a = 2 × (–p)

⇒ b – a ∈ R

⇒ R is symmetric

Transitive : : For Transitivity, we need to prove that-

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let a, b, c ∈ Z and such that (a, b) ∈ R and (b, c) ∈ R

⇒ a – b = 2p(say) and b – c = 2q(say) , For some p, q ∈ Z

⇒ a – c = 2 (p + q)

⇒ a – c is divisible by 2

⇒ (a, c) ∈ R

⇒ R is transitive

Now, since R is symmetric, reflexive as well as transitive-

⇒ R is an equivalence relation.

(b) Reflexive and symmetric and not transitive

• Let x ∈W. 

(x, x) ∈R, since the words ‘x’ and ‘x’ have all letters in common 

⇒ ‘x’ and ‘x’ have at least one letter in common 

⇒ R is reflexive. 

• Let x, y, z ∈W. 

Then (x, y) ∈R ⇒ ‘x’ and ‘y’ have at least one letter in common 

⇒ ‘y’ and ‘x’ have at least one letters common 

⇒ (y, x) ∈R 

⇒ R is symmetric. 

• Let x, y, z ∈W 

Then (x, y) ∈ R and (y, z) ∈R 

⇒ ‘x’ and ‘y‘ have at least one letter common and 

‘y‘ and ‘z’ have at least one letter common 

which does not necessarily mean that ‘x’ and ‘z’ have at lest one letter common. 

∴ R is not transitive 

For example, let x = ‘AND’, y = ‘NOT’, z = ‘PET’ 

x and y have ‘N’ common 

y and z have ‘T’ common 

but x and z have no common letter.

(d) 7

A = {1, 2, 3}. 

R = {(1, 2), (2, 3)} 

To make R an equivalence relation, it should be: 

(i) Reflexive: So three more ordered pairs (1, 1), (2, 2), (3, 3) should be added to R to make it reflexive. 

(ii) Symmetric : As R contains (1, 2) and (2, 3) so two more ordered pairs (2, 1) and (3, 2) should be added to make it symmetric. 

(iii) Transitive: (1, 2) ∈ R, (2, 3) ∈ R. So to make R transitive (1, 3) should be added to R. Also to maintain the symmetric property (3, 1) should then be added to R. 

So, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (2, 1), (3, 2), (1, 3), (3, 1)} is an equivalence relation. So minimum 7 ordered pairs are to be added.

(c) Transitive only

• There exists a real number –2, such that | –2 | is not less than –2 as 2 \(\not\leq\)- 2.

Thus for all negative, real numbers \(x\), | \(x\) | \(\not\leq\) \(x\). 

Hence (\(x\), \(x\)) ∉ R V real numbers. 

Hence R is not reflexive. 

• R is not symmetric since there exist real numbers –2 and 3 such that | –2 | ≤ 3 but | 3 | \(\not\leq\) - 2, 

i.e., (–2, 3) ∈ R \(\not\Rightarrow\) (3, –2) ∈ R. 

• R is transitive since V real numbers a, b, c 

(a, b) ∈ R, (b, c) ∈ R ⇒ | a | ≤ b and | b | ≤ c 

⇒ | a | ≤ b ≤ | b | ≤ c                       

(∵ For all \(x\), | \(x\) | ≤ \(x\)) 

⇒ | a | ≤ c 

⇒ (a, c) ∈ R

(a) An equivalence relation

• For all a ∈I, a – a = 0, which is divisible by 11. 

Thus, (a, a) ∈R for all a ∈N ⇒ R is reflexive 

• Let (a, b) ∈R (a – b) is divisible by 11 

⇒ – (a – b) is divisible by 11 

⇒ (b – a) is divisible by 11 

⇒ (b, a) ∈R 

⇒ R is symmetric. 

• Let (x, y) ∈R and (y, z) ∈R 

⇒ (x – y) is divisible by 11 and (y – z) is divisible by 11 

⇒ (x – y) + (y – z) is divisible by 11 

⇒ (x – z) is divisible by 11 

⇒ (x, z) ∈R 

⇒ R is transitive 

∴ R is an equivalence relation.

(c) An equivalence relation

R is reflexive as (a, a) ∈ R. a + a = 2a is even. 

R is symmetric as (a, b) ∈ R ⇒ (b, a) ∈ R as a + b = b + a = even (Commutative law) 

R is transitive as (a, b) ∈ R and (b, c) ∈ R 

⇒ (a + b) is even and (b + c) is even 

⇒ (a + b) + (b + c) is even 

⇒ (a + 2b + c) is even 

⇒ (a + c) is even as (2b is even) 

⇒ (a, c) ∈ R 

∴ R is an equivalence relation on the set of integers.

Reflexive: Since 5 divides a – a for all a ∈ Z, therefore, R is reflexive. 

Symmetric: (a, b) ∈ R ⇒ 5 divides a – b 

⇒ 5 divides b – a ⇒ b – a ∈ R 

∴ R is symmetric. 

Transitive: (a, b) ∈ R and (b, c) ∈ R 

⇒ a – b and b – c are both divisible by 5 

⇒ a – b + b – c is divisible by 5 

⇒ (a – c) is divisible by 5 

⇒ (a, c) ∈ R 

∴ R is transitive. 

Since R is reflexive, symmetric and transitive, therefore, R is an equivalence relation.

(d) R is an equivalence relation

R = {(x, y) : xy > 0 x, y ∈N} 

• x, x ∈N ⇒ x2 > 0 ⇒ R is reflexive 

• x, y ∈N and (x, y) ∈R ⇒ xy > 0 

⇒ yx > 0 (y, x) ∈R 

⇒ R is symmetric 

• x, y, z ∈N and (x, y) ∈R, (y, z) ∈R 

⇒ xy > 0 and yz > 0 

⇒ xz > 0 ⇒ (x, z) ∈R 

⇒ R is transitive. 

∴ R is an equivalence relation.

We have, L is the set of lines.

R = {(L₁, L₂) : L₁ is parallel to L₂} be a relation on L

Now,

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive.

Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Since a line is always parallel to itself.

∴ (L₁, L₂) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that-

If (a, b) ∈ R, then (b, a) ∈ R

Let L₁, L₂∈ L and (L₁, L₂) ∈ R

⇒ L₁ is parallel to L₂

⇒ L₂ is parallel to L₁

⇒ (L₁, L₂) ∈ R

⇒ R is symmetric

Transitive: For Transitivity, we need to prove that-

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let L₁, L₂ and L₃∈ L such that (L₁, L₂) ∈ R and (L₂, L₃) ∈ R

⇒ L₁ is parallel to L₂ and L₂ is parallel to L₃

⇒ L₁ is parallel to L₃

⇒ (L₁, L₃) ∈ R

⇒ R is transitive

Since, R is reflexive, symmetric and transitive, so R is an equivalence relation.

And, the set of lines parallel to the line y = 2x + 4 is y = 2x + c For all c ∈ R

where R is the set of real numbers.

We have been given that,

A is the set of all human beings in a town at a particular time.

Here, R is the binary relation on set A.

So, recall that

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Using these criteria, we can solve these.

We have,

R = {(x, y): x is father of y}

Check for Reflexivity:

Since x and x are the same people.

Then, x cannot be the father of itself.

A person cannot be a father of itself.

Leo is the father of Thiago

So, ∀ x ∈ A, then (x, x) ∉ R.

∴ R is not reflexive.

Check for Symmetry:

If x is the father of y.

Then, y cannot be the father of x.

If Sam is the father of Mac, then Mac is the son of Sam.

Mac cannot be the father of Sam.

So, if (x, y) ∈ R, then (y, x) ∉ R.

∀ x, y ∈ A

∴ R is not symmetric.

Check for Transitivity:

If x is the father of y and y is the father of z, then, x is not the father of z.

Take Mickey, Sam, and Mac.

If Mickey is the father of Sam, and Sam is the father of Mac.

Thus, Mickey is not the father of Mac, but the grandfather of Mac.

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.

∀ x, y, z ∈ A

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

We have set,

A = {a, b, c}

Here, R₁, R₂, R_3, and R₄ are the binary relations on set A.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

So, using these results let us start determining given relations.

We have

R₁ = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}

(i). Reflexive:

For all a, b, c ∈ A. [∵ A = {a, b, c}]

Then, (a, a) ∈ R₁

(b, b) ∈ A

(c, c) ∈ A

[∵ R₁ = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]

So, ∀ a, b, c ∈ A, then (a, a), (b, b), (c, c) ∈ R.

∴ R₁ is reflexive.

(ii). Symmetric:

If (a, a), (b, b), (c, c), (a, c), (b, c) ∈ R₁

Then, clearly (a, a), (b, b), (c, c), (c, a), (c, b) ∈ R₁

∀ a, b, c ∈ A

[∵ R₁ = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]

But, we need to try to show a contradiction to be able to determine the symmetry.

So, we know (a, b) ∈ R₁

But, (b, a) ∉ R₁

So, if (a, b) ∈ R₁, then (b, a) ∉ R₁.

∀ a, b ∈ A

∴ R₁ is not symmetric.

(iii). Transitive:

If (b, c) ∈ R₁ and (c, a) ∈ R₁

But, (b, a) ∉ R₁ [Check the Relation R₁ that does not contain (b, a)]

∀ a, b ∈ A

[∵ R₁ = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]

So, if (b, c) ∈ R₁ and (c, a) ∈ R₁, then (b, a) ∉ R₁.

∀ a, b, c ∈ A

∴ R₁ is not transitive.

Now, we have

R₂ = {(a, a)}

(i). Reflexive:

Here, only (a, a) ∈ R₂

for a ∈ A. [∵ A = {a, b, c}]

[∵ R₂ = {(a, a)}]

So, for a ∈ A, then (a, a) ∈ R₂.

∴ R₂ is reflexive.

(ii). Symmetric:

For symmetry,

If (x, y) ∈ R, then (y, x) ∈ R

∀ x, y ∈ A.

Notice, in R₂ we have

R₂ = {(a, a)}

So, if (a, a) ∈ R₂, then (a, a) ∈ R₂.

Where a ∈ A.

∴ R₂ is symmetric.

(iii). Transitive:

Here,

(a, a) ∈ R₂ and (a, a) ∈ R₂

Then, obviously (a, a) ∈ R₂

Where a ∈ A.

[∵ R₂ = {(a, a)}]

So, if (a, a) ∈ R₂ and (a, a) ∈ R₂, then (a, a) ∈ R₂, where a ∈ A.

∴ R₂ is transitive.

Now, we have

R₃ = {(b, a)}

(i). Reflexive:

∀ a, b ∈ A [∵ A = {a, b, c}]

But, (a, a) ∉ R₃

Also, (b, b) ∉ R₃

[∵ R₃ = {(b, a)}]

So, ∀ a, b ∈ A, then (a, a), (b, b) ∉ R₃

∴ R₃ is not reflexive.

(ii). Symmetric:

If (b, a) ∈ R₃

Then, (a, b) should belong to R₃.

∀ a, b ∈ A. [∵ A = {a, b, c}]

But, (a, b) ∉ R₃

[∵ R₃ = {(b, a)}]

So, if (a, b) ∈ R₃, then (b, a) ∉ R₃

∀ a, b ∈ A

∴ R₃ is not symmetric.

(iii). Transitive:

We have (b, a) ∈ R₃ but do not contain any other element in R₃.

Transitivity can’t be proved in R₃.

[∵ R₃ = {(b, a)}]

So, if (b, a) ∈ R₃ but since there is no other element.

∴ R₃ is not transitive.

Now, we have

R₄ = {(a, b) (b, c) (c, a)}

(i). Reflexive:

∀ a, b, c ∈ A [∵ A = {a, b, c}]

But, (a, a) ∉ R₄

Also, (b, b) ∉ R₄ and (c, c) ∉ R₄

[∵ R₄ = {(a, b) (b, c) (c, a)}]

So, ∀ a, b, c ∈ A, then (a, a), (b, b), (c, c) ∉ R₄

∴ R₄ is not reflexive.

(ii). Symmetric:

If (a, b) ∈ R₄, then (b, a) ∈ R₄

But (b, a) ∉ R₄

[∵ R₄ = {(a, b) (b, c) (c, a)}]

So, ∀ a, b ∈ A, if (a, b) ∈ R₄, then (b, a) ∉ R₄.

⇒ R₄ is not symmetric.

It is sufficient to show only one case of ordered pairs violating the definition.

∴ R₄ is not symmetric.

(iii). Transitivity:

We have,

(a, b) ∈ R₄ and (b, c) ∈ R₄

⇒ (a, c) ∈ R₄

But, is it so?

No, (a, c) ∉ R₄

So, it is enough to determine that R₄ is not transitive.

∀ a, b, c ∈ A, if (a, b) ∈ R₄ and (b, c) ∈ R₄, then (a, c) ∉ R₄.

∴ R₄ is not transitive.

To prove : (a, b) R₁ and (b,c) R₁ (a, c) R₁ is not true for all a, b, c R. 

Given,

R₁ = {(a, b): 1 + ab > 0} 

Let a = 1, b = - 0.5, c = - 4 

Here, 

(1, -0.5) R₁ 

[∵ 1+(1×-0.5) = 0.5 > 0] 

And, 

(-0.5, -4) R₁ 

[∵ 1+(-0.5×-4) = 3 > 0] 

But, 

(1, -4) ∉ R₁ 

[∵ 1+(1×-4) = - 3 < 0] 

∴ (a, b) R₁ and (b,c) R₁ (a, c) R₁ is not true for all a, b, c R 

Hence Proved.

NOTE : 

Here R₁ is a relation whereas R denotes a real number.

B. reflexive, transitive but not symmetric

S: a S b ⟺ a ≥ b

Since a=a ∀a ∈ R, therefore a ≥ a always. Hence (a, a) always belongs to S ∀a ∈ R. Therefore, S is reflexive.

If a ≥ b then b ≤ a ⇏ b ≥ a. Hence if (a, b) belongs to S, then (b, a) does not always belongs to S. Hence S is not symmetric.

If a ≥ b and b ≥ c, therefore a ≥ c. Hence if (a, b) and (b, c) belongs to S, then (a, c) will belong to S ∀a, b, c∈R. Hence, S is transitive.

C. equivalence

R: a R b ⟺ a ≅ b

Since, every triangle a∈T is congruent to itself, therefore (a, a)∈R ∀a∈T. Hence, R is reflexive.

If a ≅ b, then b ≅ a. Hence if (a, b)∈R, then (b, a)∈R ∀a, b∈T. Hence, R is symmetric.

If a ≅ b and b ≅ c, then a ≅ c. Hence if (a, b) and (b, c) belongs to R, then (a, c) will belong to R ∀a, b, c∈T. Hence, R is transitive.

Since R is reflexive, symmetric and transitive, therefore R is equivalence relation.

N is the set of all the natural numbers.

N = {1, 2, 3, 4, 5, 6, 7…..}

R = {(a, b) : a, b, ϵ N and a < b}

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

For reflexivity,

A relation R on N is said to be reflexive if (a, a) є R for all a є N.

But, here we see that a < b, so the two co-ordinates are never equal. Thus, the relation is not reflexive.

For symmetry,

A relation R on N is said to be symmetrical if (a, b) є R è(b, a) є R

Here, (a, b) є R does not imply (b, a) є R . Thus, it is not symmetric.

For transitivity,

A relation R on A is said to be transitive if (a, b) є R and (b, c) є R è (a, c) є R for all (a, b, c) є N. 

Let’s take three values a, b and c such that a < b < c. So, (a, b) є R and (b, c) є R è (a, c) є R. 

Thus, it is transitive.

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

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

(ii) R = {(x, y) : x, y ϵ N, x + 2y = 8}.

y = \(\frac{8-x}{2}\)

Put x = 2, y = 3

Put x = 4, y = 2 

Put x = 6, y = 1

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

R^–1 = {(3, 2), (2, 4), (1, 6)}

Relatively prime numbers are also known as co-prime numbers. If there is no integer greater than one that divides both 

(that is, their greatest common divisor is one). 

For example,

12 and 13 are relatively prime, but 12 and14 are not as their greatest common divisor is two. 

Given, 

(x, y) R x is relatively prime to y.

Here, 

2 is co-prime to 3, 5 and 7. 

3 is co-prime to 2, 4, 5 and 7. 

4 is co-prime to 3, 5 and 7. 

5 is co-prime to 2, 3, 4, 6 and 7. 

6 is co-prime to 5 and 7. 

7 is co-prime to 2, 3, 4, 5 and 6. 

∴ R = {(2,3), (2,5), (2,7), (3,2), (3,4), (3,5), (3,7), (4,3), (4.5), (4,7), (5,2), (5,3), (5,4), (5,6), (5,7), (6,5), (6,7), (7,2), (7,3), (7,4), (7,5), (7,6)}

D. 5

A = {1, 2, 3}

Then the equivalence relations would be,

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

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

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

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

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

Hence, total 5 equivalence relations.

b = 2a – 4

a = \(\frac{b+4}{2}\)

Put b = -2 , a = 1 

Put a = 4 , b = 4 

a = 1 , b = 4

Let A = {2, 3, 4, 5, 6} and B = {1, 2, 3} 

Given, 

x = 2y where x {2, 3, 4, 5, 6} and y {1, 2, 3} 

On putting y = 1, 

x = 2(1) = 2A 

On putting y = 2, 

x = 2(2) = 4A 

On putting y = 3, 

x = 2(3) = 6A 

∴ R = {(2, 1), (4, 2), (6, 3)}

x² + y² = 9 

We can have only integral values of x and y. 

Put x = 0 , y = 3 , 0² + 3² = 9 

Put x = 3 , y = 0 , 3² + 0² = 9 

R = {(0, 3) , (3, 0) , (0 , -3) , (-3 , 0)} 

The domain of R is the set of first co-ordinates of R 

Dom(R) = {-3 , 0 , 3}

The range of R is the set of second co-ordinates of R 

Range(R) = {-3 , 0 , 3}

A = {0, 1, 2, 3, 4, 5, 6, 7, 8} 

2a + 3b = 12

b = \(\frac{12-2a}{3}\)

a=0 è b=4

a=3 è b=2

a=6 è b=0

R = {(0, 4), (3, 2), (6, 0)}

Since, R is a subset of A × A, it a relation to A. 

The domain of R is the set of first co-ordinates of R 

Dom(R) = {0, 3, 6}

The range of R is the set of second co-ordinates of R 

Range(R) = {4, 2, 0}

(d) Symmetric and transitive

| a – a | = | 0 | = 0 so (a, a) ∉R ⇒ R is not reflexive

(a, b) ∈ R ⇒ | a – b | > 0 ⇒ | b – a | > 0 ⇒ (b, a) ∈R           (∵ | a – b | = | b – a |) 

⇒ R is symmetric 

(a, b) ∈ R ⇒ | a – b | > 0 and (b, c) ∈ R ⇒ | b – c | > 0

V real numbers a, b, c. 

∴ | a – b | > 0 and | b – c | > 0 ⇒ | a – c | > 0 

⇒ (a, c) ∈ R ⇒ R is transitive.