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In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, ∀ a, b ∈ S, R = {(a, b) : a = ± b }

Now,

R is Reflexive if (a,a) ∈ R ∀ a ∈ S

For any a ∈ S, we have

a = ±a

⇒ (a,a) ∈ R

Thus, R is reflexive.

R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ S

(a,b) ∈ R

⇒ a = ± b

⇒ b = ± a

⇒ (b,a) ∈ R

Thus, R is symmetric .

R is Transitive if (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a,b,c ∈ S

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ S

⇒ a = ± b and b = ± c

⇒ a = ± c

⇒ (a, c) ∈ R

Thus, R is transitive.

Hence, R is an equivalence relation.

(i) Consider aRb if a – b > 0

Let us check for this relation whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R

But a − a = 0 ≯ 0

Thus, this relation is not reflexive.

Symmetry:

Let (a, b) ∈ R ⇒ a − b > 0

⇒ − (b − a) > 0 ⇒ b − a < 0

Therefore, the given relation is not symmetric.

Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R.

Then, a − b > 0 and b − c > 0

Adding the two, we get a – b + b − c > 0

⇒ a – c > 0 ⇒ (a, c) ∈ R.

Clearly, the given relation is transitive.

(ii) Consider aRb iff 1 + a b > 0

Now for this relation we have to check whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R

⇒ 1 + a × a > 0

i.e. 1 + a² > 0 [Since, square of any number is positive]

So, the given relation is reflexive.

Symmetry:

Let (a, b) ∈ R ⇒ 1 + ab > 0

⇒ 1 + ba > 0 ⇒ (b, a) ∈ R

So, the given relation is symmetric.

Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R

⇒1 + ab > 0 and 1 + bc >0

But 1 + ac ≯ 0 ⇒ (a, c) ∉ R

Thus, the given relation is not transitive.

(iii) Consider aRb if |a| ≤ b.

Let us check for this relation whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R [Since, |a| = a] ⇒ |a| ≮ a

Clearly, R is not reflexive.

Symmetry:

Let (a, b) ∈ R ⇒ |a| ≤ b ⇒ |b| ≰ a for all a, b ∈ R

⇒ (b, a) ∉ R

Thus, R is not symmetric.

Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R

⇒ |a| ≤ b and |b| ≤ c

Multiplying the corresponding sides, we get

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

So, R is transitive.

(i) The relation on A having properties of being reflexive, transitive, but not symmetric is

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

Relation R satisfies reflexivity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R 

And (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R

However, (2, 1) ∈ R, but (1, 2) ∉ R

(ii)  The relation on A having properties of being reflexive, transitive, but not symmetric is

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

Relation R satisfies reflexivity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R 

And (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R

However, (2, 1) ∈ R, but (1, 2) ∉ R

(iii) The relation on A having properties of being symmetric, reflexive and transitive is

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

So, the R is an equivalence relation on A.

Given that, A = {1, 2, 3} and R = {1, 1), (2, 2), (3, 3),(4, 4), (1, 2), (1, 3), (3, 2)}.

Now,

R is reflexive ∵ (1,1),(2,2),(3,3),(4,4) ∈ R

R is not symmetric ∵ (1,2),(1,3),(3,2) ∈ R but (2,1),(3,1),(2,3) ∉ R

R is transitive ∵ (1,3) ∈ R and (3,2) ∈ R ⇒ (1,2) ∈ R

Thus, R is reflexive and transitive but not symmetric.

Given that, ∀ A, B ∈ S, R = {(A, B) : d(A, B) < 2 units}.

Now,

R is Reflexive if (A,A) ∈ R ∀ A ∈ S

For any A ∈ S, we have

d(A,A) = 0, which is less than 2 units

⇒ (A,A) ∈ R

Thus, R is reflexive.

R is Symmetric if (A, B) ∈ R ⇒ (B,A) ∈ R ∀ A,B ∈ S

(A, B) ∈ R

⇒ d(A, B) < 2 units

⇒ d(B, A) < 2 units

⇒ (B,A) ∈ R

Thus, R is symmetric .

R is Transitive if (A, B) ∈ R and (B,C) ∈ R ⇒ (A,C) ∈ R ∀ A,B,C ∈ S

Consider points A(0,0),B(1.5,0) and C(3.2,0).

d(A,B)=1.5 units < 2 units and d(B,C)=1.7 units < 2 units

d(A,C)= 3.2 ≮ 2

⇒ (A, B) ∈ R and (B,C) ∈ R ⇒ (A,C) ∉ R

Thus, R is not transitive.

Thus, R is reflexive, symmetric but not transitive.

Given that, A = {1, 2, 3} and R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}.

Now,

R is reflexive ∵ (1,1),(2,2),(3,3) ∈ R

R is not symmetric ∵ (1,2),(2,3) ∈ R but (2,1),(3,2) ∉ R

R is not transitive ∵ (1,2) ∈ R and (2,3) ∈ R ⇒ (1,3) ∉ R

Thus, R is reflexive but neither symmetric nor transitive.

Given as R = {(a, b): b = a + 1}

Let us check for this relation whether it is reflexive, transitive and symmetric

Reflexivity:

Let a be an arbitrary element of R.

Then, a = a + 1 cannot be true for all a ∈ A.

⇒ (a, a) ∉ R

Therefore, R is not reflexive on A.

Symmetry:

Let (a, b) ∈ R

⇒ b = a + 1

⇒ a = b − 1

Thus, (b, a) ∉ R

Clearly, R is not symmetric on A.

Transitivity:

Let (1, 2) and (2, 3) ∈ R

⇒ 2 = 1 + 1 and 3

2 + 1  is true.

But 3 ≠ 1+1 ⇒ (1, 3) ∉ R

Therefore, R is not transitive on A.

Given that, ∀ a, b ∈ S, R = {(a, b) : a² + b² = 1 }

Now,

R is Reflexive if (a,a) ∈ R ∀ a ∈ S

For any a ∈ S, we have

a²+a² = 2 a² ≠ 1

⇒ (a,a) ∉ R

Thus, R is not reflexive.

R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ S

(a,b) ∈ R

⇒ a² + b² = 1

⇒ b² + a² = 1

⇒ (b,a) ∈ R

Thus, R is symmetric .

R is Transitive if (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a,b,c ∈ S

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ S

⇒ a² + b² = 1 and b² + c² = 1

Adding both, we get

a²+ c²+2b² = 2

⇒ (a, c) ∉ R

Thus, R is not transitive.

Thus, R is symmetric but neither reflexive nor transitive.

We have, R = {(a, b) : a > b} relation defined on N.

Now,

We observe that, any element a ∈ N cannot be greater than itself.

⇒ (a,a) ∉ R ∀ a ∈ N

⇒ R is not reflexive.

Let (a,b) ∈ R ∀ a, b ∈ N

⇒ a is greater than b

But b cannot be greater than a if a is greater than b.

⇒ (b,a) ∉ R

For e.g., we observe that (5,2) ∈ R i.e 5 > 2 but 2 ≯ 5 ⇒ (2,5) ∉ R

⇒ R is not symmetric

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ N

⇒ a > b and b > c

⇒ a > c

⇒ (a,c) ∈ R

For e.g., we observe that

(5,4) ∈ R ⇒ 5 > 4 and (4,3) ∈ R ⇒ 4 > 3

And we know that 5 > 3 ∴ (5,3) ∈ R

⇒ R is transitive.

Thus, R is transitive but not reflexive not symmetric.

We have, R = {(a, b) : a = b²} relation defined on N.

Now,

We observe that, any element a ∈ N cannot be equal to its square except 1.

⇒ (a,a) ∉ R ∀ a ∈ N

For e.g. (2,2) ∉ R ∵ 2 ≠ 2²

⇒ R is not reflexive.

Let (a,b) ∈ R ∀ a, b ∈ N

⇒ a = b²

But b cannot be equal to square of a if a is equal to square of b.

⇒ (b,a) ∉ R

For e.g., we observe that (4,2) ∈ R i.e 4 = 2² but 2 ≠ 4²⇒ (2,4) ∉ R

⇒ R is not symmetric

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ N

⇒ a = b² and b = c²

⇒ a ≠ c²

⇒ (a,c) ∉ R

For e.g., we observe that

(16,4) ∈ R ⇒ 16 = 4² and (4,2) ∈ R ⇒ 4 = 2²

But 16 ≠ 2²

⇒ (16,2) ∉ R

⇒ R is not transitive.

Thus, R is neither reflexive nor symmetric nor transitive.

Cartesian product: Let A and B be two non-empty sets. Then the set of all possible ordered pairs (x, y) such that the first component x of the ordered pairs is an element of set A, and the second component y is an element of set B, is called the cartesian product of the sets A and B. It is denoted by A × B read as “A cross B”. 

A × B = {(a, b), a ∈ A and b ∈ B} 

Also, n(A × B) = n(A) × n(B) = pq if set A has p elements and set B has q elements.

Notes: 1. The cartesian product A × B is not the same as B × A. 

In A × B, the set A is named first so its elements will appear as the first components of the ordered pairs. 

In B × A, the set B is named first, so its elements will appear as the first components of the ordered pairs. 

2. If either A or B is a null set, then we define A × B to be a null set. 

If A = {a, b} and B = ϕ then A × B = ϕ 

3. If either A or B is an infinite set and the other is a non-empty set, then A × B is also an infinite set. 

4. If A and B are two non-empty sets having n-elements in common, then A × B and B × A have n² elements in common. 

5. If A = B, then A × B = A x A and is denoted by A². 

Examples: 

Ex. 1. If A = {a, b} and B = {1, 2, 3}, then 

A × B = {(a, 1), (b, 1), (a, 2), (b, 2), (a, 3), (b, 3)} 

B × A = {(1, a), (2, a), (3, a), (1, b), (2, b), (3, b)}. 

Ex. 2. If A = {2, 4, 6}, then 

A × A = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}.

Relation: 

If A and B are any two non-empty sets, then any subset of A × B is defined as a relation from A to B. 

For example, 

Suppose A = {1, 2, 3} and B = {1, 2, 3, 4}. Then {(2, 3), (2, 4), (1, 3)} is a relation in A × B. Many more relations (subsets) can be selected at random from our product set A × B.