RD Chapter 1 Relations Ex 1.1 Solutions

Question - 1 : -
Let A be the set of all human beings in a town at a particular time. Determine whether of the following relation is reflexive, symmetric and transitive:
(i) R = {(x, y): x and y work at the same place}
(ii) R = {(x, y): x and y live in the same locality}
(iii) R = {(x, y): x is wife of y}
(iv) R = {(x, y): x is father of y}

Answer - 1 : -

(i) Given R = {(x, y): x and y work at the same place}

Now we have to check whether the relation is reflexive:

Let x be an arbitrary element of R.

Then, x ∈R

⇒ x and x work at the same place is true since they are the same.

⇒(x, x) ∈R [condition for reflexive relation]

So, R is a reflexive relation.

Now let us check Symmetric relation:

Let (x, y) ∈R

⇒x and y work at the same place [given]

⇒y and x work at the same place

⇒(y, x) ∈R

So, R is a symmetric relation.

Transitive relation:

Let (x, y) ∈R and (y, z) ∈R.

Then, x and y work at the same place. [Given]

y and z also work at the same place. [(y, z) ∈R]

⇒ x, y and z all work at the same place.

⇒x and z work at the same place.

⇒ (x, z) ∈R

So, R is a transitive relation.

Hence R is reflexive, symmetric and transitive.

(ii) Given R = {(x, y): x and y live in the same locality}

Now we have to check whether the relation R is reflexive, symmetric and transitive.

Let x be an arbitrary element of R.

Then, x ∈R

It is given that x and x live in the same locality is true since they are the same.

So, R is a reflexive relation.

Symmetry:

Let (x, y) ∈ R

⇒ x and y live in the same locality [given]

⇒ y and x live in the same locality

⇒ (y, x) ∈ R

So, R is a symmetric relation.

Transitivity:

Let (x, y) ∈R and (y, z) ∈R.

Then,

x and y live in the same locality and y and z live in the same locality

⇒ x, y and z all live in the same locality

⇒ x and z live in the same locality

⇒ (x, z) ∈ R

So, R is a transitive relation.

Hence R is reflexive, symmetric and transitive.

(iii) Given R = {(x, y): x is wife of y}

Now we have to check whether the relation R is reflexive, symmetric and transitive.

First let us check whether the relation is reflexive:

Let x be an element of R.

Then, x is wife of x cannot be true.

⇒ (x, x) ∉R

So, R is not a reflexive relation.

Symmetric relation:

Let (x, y) ∈R

⇒ x is wife of y

⇒ x is female and y is male

⇒ y cannot be wife of x as y is husband of x

⇒ (y, x) ∉R

So, R is not a symmetric relation.

Transitive relation:

Let (x, y) ∈R, but (y, z) ∉R

Since x is wife of y, but y cannot be the wife of z, y is husband of x.

⇒ x is not the wife of z

⇒(x, z) ∈R

So, R is a transitive relation.

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

Now we have to check whether the relation R is reflexive, symmetric and transitive.

Reflexivity:

Let x be an arbitrary element of R.

Then, x is father of x cannot be true since no one can be father of himself.

So, R is not a reflexive relation.

Symmetry:

Let (x, y) ∈R

⇒ x is father of y

⇒ y is son/daughter of x

⇒ (y, x) ∉R

So, R is not a symmetric relation.

Transitivity:

Let (x, y) ∈R and (y, z) ∈R.

Then, x is father of y and y is father of z

⇒ x is grandfather of z

⇒ (x, z) ∉R

So, R is not a transitive relation.

Question - 2 : -
Three relations R1, R2 and R3 are defined on a set A = {a, b, c} as follows:
R1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}
R2 = {(a, a)}
R3 = {(b, c)}
R4 = {(a, b), (b, c), (c, a)}.
Find whether or not each of the relations R1, R2, R3, R4 on A is (i) reflexive (ii) symmetric and (iii) transitive.

Answer - 2 : -

(i) Consider R₁

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

Now we have check R₁ isreflexive, symmetric and transitive

Reflexive:

Given (a, a), (b, b)and (c, c) ∈ R₁

So, R₁ isreflexive.

Symmetric:

We see that the ordered pairs obtained by interchanging the components of R₁ arealso in R₁.

So, R₁ is symmetric.

Transitive:

Here, (a, b) ∈R₁, (b, c)∈R₁ and also (a, c)∈R₁

So, R₁ istransitive.

(ii) Consider R₂

Given R₂ ={(a, a)}

Reflexive:

Clearly (a,a) ∈R₂.

So, R₂ isreflexive.

Symmetric:

Clearly (a,a) ∈R

⇒ (a, a) ∈R.

So, R₂ issymmetric.

Transitive:

R₂ isclearly a transitive relation, since there is only one element in it.

(iii) Consider R₃

Given R₃ = {(b, c)}

Reflexive:

Here,(b, b)∉ R₃ neither (c, c) ∉ R₃

So, R₃ isnot reflexive.

Symmetric:

Here, (b, c) ∈R₃, but (c,b) ∉R₃

So, R₃isnot symmetric.

Transitive:

Here, R₃ has only two elements.

Hence, R₃ istransitive.

(iv) Consider R₄

Given R₄ = {(a, b), (b, c), (c, a)}.

Reflexive:

Here, (a, a) ∉ R₄, (b, b)∉ R₄ (c, c)∉ R₄

So, R₄ is not reflexive.

Symmetric:

Here, (a, b) ∈ R₄, but (b,a) ∉ R₄.

So, R₄ is not symmetric

Transitive:

Here, (a, b) ∈R₄, (b, c)∈R₄, but (a, c)∉R₄

So, R₄ is not transitive.

Question - 3 : -
Test whether the following relation R₁, R₂,and R₃are (i) reflexive (ii) symmetric and (iii) transitive:
(i) R₁ on Q₀ defined by (a, b) ∈ R₁ ⇔ a = 1/b.
(ii) R₂ on Z defined by (a, b) ∈ R₂ ⇔ |a – b| ≤ 5
(iii) R₃on R defined by (a, b) ∈ R₃ ⇔ a²–4ab + 3b²= 0.

Answer - 3 : -

(i) Given R₁ on Q₀ definedby (a, b) ∈ R₁ ⇔ a = 1/b.

Reflexivity:

Let a be an arbitrary element of R₁.

Then, a ∈ R₁

⇒ a ≠1/a for all a ∈ Q₀

So, R₁ is not reflexive.

Symmetry:

Let (a, b) ∈ R₁

Then,(a, b) ∈ R₁

Therefore we can write‘a’ as a =1/b

⇒ b = 1/a

⇒ (b, a) ∈ R₁

So, R₁ is symmetric.

Transitivity:

Here, (a, b) ∈R₁ and (b, c) ∈R₂

⇒ a = 1/b and b = 1/c

⇒ a = 1/ (1/c) = c

⇒ a ≠ 1/c

⇒ (a, c) ∉ R₁

So, R₁ is not transitive.

(ii) Given R₂ on Z definedby (a, b) ∈ R₂ ⇔ |a – b| ≤ 5

Now we have checkwhether R₂ is reflexive, symmetric and transitive.

Reflexivity:

Let a be anarbitrary element of R₂.

Then, a ∈ R₂

On applying the givencondition we get,

⇒ | a−a | = 0 ≤ 5

So, R₁ is reflexive.

Symmetry:

Let (a, b) ∈ R₂

⇒ |a−b| ≤ 5 [Since, |a−b| = |b−a|]

⇒ |b−a| ≤ 5

⇒ (b, a) ∈ R₂

So, R2 is symmetric.

Transitivity:

Let (1, 3) ∈ R₂ and (3, 7) ∈R₂

⇒|1−3|≤5 and |3−7|≤5

But |1−7| ≰5

⇒ (1, 7) ∉ R₂

So, R₂ is not transitive.

(iii) Given R₃on R definedby (a, b) ∈ R₃ ⇔ a²–4ab + 3b²= 0.

Now we have checkwhether R₂ is reflexive, symmetric and transitive.

Reflexivity:

Let a be anarbitrary element of R₃.

Then, a ∈ R₃

⇒ a²− 4a × a+ 3a²= 0

So, R₃ is reflexive

Symmetry:

Let (a, b) ∈ R₃

⇒ a²−4ab+3b²=0

But b²−4ba+3a²≠0 for all a, b ∈ R

So, R₃ is not symmetric.

Transitivity:

Let (1, 2) ∈ R₃ and (2, 3) ∈ R₃

⇒ 1 − 8 + 6 =0 and 4 – 24 + 27 = 0

But 1 – 12+ 9 ≠ 0

So, R₃ is not transitive.

Question - 4 : - Let A = {1, 2, 3}, and let R₁ ={(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}, R₂ ={(2, 2), (3, 1), (1, 3)}, R₃ = {(1, 3), (3, 3)}. Findwhether or not each of the relations R₁, R₂, R₃ onA is (i) reflexive (ii) symmetric (iii) transitive.

Answer - 4 : -

Consider R₁

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

Reflexivity:

Here, (1, 1), (2, 2), (3, 3) ∈R

So, R₁ is reflexive.

Symmetry:

Here, (2, 1) ∈ R₁,

But (1, 2) ∉ R₁

So, R₁ is not symmetric.

Transitivity:

Here, (2, 1) ∈R₁ and (1, 3) ∈R₁,

But (2, 3) ∉R₁

So, R₁ is not transitive.

Now consider R₂

Given R₂ = {(2, 2), (3, 1), (1, 3)}

Reflexivity:

Clearly, (1, 1) and (3, 3)∉R₂

So, R₂ is not reflexive.

Symmetry:

Here, (1, 3) ∈ R₂ and (3, 1) ∈ R₂

So, R₂ is symmetric.

Transitivity:

Here, (1,3) ∈ R₂ and (3,1) ∈ R₂

But (3, 3) ∉R₂

So, R₂ is not transitive.

Consider R₃

Given R₃ = {(1, 3), (3, 3)}

Reflexivity:

Clearly, (1,1) ∉ R₃

So, R₃ is not reflexive.

Symmetry:

Here, (1, 3) ∈ R₃, but (3, 1) ∉ R₃

So, R₃ is not symmetric.

Transitivity:

Here, (1, 3) ∈ R₃ and (3, 3) ∈ R₃

Also, (1, 3) ∈ R₃

So, R₃ is transitive.

Question - 5 : -
The following relation is defined on the set of real numbers.
(i) aRb if a – b > 0
(ii) aRb iff 1 + a b > 0
(iii) aRb if |a| ≤ b.
Find whether relation is reflexive, symmetric or transitive.

Answer - 5 : -

(i) ConsideraRb if a – b > 0

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

Reflexivity:

Let a be anarbitrary element of R.

Then, a ∈ R

But a −a = 0 ≯ 0

So, this relation is not reflexive.

Symmetry:

Let (a, b) ∈ R

⇒ a −b > 0

⇒ − (b −a) >0

⇒ b −a < 0

So, 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.

So, the given relation is transitive.

(ii) Consider aRb iff1 + a b > 0

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

Reflexivity:

Let a be anarbitrary 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 + b c >0

But 1+ ac ≯ 0

⇒ (a, c) ∉ R

So, the given relation is not transitive.

(iii) Consider aRb if|a| ≤ b.

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

Reflexivity:

Let a be anarbitrary element of R.

Then, a ∈ R [Since, |a|=a]

⇒ |a|≮ a

So, R is not reflexive.

Symmetry:

Let (a, b) ∈ R

⇒ |a| ≤ b

⇒ |b| ≰ a for all a, b ∈ R

⇒ (b, a) ∉ R

So, 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| ≤ bc

⇒ |a| ≤ c

⇒ (a, c) ∈ R

Thus, R is transitive.

Question - 6 : -
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6}as R = {(a, b): b = a + 1} is reflexive,symmetric or transitive.

Answer - 6 : -

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

Now for this relationwe have to check whether it is reflexive, transitive and symmetric Reflexivity:

Let a be an arbitraryelement of R.

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

⇒ (a, a) ∉ R

So, R is not reflexive on A.

Symmetry:

Let (a, b) ∈ R

⇒ b = a + 1

⇒ −a = −b + 1

⇒ a = b − 1

Thus, (b, a) ∉ R

So, 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

So, R is not transitive on A.

Question - 7 : -
Check whether the relation R on R defined as R = {(a, b): a ≤ b³}is reflexive, symmetric or transitive.

Answer - 7 : -

Given R = {(a, b): a ≤ b³}

It is observed that(1/2, 1/2) in R as 1/2 > (1/2)³ = 1/8

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2³ = 8)

But,

(2, 1) ∉ R (as 2 > 1³ = 1)

∴ R is not symmetric.

We have (3, 3/2),(3/2, 6/5) in “R as” 3 < (3/2)³ and 3/2 < (6/5)³

But (3,6/5) ∉ R as 3 > (6/5)³

∴ R is not transitive.

Hence, Ris neither reflexive, nor symmetric, nor transitive.

Question - 8 : - Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.

Answer - 8 : -

Let A be a set.

Then, Identity relation IA=IA is reflexive, since (a, a) ∈ A ∀a

The converse of it need not be necessarily true.

Consider the set A = {1, 2, 3}

Here,

Relation R = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on A.

However, R is not an identity relation.

Question - 9 : -
If A = {1, 2, 3, 4} define relations on A which have properties of being
(i) Reflexive, transitive but not symmetric
(ii) Symmetric but neither reflexive nor transitive.
(iii) Reflexive, symmetric and transitive.

Answer - 9 : -

(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)}

The relation R is an equivalence relation on A.

Question - 10 : -

Answer - 10 : -

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RD Chapter 1 Relations Ex 1.1 Solutions

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