# RD Sharma Solutions For Class 12 Maths Exercise 1.1 Chapter 1 Relation

The pdf of RD Sharma Solutions for Class 12 Exercise 1.1 of Chapter 1 Relations can be downloaded from the given links. The solutions for the exercise wise answers are prepared by experts at BYJUâ€™S in the best possible way and are easily understandable by students. In this exercise, students will gain knowledge about relations and types. The RD Sharma Solutions for Class 12 are stepwise and detailed to make learning easy for students. This exercise consists of two levels according to the increasing order of difficulties. Let us have a look at important topics covered in this exercise.

• Types of relations
• Void relation
• Universal relation
• Identity relation
• Reflexive relation
• Symmetric relation
• Transitive relation
• Antisymmetric relation

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Exercise 1.2 Solutions

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1. LetÂ AÂ 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)Â RÂ = {(x,Â y):Â xÂ andÂ yÂ work at the same place}

(ii) RÂ = {(x,Â y):Â xÂ andÂ yÂ live in the same locality}

(iii) RÂ = {(x,Â y):Â xÂ is wife of y}

(iv) RÂ = {(x,Â y):Â xÂ is father of y}

Solution:

(i) Given RÂ = {(x,Â y):Â xÂ andÂ yÂ work at the same place}

Now we have to check whether the relation is reflexive:

LetÂ xÂ beÂ anÂ arbitraryÂ elementÂ ofÂ R.

Then, x âˆˆR

â‡’ xÂ andÂ xÂ workÂ atÂ theÂ sameÂ placeÂ isÂ trueÂ sinceÂ theyÂ areÂ theÂ same.

â‡’(x,Â x)Â âˆˆR [condition for reflexive relation]

So,Â RÂ isÂ aÂ reflexiveÂ relation.

Now let us check Symmetric relation:

LetÂ (x,Â y) âˆˆR

â‡’xÂ andÂ yÂ workÂ atÂ theÂ sameÂ placeÂ [given]

â‡’yÂ andÂ xÂ workÂ atÂ theÂ sameÂ place

â‡’(y,Â x) âˆˆR

So,Â RÂ isÂ aÂ symmetricÂ relation.

Transitive relation:

LetÂ (x,Â y) âˆˆRÂ andÂ (y,Â z) âˆˆR.

Then, xÂ andÂ yÂ workÂ atÂ theÂ sameÂ place. [Given]

yÂ andÂ zÂ alsoÂ workÂ atÂ theÂ sameÂ place. [(y, z) âˆˆR]

â‡’Â x,Â yÂ andÂ zÂ allÂ workÂ atÂ theÂ sameÂ place.

â‡’xÂ andÂ zÂ workÂ atÂ theÂ sameÂ place.

â‡’Â (x,Â z) âˆˆR

So,Â RÂ isÂ aÂ transitiveÂ relation.

Hence R is reflexive, symmetric and transitive.

(ii) Given RÂ = {(x,Â y):Â xÂ andÂ yÂ live in the same locality}

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

LetÂ xÂ beÂ anÂ arbitraryÂ elementÂ ofÂ R.

Then, x âˆˆR

It is given that xÂ andÂ xÂ liveÂ inÂ theÂ sameÂ localityÂ isÂ trueÂ sinceÂ theyÂ areÂ theÂ same.

So,Â RÂ isÂ aÂ reflexiveÂ relation.

Symmetry:

Let (x, y) âˆˆ R

â‡’Â xÂ andÂ yÂ liveÂ inÂ theÂ sameÂ locality [given]

â‡’Â yÂ andÂ xÂ liveÂ inÂ theÂ sameÂ locality

â‡’ (y, x) âˆˆ R

So,Â RÂ isÂ aÂ symmetricÂ relation.

Transitivity:

LetÂ (x,Â y) âˆˆRÂ andÂ (y,Â z) âˆˆR.

Then,

xÂ andÂ yÂ liveÂ inÂ theÂ sameÂ localityÂ andÂ yÂ andÂ zÂ liveÂ inÂ theÂ sameÂ locality

â‡’Â x,Â yÂ andÂ zÂ allÂ liveÂ inÂ theÂ sameÂ locality

â‡’Â xÂ andÂ zÂ liveÂ inÂ theÂ sameÂ locality

â‡’ (x, z) âˆˆ R

So, R is a transitive relation.

Hence R is reflexive, symmetric and transitive.

(iii) Given RÂ = {(x,Â y):Â xÂ 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:

LetÂ xÂ beÂ anÂ elementÂ ofÂ R.

Then, xÂ isÂ wifeÂ ofÂ xÂ cannotÂ beÂ true.

â‡’ (x,Â x) âˆ‰R

So,Â RÂ isÂ notÂ aÂ reflexiveÂ relation.

Symmetric relation:

LetÂ (x,Â y) âˆˆR

â‡’ xÂ isÂ wifeÂ ofÂ y

â‡’ xÂ isÂ femaleÂ andÂ yÂ isÂ male

â‡’ yÂ cannotÂ beÂ wifeÂ ofÂ xÂ asÂ yÂ isÂ husbandÂ ofÂ x

â‡’ (y,Â x) âˆ‰R

So,Â RÂ isÂ notÂ aÂ symmetricÂ relation.

Transitive relation:

LetÂ (x,Â y) âˆˆR,Â butÂ (y,Â z) âˆ‰R

SinceÂ xÂ isÂ wifeÂ ofÂ y,Â butÂ yÂ cannotÂ beÂ theÂ wifeÂ ofÂ z,Â yÂ isÂ husbandÂ ofÂ x.

â‡’ xÂ isÂ notÂ theÂ wifeÂ ofÂ z

â‡’(x,Â z) âˆˆR

So,Â RÂ isÂ aÂ transitiveÂ relation.

(iv) Given RÂ = {(x,Â y):Â xÂ is father of y}

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

Reflexivity:

LetÂ xÂ beÂ anÂ arbitraryÂ elementÂ ofÂ R.

Then, xÂ isÂ fatherÂ ofÂ xÂ cannotÂ beÂ trueÂ sinceÂ noÂ oneÂ canÂ beÂ fatherÂ ofÂ himself.

So,Â RÂ isÂ notÂ aÂ reflexiveÂ relation.

Symmetry:

LetÂ (x,Â y) âˆˆR

â‡’ xÂ isÂ fatherÂ ofÂ y

â‡’ yÂ isÂ son/daughterÂ ofÂ x

â‡’ (y,Â x) âˆ‰R

So,Â RÂ isÂ notÂ aÂ symmetricÂ relation.

Transitivity:

LetÂ (x,Â y) âˆˆRÂ andÂ (y,Â z) âˆˆR.

Then,Â xÂ isÂ fatherÂ ofÂ yÂ andÂ yÂ isÂ fatherÂ ofÂ z

â‡’ xÂ isÂ grandfatherÂ ofÂ z

â‡’ (x,Â z) âˆ‰R

So,Â RÂ isÂ notÂ aÂ transitiveÂ relation.

2. Three relationsÂ R1,Â R2Â andÂ R3Â are defined on a setÂ AÂ = {a, b, c} as follows:
R1Â = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}
R2Â = {(a, a)}
R3Â = {(b, c)}
R4Â = {(a, b), (b, c), (c, a)}.

Find whether or not each of the relationsÂ R1,Â R2,Â R3,Â R4Â onÂ AÂ is (i) reflexive (ii) symmetric and (iii) transitive.

Solution:

(i) Consider R1

Given R1Â = {(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) âˆˆ R1

So, R1Â is reflexive.

Symmetric:

We see that the ordered pairs obtained by interchanging the components of R1Â are also in R1.

So, R1Â is symmetric.

Transitive:

Here, (a,Â b) âˆˆR1,Â (b,Â c) âˆˆR1Â andÂ alsoÂ (a,Â c) âˆˆR1

So, R1Â is transitive.

(ii) Consider R2

Given R2Â = {(a, a)}

Reflexive:

ClearlyÂ (a, a)Â âˆˆR2.

So, R2Â is reflexive.

Symmetric:

ClearlyÂ (a, a)Â âˆˆR

â‡’Â (a, a)Â âˆˆR.

So, R2Â is symmetric.

Transitive:

R2Â is clearly a transitive relation, since there is only one element in it.

(iii) Consider R3

Given R3Â = {(b, c)}

Reflexive:

Here,(b,Â b)âˆ‰Â R3Â neitherÂ (c,Â c)Â âˆ‰Â R3

So, R3Â is not reflexive.

Symmetric:

Here, (b,Â c)Â âˆˆR3,Â butÂ (c, b)Â âˆ‰R3

So, R3 is not symmetric.

Transitive:

Here, R3Â has only two elements.

Hence, R3Â is transitive.

(iv) Consider R4

Given R4Â = {(a, b), (b, c), (c, a)}.

Reflexive:

Here, (a,Â a)Â âˆ‰Â R4,Â (b,Â b) âˆ‰Â R4Â (c,Â c) âˆ‰Â R4

So,Â R4Â isÂ notÂ reflexive.

Symmetric:

Here, (a,Â b) âˆˆÂ R4,Â butÂ (b, a)Â âˆ‰Â R4.

So,Â R4Â isÂ notÂ symmetric

Transitive:

Here, (a,Â b) âˆˆR4,Â (b,Â c) âˆˆR4,Â butÂ (a,Â c) âˆ‰R4

So,Â R4Â isÂ notÂ transitive.

3. Â Test whether the following relationÂ R1, R2, and R3 are (i) reflexive (ii) symmetric and (iii) transitive:

(i) R1Â onÂ Q0Â defined by (a, b) âˆˆ R1Â â‡”Â aÂ = 1/b.

(ii) R2Â onÂ ZÂ defined by (a, b) âˆˆ R2Â â‡” |a â€“ b| â‰¤ 5

(iii) R3Â onÂ RÂ defined by (a, b) âˆˆÂ R3Â â‡”Â a2Â â€“ 4abÂ + 3b2Â = 0.

Solution:

(i) Given R1Â onÂ Q0Â defined by (a, b) âˆˆ R1Â â‡”Â aÂ = 1/b.

Reflexivity:

LetÂ aÂ be an arbitrary element ofÂ R1.

Then, aÂ âˆˆÂ R1

â‡’Â aÂ â‰ 1/aÂ forÂ allÂ aÂ âˆˆÂ Q0

So,Â R1Â isÂ notÂ reflexive.

Symmetry:

Let (a,Â b)Â âˆˆÂ R1

Then, (a,Â b)Â âˆˆÂ R1

Therefore we can write â€˜aâ€™ as a =1/b

â‡’ b = 1/a

â‡’ (b, a)Â âˆˆ R1

So,Â R1Â isÂ symmetric.

Transitivity:

Here, (a,Â b)Â âˆˆR1Â andÂ (b,Â c)Â âˆˆR2

â‡’ a = 1/b and b = 1/c

â‡’ a = 1/ (1/c) = c

â‡’ aÂ â‰  1/c

â‡’ (a, c)Â âˆ‰ R1

So,Â R1Â isÂ notÂ transitive.

(ii) Given R2Â onÂ ZÂ defined by (a, b) âˆˆ R2Â â‡” |a â€“ b| â‰¤ 5

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

Reflexivity:

LetÂ aÂ be an arbitrary element ofÂ R2.

Then, aÂ âˆˆÂ R2

On applying the given condition we get,

â‡’Â |Â aâˆ’aÂ |Â =Â 0Â â‰¤Â 5

So,Â R1Â isÂ reflexive.

Symmetry:

LetÂ (a,Â b) âˆˆÂ R2

â‡’Â |aâˆ’b|Â â‰¤Â 5Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Â [Since,Â |aâˆ’b|Â =Â |bâˆ’a|]

â‡’Â |bâˆ’a|Â â‰¤Â 5

â‡’Â (b,Â a)Â âˆˆÂ R2

So,Â R2Â isÂ symmetric.

Transitivity:

LetÂ (1,Â 3)Â âˆˆÂ R2Â andÂ (3,Â 7)Â âˆˆR2

â‡’|1âˆ’3|â‰¤5Â andÂ |3âˆ’7|â‰¤5

ButÂ |1âˆ’7|Â â‰°5

â‡’Â (1, 7)Â âˆ‰Â R2

So,Â R2Â isÂ notÂ transitive.

(iii) Given R3Â onÂ RÂ defined by (a, b) âˆˆÂ R3Â â‡”Â a2Â â€“ 4abÂ + 3b2Â = 0.

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

Reflexivity:

Let aÂ be an arbitrary element ofÂ R3.

Then, aÂ âˆˆÂ R3

â‡’Â a2Â âˆ’Â 4aÂ Ã—Â a+Â 3a2=Â 0

So,Â R3 isÂ reflexive

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R3

â‡’Â a2âˆ’4ab+3b2=0

ButÂ b2âˆ’4ba+3a2â‰ 0Â forÂ allÂ a,Â bÂ âˆˆÂ R

So,Â R3Â isÂ notÂ symmetric.

Transitivity:

Let (1,Â 2)Â âˆˆÂ R3Â andÂ (2,Â 3)Â âˆˆÂ R3

â‡’Â 1 âˆ’ 8 + 6 = 0Â andÂ 4 â€“ 24 + 27 = 0

ButÂ 1Â â€“Â 12 + 9 â‰  0

So,Â R3Â isÂ notÂ transitive.

4. LetÂ AÂ = {1, 2, 3}, and letÂ R1Â = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)},Â R2Â = {(2, 2), (3, 1), (1, 3)},Â R3Â = {(1, 3), (3, 3)}. Find whether or not each of the relationsÂ R1,Â R2,Â R3Â on A is (i) reflexive (ii) symmetric (iii) transitive.

Solution:

Consider R1

Given R1Â = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

Reflexivity:

Here, (1,Â 1),Â (2,Â 2),Â (3,Â 3)Â âˆˆR

So,Â R1Â isÂ reflexive.

Symmetry:

Here, (2, 1)Â âˆˆÂ R1,

ButÂ (1, 2)Â âˆ‰Â R1

So,Â R1Â isÂ notÂ symmetric.

Transitivity:

Here,Â (2,Â 1)Â âˆˆR1Â andÂ (1,Â 3) âˆˆR1,

ButÂ (2,Â 3) âˆ‰R1

So,Â R1Â isÂ notÂ transitive.

Now consider R2

Given R2Â = {(2, 2), (3, 1), (1, 3)}

Reflexivity:

Clearly,Â (1,Â 1)Â andÂ (3,Â 3) âˆ‰R2

So,Â R2Â isÂ notÂ reflexive.

Symmetry:

Here,Â (1,Â 3)Â âˆˆÂ R2Â andÂ (3,Â 1)Â âˆˆÂ R2

So,Â R2Â isÂ symmetric.

Transitivity:

Here,Â (1, 3)Â âˆˆÂ R2Â andÂ (3, 1)Â âˆˆÂ R2Â

ButÂ (3,Â 3) âˆ‰R2

So,Â R2Â isÂ notÂ transitive.

Consider R3

Given R3Â = {(1, 3), (3, 3)}

Reflexivity:

Clearly,Â (1, 1)Â âˆ‰Â R3

So,Â R3Â isÂ notÂ reflexive.

Symmetry:

Here,Â (1,Â 3)Â âˆˆÂ R3,Â butÂ (3,Â 1)Â âˆ‰Â R3

So,Â R3Â isÂ notÂ symmetric.

Transitivity:

Here,Â (1,Â 3)Â âˆˆÂ R3Â andÂ (3,Â 3)Â âˆˆÂ R3

Also,Â (1,Â 3)Â âˆˆÂ R3

So,Â R3Â isÂ transitive.

5. The following relation is defined on the set of real numbers.
(i) aRbÂ ifÂ aÂ â€“Â bÂ > 0

(ii) aRb iff 1 + a b > 0

(iii) aRb if |a| â‰¤ b.

Find whether relation is reflexive, symmetric or transitive.

Solution:

(i) Consider aRbÂ ifÂ aÂ â€“Â bÂ > 0

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

Reflexivity:

LetÂ aÂ be an arbitrary element ofÂ R.

Then, aÂ âˆˆÂ R

ButÂ a âˆ’ aÂ =Â 0Â â‰¯Â 0

So,Â thisÂ relationÂ isÂ notÂ reflexive.

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R

â‡’Â a âˆ’ bÂ >Â 0

â‡’Â âˆ’ (b âˆ’ a)Â >0

â‡’Â b âˆ’ aÂ <Â 0

So,Â theÂ givenÂ relationÂ isÂ notÂ symmetric.

Transitivity:

LetÂ (a,Â b) âˆˆRÂ andÂ (b,Â c) âˆˆR.

Then, a âˆ’ bÂ >Â 0Â andÂ b âˆ’ cÂ > 0

aÂ â€“Â b + bÂ âˆ’Â cÂ >Â 0

â‡’Â aÂ â€“Â c >Â 0

â‡’Â (a,Â c)Â âˆˆÂ R.

So,Â theÂ givenÂ relationÂ isÂ 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:

LetÂ aÂ be an arbitrary element ofÂ R.

Then, aÂ âˆˆÂ R

â‡’ 1Â +Â aÂ Ã—Â aÂ >Â 0

i.e.Â 1Â +Â a2Â >Â 0Â  Â  Â  Â  Â  Â  Â [Since,Â squareÂ ofÂ anyÂ numberÂ isÂ positive]

So,Â theÂ givenÂ relationÂ isÂ reflexive.

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R

â‡’Â 1 +Â a bÂ >Â 0

â‡’Â 1 + b aÂ >Â 0

â‡’Â (b,Â a)Â âˆˆÂ R

So,Â theÂ givenÂ relationÂ isÂ symmetric.

Transitivity:

LetÂ (a,Â b) âˆˆRÂ andÂ (b,Â c) âˆˆR

â‡’1 +Â a bÂ >Â 0Â andÂ 1 +Â b cÂ >0

ButÂ 1+Â acÂ â‰¯Â 0

â‡’Â (a,Â c)Â âˆ‰Â R

So,Â theÂ givenÂ relationÂ isÂ notÂ transitive.

(iii) Consider aRb if |a| â‰¤ b.

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

Reflexivity:

LetÂ aÂ be an arbitrary element ofÂ R.

Then,Â aÂ âˆˆÂ RÂ Â  Â  Â  Â  Â  Â  Â  Â Â Â [Since,Â |a|=a]

â‡’Â |a|â‰®Â a

So,Â RÂ isÂ notÂ reflexive.

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R

â‡’Â |a|Â â‰¤Â b

â‡’Â |b|Â â‰°Â aÂ forÂ allÂ a,Â bÂ âˆˆÂ R

â‡’Â (b,Â a)Â âˆ‰Â R

So,Â RÂ isÂ notÂ symmetric.

Transitivity:

LetÂ (a,Â b)Â âˆˆÂ RÂ andÂ (b,Â c)Â âˆˆÂ R

â‡’Â |a|Â â‰¤Â bÂ andÂ |b|Â â‰¤Â c

MultiplyingÂ theÂ correspondingÂ sides,Â weÂ get

|a|Â Ã—Â |b|Â â‰¤Â b c

â‡’Â |a|Â â‰¤Â c

â‡’Â (a,Â c)Â âˆˆÂ R

Thus,Â RÂ isÂ transitive.

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

Â

Solution:

Given R = {(a,Â b):Â bÂ =Â aÂ + 1}

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

Let a be an arbitrary element ofÂ R.

Then, aÂ =Â aÂ +Â 1Â cannotÂ beÂ trueÂ forÂ allÂ aÂ âˆˆÂ A.

â‡’Â (a,Â a)Â âˆ‰Â R

So,Â RÂ isÂ notÂ reflexiveÂ onÂ A.

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R

â‡’Â bÂ =Â aÂ +Â 1

â‡’Â âˆ’aÂ =Â âˆ’bÂ +Â 1

â‡’Â aÂ =Â bÂ âˆ’Â 1

Thus,Â (b,Â a)Â âˆ‰Â R

So,Â RÂ isÂ notÂ symmetricÂ onÂ A.

Transitivity:

LetÂ (1,Â 2)Â andÂ (2,Â 3)Â âˆˆÂ R

â‡’Â 2Â =Â 1Â +Â 1Â andÂ 3

2Â +Â 1Â Â isÂ true.

ButÂ 3Â â‰ Â 1+1

â‡’Â (1,Â 3)Â âˆ‰Â R

So,Â RÂ isÂ notÂ transitiveÂ onÂ A.

7. Check whether the relation R onÂ RÂ defined as R = {(a,Â b):Â aÂ â‰¤Â b3} is reflexive, symmetric or transitive.

Solution:

Given R = {(a,Â b):Â aÂ â‰¤Â b3}

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

âˆ´ R is not reflexive.

Now,

(1, 2) âˆˆ R (as 1 < 23Â = 8)

But,

(2, 1) âˆ‰ R (as 2 > 13Â = 1)

âˆ´ R is not symmetric.

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

ButÂ (3, 6/5)Â âˆ‰ R as 3 > (6/5)3

âˆ´ R is not transitive.

Hence, R isÂ 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:

LetÂ AÂ be a set.

Then, IdentityÂ relationÂ IA=IAÂ isÂ reflexive,Â sinceÂ (a,Â a)Â âˆˆ A âˆ€a

The converse of it need not be necessarily true.

Consider the setÂ AÂ = {1, 2, 3}

Here,

RelationÂ RÂ = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive onÂ A.

However,Â RÂ is not an identity relation.

9. IfÂ AÂ = {1, 2, 3, 4} define relations onÂ AÂ 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 onÂ AÂ having properties of being reflexive, transitive, but not symmetric is

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

RelationÂ RÂ satisfiesÂ reflexivityÂ andÂ transitivity.

â‡’ (1,Â 1),Â (2,Â 2),Â (3,Â 3)Â âˆˆÂ R

AndÂ (1,Â 1),Â (2,Â 1)Â âˆˆÂ RÂ â‡’ (1,Â 1)Â âˆˆÂ R

However,Â (2,Â 1)Â âˆˆÂ R,Â butÂ (1,Â 2)Â âˆ‰Â R

(ii) Â The relation onÂ AÂ having properties of being reflexive, transitive, but not symmetric is

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

RelationÂ RÂ satisfiesÂ reflexivityÂ andÂ transitivity.

â‡’ (1,Â 1),Â (2,Â 2),Â (3,Â 3)Â âˆˆÂ R

AndÂ (1,Â 1),Â (2,Â 1)Â âˆˆÂ RÂ â‡’ (1,Â 1)Â âˆˆÂ R

However,Â (2,Â 1)Â âˆˆÂ R,Â butÂ (1,Â 2)Â âˆ‰Â R

(iii) The relation onÂ AÂ having properties of being symmetric, reflexive and transitive is

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

The relationÂ RÂ is an equivalence relation onÂ A.

#### 1 Comment

1. sahil slathia

lots of love to this app