Class 12 Mathmatics Relations Exercise and Solutions

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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}

Solution:

(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.

2. Three relations R1, R2 and R3 are defined on a set A = {abc} 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 is (i) reflexive (ii) symmetric and (iii) transitive.

Solution:

(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 is reflexive, symmetric and transitive

Reflexive:

Given (a, a), (b, b) and (c, c) ∈ R1

So, R1 is reflexive.

Symmetric:

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

So, R1 is symmetric.

Transitive:

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

So, R1 is transitive.

(ii) Consider R2

Given R2 = {(a, a)}

Reflexive:

Clearly (a, a) ∈R2.

So, R2 is reflexive.

Symmetric:

Clearly (a, a) ∈R

⇒ (a, a) ∈R.

So, R2 is symmetric.

Transitive:

R2 is clearly 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 is not reflexive.

Symmetric:

Here, (b, c) ∈R3, but (c, b) ∉R3

So, Ris not symmetric.

Transitive:

Here, R3 has only two elements.

Hence, R3 is transitive.

(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.

3.  Test whether the following relation R1, R2, and Rare (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) Ron R defined by (a, b) ∈ R3 ⇔ a2 – 4ab + 3b2 = 0.

Solution:

(i) Given R1 on Q0 defined by (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 defined by (a, b) ∈ R2 ⇔ |a – b| ≤ 5

Now we have check whether R2 is reflexive, symmetric and transitive.

Reflexivity:

Let a be an arbitrary element of R2.

Then, a ∈ R2

On applying the given condition 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 Ron R defined by (a, b) ∈ R3 ⇔ a2 – 4ab + 3b2 = 0.

Now we have check whether R2 is reflexive, symmetric and transitive.

Reflexivity:

Let a be an arbitrary element of R3.

Then, a ∈ R3

⇒ a− 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.

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)}. Find whether or not each of the relations R1, R2, R3 on A is (i) reflexive (ii) symmetric (iii) transitive.

Solution:

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) ∈ R

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.

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.

Solution:

(i) Consider aRb if 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

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 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 + a b > 0

⇒ 1 + b a > 0

⇒ (b, a) ∈ R

So, the given relation is symmetric.

Transitivity:

Let (a, b) ∈R and (b, c) ∈R

⇒1 + a b > 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 relation we have to check whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary 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| ≤ b c

⇒ |a| ≤ c

⇒ (a, c) ∈ R

Thus, R is transitive.

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.

Solution:

Given R = {(a, b): b = a + 1}

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 = 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.

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

Solution:

Given R = {(ab): ≤ 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, R is neither reflexive, nor symmetric, nor transitive.

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

Solution:

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.

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.

Solution:

(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.

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