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Relation R on a set A is said to be a symmetric relation on A if:

(a,b) ∈ R

⇒ (b,a) ∈ R ∀ a,b∈ A

eg A = {1,2,3}

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

(i) Yes. 

(ii) No, because in the ordered pair (a, d), a ∈ A and d ∉ B. 

(iii) No, because in (y, d), y ∈ B. 

(iv) No. because here the first entries in all the ordered pairs are in the set B. 

(v) No. 

(vi) No, because the element z is not an ordered pair. 

(vii) No, because the elements of the set are not ordered pairs.

Relation R on a set A is said to be a transitive relation on A if:

(a,b) ∈ R and (b,c) ∈ R

⇒ (a,c) ∈ R ∀

a,b,c ∈ A

A={1,2,3}

⇒ R={(1,2),(2,3),(1,3)}

The domain of the relation

dom (R) = {-1, 1, -2, 2}

The range of the relation

range (R) = {1, 4}

By substituting the values of x and y we get

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

The range of R for the relation on N

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

By substituting a as 2 and 3 we get

R = {(2, 8), (3, 27)

We know that range of R

range (R) = {8, 27}

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

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

R₂ on Z defined by (a, b) ∈ R₂⇔ |a – b| ≤ 5

Check for Reflexivity:

∀ a ∈ Z,

(a, a) ∈ R₂ needs to be proved for reflexivity.

If (a, b) ∈ R₂

Then, |a – b| ≤ 5 …(1)

So, for (a, a) ∈ R₁

Replace b by a in equation (1), we get

|a – a| ≤ 5

⇒ 0 ≤ 5

⇒ (a, a) ∈ R₂

So, ∀ a ∈ Z, then (a, a) ∈ R₂

∴ R₂ is reflexive.

Check for Symmetry:

∀ a, b ∈ Z

If (a, b) ∈ R₂

We have, |a – b| ≤ 5 …(2)

Replace a by b & b by a in equation (2), we get

|b – a| ≤ 5

Since, the value is in mod, |b – a| = |a – b|

⇒ The statement |b – a| ≤ 5 is true.

⇒ (b, a) ∈ R₂

So, if (a, b) ∈ R₂, then (b, a) ∈ R₂

∀ a, b ∈ Q₀

∴ R₁ is symmetric.

Check for Transitivity:

∀ a, b, c ∈ Z

If (a, b) ∈ R₂ and (b, c) ∈ R₂

⇒ |a – b| ≤ 5 and |b – c| ≤ 5

Since, inequalities cannot be added or subtract. We need to take example to check for,

|a – c| ≤ 5

Take values a = 18, b = 14 and c = 10

Check: |a – b| ≤ 5

⇒ |18 – 14| ≤ 5

⇒ |4| ≤ 5 is true.

Check: |b – c| ≤ 5

⇒ |14 – 10| ≤ 5

⇒ |4| ≤ 5

Check: |a – c| ≤ 5

⇒ |18 – 10| ≤ 5

⇒ |8| ≤ 5 is not true.

⇒ (a, c) ∉ R₂

So, if (a, b) ∈ R₂ and (b, c) ∈ R₂, then (a, c) ∉ R₁

∀ a, b, c ∈ Z

∴ R₂ is not transitive.

We know that a is an integer where – 3 < a < 3

So we get

R = {(-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1)}

dom (R) = {-2, -1, 0, 1, 2}

range (R) = {3, 2, 1, 0}

By substituting values we get

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

So we get

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

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

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

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

R₃ on R defined by (a, b) ∈ R₃⇔ a² – 4ab + 3b² = 0

Check for Reflexivity:

∀ a ∈ R,

(a, a) ∈ R₃ needs to be proved for reflexivity.

If (a, b) ∈ R₃, then we have

a² – 4ab + 3b² = 0

Replace b by a, we get

a² – 4aa + 3a² = 0

⇒ a² – 4a² + 3a² = 0

⇒ –3a² + 3a² = 0

⇒ 0 = 0, which is true.

⇒ (a, a) ∈ R₃

So, ∀ a ∈ R, (a, a) ∈ R₃

∴ R₃ is reflexive.

Check for Symmetry:

∀ a, b ∈ R

If (a, b) ∈ R₃, then we have

a² – 4ab + 3b² = 0

⇒ a² – 3ab – ab + 3b² = 0

⇒ a (a – 3b) – b (a – 3b) = 0

⇒ (a – b) (a – 3b) = 0

⇒ (a – b) = 0 or (a – 3b) = 0

⇒ a = b or a = 3b …(1)

Replace a by b and b by a in equation (1), we get

⇒ b = a or b = 3a …(2)

Results (1) and (2) does not match.

⇒ (b, a) ∉ R₃

∴ R₃ is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ R

If (a, b) ∈ R₃ and (b, c) ∈ R₃

⇒ a² – 4ab + 3b² = 0 and b² – 4bc + 3c² = 0

⇒ a² – 3ab – ab + 3b² = 0 and b² – 3bc – bc + 3c²

⇒ a (a – 3b) – b (a – 3b) = 0 and b (b – 3c) – c (b – 3c) = 0

⇒ (a – b) (a – 3b) = 0 and (b – c) (b – 3c) = 0

⇒ (a – b) = 0 or (a – 3b) = 0

And (b – c) = 0 or (b – 3c) = 0

⇒ a = b or a = 3b And b = c or b = 3c

What we need to prove here is that, a = c or a = 3c

Take a = b and b = c

Clearly implies that a = c.

[∵ if a = b, just substitute a in place of b in b = c. We get, a = c]

Now, take a = 3b and b = 3c

If a = 3b

b = a/3

Substitute b = a/3 in b = 3c. We get

a/3 = 3c

⇒ a = 9c, which is not the desired result.

⇒ (a, c) ∉ R₃

∴ R₃ is not transitive.

⇒ Identity relation of a set refers to the relation in which every element on the set is related to itself.

Thus the Identity relation of set A is as under:

⇒ R = {(a,a),(b,b),(c,c)}

We have been given,

A = {1, 2, 3}

Here, R₁, R_2, and R₃ are the binary relations on 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.

Let us take R₁.

R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

(i). Reflexive:

∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∈ R₁

(2, 2) ∈ R₂

(3, 3) ∈ R₃

So, for a ∈ A, (a, a) ∈ R₁

∴ R₁ is reflexive.

(ii). Symmetric:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₁, then (3, 1) ∈ R₁

[∵ R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]

But if (2, 1) ∈ R₁, then (1, 2) ∉ R₁

[∵ R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]

So, if (a, b) ∈ R₁, then (b, a) ∉ R₁

∀ a, b ∈ A

∴ R₁ is not symmetric.

(iii). Transitivity:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₁ and (3, 3) ∈ R₁

Then, (1, 3) ∈ R₁

[∵ R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]

But, if (2, 1) ∈ R₁ and (1, 3) ∈ R₁

Then, (2, 3) ∉ R₁

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

∀ a, b, c ∈ A

∴ R₁ is not transitive.

Now, take R₂.

R₂ = {(2, 2), (3, 1), (1, 3)}

(i). Reflexive:

∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∉ R₂

(2, 2) ∈ R₂

(3, 3) ∉ R₂

So, for a ∈ A, (a, a) ∉ R₂

∴ R₂ is not reflexive.

(ii). Symmetric:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₂, then (3, 1) ∈ R₂

[∵ R₂ = {(2, 2), (3, 1), (1, 3)}]

If (2, 2) ∈ R₂, then (2, 2) ∈ R₂

[∵ R₂ = {(2, 2), (3, 1), (1, 3)}]

So, if (a, b) ∈ R₂, then (b, a) ∈ R₂

∀ a, b ∈ A

∴ R₂ is symmetric.

(iii). Transitivity:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₂ and (3, 1) ∈ R₂

Then, (1, 1) ∉ R₂

[∵ R₂ = {(2, 2), (3, 1), (1, 3)}]

So, if (a, b) ∈ R₂ and (b, c) ∈ R₂, then (a, c) ∉ R₂

∀ a, b, c ∈ A

∴ R₂ is not transitive.

Now take R₃.

R₃ = {(1, 3), (3, 3)}

(i). Reflexive:

∀ 1, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∉ R₃

(3, 3) ∈ R₃

So, for a ∈ A, (a, a) ∉ R₃

∴ R₃ is not reflexive.

(ii). Symmetric:

∀ 1, 3 ∈ A

If (1, 3) ∈ R₃, then (3, 1) ∉ R₃

[∵ R₃ = {(1, 3), (3, 3)}]

So, if (a, b) ∈ R₃, then (b, a) ∉ R₃

∀ a, b ∈ A

∴ R₃ is not symmetric.

(iii). Transitivity:

∀ 1, 3 ∈ A

If (1, 3) ∈ R₃ and (3, 3) ∈ R₃

Then, (1, 3) ∈ R₃

[∵ R₃ = {(1, 3), (3, 3)}]

So, if (a, b) ∈ R₃ and (b, c) ∈ R₃, then (a, c) ∈ R₃

∀ a, b, c ∈ A

∴ R₃ is transitive.

The smallest reflexive relation of set A = {1, 2, 3, 4} is as under:

As Relation R on a set A is said to be a reflexive relation on A if:

⇒ (a,a) ∈ R ∀ a ∈ A

⇒ R = {(1,1),(2,2),(3,3),(4,4)}

The relation between R and R^–1 on a Set A is as under:

⇒ R^-1 = {(b,a):(a,b)∈ R}

⇒ Clearly (a,b) ∈ R

⇒ (b,a) ∈ R^-1

i.e Domain(R) = Range(R^-1)

and Domain(R^-1) = Range(R)

By substituting values of a and b we get

R = {(3, 3), (6, 2), (9, 1)}

So we get

dom (R) = {3, 6, 9}

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

Given x and y are natural numbers, i.e. x,y ∈ N.

∴ Range of R = Set of all second elements in the relation.

= Values of ‘y’ which are in relation.

= N (Natural Numbers)

Since, x+ 2y = 8 and x,y are Natural numbers;

⇒ R = {(2,3), (4,2), (6,1)}

⇒ Range of R = {1,2,3}

NOTE: 0 is a whole number that’s why it is not considered in this set.

Given a relation R = {(a, a³) : a is a prime number less than 5}

Now according to the question ‘a’ is a prime number < 5:

⇒ a = {2,3}

⇒ R={(2,8),(3,27)}

⇒ Range of R={8,27)

Given set {1,2,3} and relation R = {(1, 2), (2, 1)}.

For relation R to be transitive:

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

But in the given relation:

⇒ (2,3) and (1,3) ∉ R

Hence, Given Relation R is not transitive.

Given A = {2, 3, 4, 5} and B = {1, 3, 4} and R is a relation from A to B which is true only iff “a is a divisor of b” i.e b is divisible by a.

⇒ R =A × B

⇒ R={(2,4),(4,4),(3,3)}

(i) R = {(2, 8), (3, 27), (5, 125), (7, 343)}

(ii) dom (R) = {2, 3, 5, 7}

(iii) range (R) = {8, 27, 125, 343}

A relation R on a set A is said to be an equivalence relation on A if:

1. Its reflexive i.e (a,a) ∈ R ∀ a ∈ A

2. Its symmetric i.e (a,b) ∈ R → (b,a) ∈ R ∀ a,b ∈ A

3. Its transitive i.e (a,b) ∈ R and (b,c) ∈ R → (a,c) ∈ R ∀

a,b,c ∈ A

Given A = {1, 2, 3, 4, 5, 6, 7, 8} and a relation R = {(x,y) : y is one half of x; x, y ϵA}

Now according to the condition y = x/2

⇒ R={(2,1),(4,2),(6,3),(8,4)}

Given A = {3, 5, 7} and B = {2, 4, 9}

Now according to the condition:

R is a relation given by “is less than.”

⇒ R= A × B

⇒ R={(3,4),(3,9),(5,9),(7,9)}

R = {a, b : a < b, a ∈ A, b ∈ B}, where A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5}. 

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

Domain of R = {1, 2, 3, 4} 

Range of R = {3, 5} 

Codomain of R = {1, 3, 5}.

(i) Given as

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

Therefore, R^‑1 = {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}

(ii) Given as

R= {(x, y): x, y ∈ N; x + 2y = 8}

Since, x + 2y = 8

x = 8 – 2y

As y ∈ N, Putting the values of y = 1, 2, 3,…… till x ∈ N

If, y = 1, x = 8 – 2(1) = 8 – 2 = 6

If, y = 2, x = 8 – 2(2) = 8 – 4 = 4

If, y = 3, x = 8 – 2(3) = 8 – 6 = 2

If, y = 4, x = 8 – 2(4) = 8 – 8 = 0

Then, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.

∴ R = {(2, 3), (4, 2), (6, 1)}

R^‑1 = {(3, 2), (2, 4), (1, 6)}

(iii) Given as

R is the relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3

Since,

x = {11, 12, 13} and y = (8, 10, 12}

y = x – 3

If, x = 11, y = 11 – 3 = 8 ∈ (8, 10, 12}

If, x = 12, y = 12 – 3 = 9 ∉ (8, 10, 12}

If, x = 13, y = 13 – 3 = 10 ∈ (8, 10, 12}

∴ R = {(11, 8), (13, 10)}

Thus, R^‑1 = {(8, 11), (10, 13)}

For any binary relation R, a second relation can be constructed by merely interchanging first and second components in every ordered pair. 

The relation thus obtained is called the inverse of the first one and designated as R^–1. 

Thus, R^–1 = {(y, x) : (x, y) ∈ R} 

For example, 

1. Then inverse of the husband-wife relation is wife-husband relation. 

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

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

So (R^–1)^–1 = R.

Let R be a relation from set A to set B. Then, the set of first element of the ordered pairs in R is called the domain and the set of second elements of the ordered pairs in R is called the range. The second set B is called the codomain of R. 

Thus for a relation R = {(a, b); a, b ∈ R }, 

Domain = {a : (a, b) ∈ R} and Range = {b : (a, b) ∈ R} 

For example, 

If A = {16, 25, 36, 49} and B = {1, 4, 5, 6} and R be the relation “is square of ” from A to B, then

R = {(a, b) : a = b², a ∈ A, b ∈ B} 

∴ R = {(16, 4), (25, 5), (36, 6)}. Then, 

Domain of R = {16, 25, 36}, Range of R = {4, 5, 6} and Codomain of R = {1, 4, 5, 6}.

Here,

R = {(x,y) : x, yϵZ, y = 2x-4} 

Now,

(a,-2)ϵ R 

⇒ -2 = 2a-4 

∴ 2a = -2+4 

∴ 2a = 2 

∴ a = 1 

Also,

(4,b²)ϵ R 

⇒ b² = 2×4-4 

∴ b² = 8 - 4 

∴ b² = 4 

∴ b = ±2