# RD Sharma Solutions For Class 12 Maths Exercise 1.2 Chapter 1 Relation

In this exercise, students will mainly get conceptual knowledge about equivalence relations. A relation is said to be an equivalence relation if it is reflexive, symmetric and transitive. Class 12 is a major turning point for the students to help them shape and achieve their future goals. The solutions for these problems are prepared in a comprehensive manner, which improves the problem-solving capacity of students. Exercise 1.2 of Chapter 1 Relations are available here in PDF format. To know more about these concepts, students can accessÂ RD Sharma Solutions for Class 12 Maths.

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Exercise 1.1 Solutions

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1. Show that the relationÂ RÂ defined byÂ RÂ = {(a, b):Â a â€“ bÂ is divisible by 3;Â a, bÂ âˆˆÂ Z} is an equivalence relation.

Solution:

Given RÂ = {(a, b):Â a â€“ bÂ is divisible by 3;Â a, bÂ âˆˆÂ Z} is a relation

To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

Let us check these properties on R.

Reflexivity:

LetÂ aÂ beÂ anÂ arbitraryÂ elementÂ ofÂ R.

Then, a â€“ a = 0 = 0 Ã— 3

â‡’Â a âˆ’ aÂ isÂ divisibleÂ byÂ 3

â‡’Â (a,Â a)Â âˆˆÂ RÂ forÂ allÂ aÂ âˆˆÂ Z

So,Â RÂ isÂ reflexiveÂ onÂ Z.

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R

â‡’Â a âˆ’ bÂ isÂ divisibleÂ byÂ 3

â‡’Â a âˆ’ bÂ = 3pÂ forÂ someÂ pÂ âˆˆÂ Z

â‡’Â b âˆ’ aÂ = 3Â (âˆ’p)

Here,Â âˆ’pÂ âˆˆÂ Z

â‡’Â b âˆ’ aÂ isÂ divisibleÂ byÂ 3

â‡’Â (b,Â a)Â âˆˆÂ RÂ forÂ allÂ a,Â bÂ âˆˆÂ Z

So,Â RÂ isÂ symmetricÂ onÂ Z.

Transitivity:

LetÂ (a,Â b)Â andÂ (b,Â c)Â âˆˆÂ R

â‡’Â a âˆ’ bÂ andÂ b âˆ’ cÂ areÂ divisibleÂ byÂ 3

â‡’Â a â€“ b = 3pÂ forÂ someÂ pÂ âˆˆÂ Z

AndÂ b âˆ’Â cÂ =Â 3qÂ forÂ someÂ qÂ âˆˆÂ Z

AddingÂ theÂ aboveÂ two equations,Â weÂ get

aÂ âˆ’Â bÂ +Â b â€“Â c =Â 3p +Â 3q

â‡’Â a âˆ’ cÂ = 3Â (p + q)

Here,Â p + qÂ âˆˆÂ Z

â‡’Â a âˆ’ cÂ isÂ divisibleÂ byÂ 3

â‡’Â (a,Â c)Â âˆˆÂ RÂ forÂ allÂ a,Â cÂ âˆˆÂ Z

So,Â RÂ isÂ transitiveÂ onÂ Z.

Therefore R is reflexive, symmetric and transitive.

Hence,Â RÂ is an equivalence relation onÂ Z.

2. Show that the relationÂ RÂ on the setÂ ZÂ of integers, given by
RÂ = {(a, b): 2 dividesÂ a â€“ b}, is an equivalence relation.

Solution:

Given RÂ = {(a, b): 2 dividesÂ a â€“ b} is a relation defined on Z.

To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

Let us check these properties on R.

Reflexivity:

LetÂ aÂ beÂ anÂ arbitraryÂ elementÂ ofÂ theÂ setÂ Z.

Then,Â aÂ âˆˆÂ R

â‡’Â a âˆ’ aÂ =Â 0Â =Â 0Â Ã—Â 2

â‡’Â 2Â dividesÂ aÂ âˆ’Â a

â‡’Â (a,Â a)Â âˆˆÂ RÂ forÂ allÂ aÂ âˆˆÂ Z

So,Â RÂ isÂ reflexiveÂ onÂ Z.

Symmetry:

LetÂ (a,Â b) âˆˆÂ R

â‡’Â 2Â dividesÂ a âˆ’ b

â‡’ (a-b)/2 = pÂ forÂ someÂ pÂ âˆˆÂ Z

â‡’ (b-a)/2 = – p

Here,Â âˆ’pÂ âˆˆÂ Z

â‡’Â 2Â dividesÂ bÂ âˆ’Â a

â‡’Â (b,Â a) âˆˆÂ RÂ forÂ allÂ a,Â bÂ âˆˆÂ Z

So,Â RÂ isÂ symmetricÂ onÂ Z

Transitivity:

LetÂ (a,Â b)Â andÂ (b,Â c)Â âˆˆÂ R

â‡’Â 2Â dividesÂ aâˆ’bÂ andÂ 2Â dividesÂ bâˆ’c

â‡’Â (a-b)/2Â =Â pÂ and (b-c)/2 =Â qÂ forÂ someÂ p,Â qÂ âˆˆÂ Z

AddingÂ theÂ aboveÂ two equations,Â weÂ get

(a – b)/2 + (b – c)/2 = p + q

â‡’Â (a – c)/2 =Â p +q

Here,Â p+Â qÂ âˆˆÂ Z

â‡’ 2Â dividesÂ aÂ âˆ’Â c

â‡’Â (a,Â c) âˆˆÂ RÂ forÂ allÂ a,Â cÂ âˆˆÂ Z

So,Â RÂ isÂ transitiveÂ onÂ Z.

Therefore R is reflexive, symmetric and transitive.

Hence,Â RÂ is an equivalence relation onÂ Z.

3. Prove that the relationÂ RÂ onÂ ZÂ defined by (a,Â b) âˆˆÂ RÂ â‡”Â aÂ âˆ’Â bÂ is divisible by 5 is an equivalence relation onÂ Z.

Solution:

Given relationÂ RÂ onÂ ZÂ defined by (a,Â b) âˆˆÂ RÂ â‡”Â aÂ âˆ’Â bÂ is divisible by 5

To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

Let us check these properties on R.

Reflexivity:

LetÂ aÂ beÂ anÂ arbitraryÂ elementÂ ofÂ R.Â Then,

â‡’Â a âˆ’ aÂ =Â 0Â =Â 0Â Ã—Â 5

â‡’Â a âˆ’ aÂ isÂ divisibleÂ byÂ 5

â‡’Â (a,Â a)Â âˆˆÂ RÂ forÂ allÂ aÂ âˆˆÂ Z

So,Â RÂ isÂ reflexiveÂ onÂ Z.

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R

â‡’Â a âˆ’ bÂ isÂ divisibleÂ byÂ 5

â‡’Â a âˆ’ bÂ =Â 5pÂ forÂ someÂ pÂ âˆˆÂ Z

â‡’Â b âˆ’ aÂ =Â 5Â (âˆ’p)

Here,Â âˆ’pÂ âˆˆÂ ZÂ Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [SinceÂ pÂ âˆˆÂ Z]

â‡’Â b âˆ’ aÂ isÂ divisibleÂ byÂ 5

â‡’Â (b,Â a)Â âˆˆÂ RÂ forÂ allÂ a,Â bÂ âˆˆÂ Z

So,Â RÂ isÂ symmetricÂ onÂ Z.

Transitivity:

LetÂ (a,Â b)Â andÂ (b,Â c)Â âˆˆÂ R

â‡’Â a âˆ’ bÂ isÂ divisibleÂ byÂ 5

â‡’Â a âˆ’ bÂ =Â 5pÂ forÂ someÂ Z

Also,Â b âˆ’ cÂ isÂ divisibleÂ byÂ 5

â‡’Â b âˆ’ cÂ =Â 5qÂ forÂ someÂ Z

AddingÂ theÂ aboveÂ two equations,Â weÂ get

aÂ âˆ’bÂ +Â b âˆ’ cÂ =Â 5pÂ +Â 5q

â‡’Â a âˆ’ cÂ =Â 5Â (Â pÂ +Â qÂ )

â‡’Â a âˆ’ cÂ isÂ divisibleÂ byÂ 5

Here,Â pÂ +Â qÂ âˆˆÂ Z

â‡’Â (a,Â c)Â âˆˆÂ RÂ forÂ allÂ a,Â cÂ âˆˆÂ Z

So,Â RÂ isÂ transitiveÂ onÂ Z.

Therefore R is reflexive, symmetric and transitive.

Hence,Â RÂ is an equivalence relation onÂ Z.

4. LetÂ nÂ be a fixed positive integer. Define a relationÂ RÂ onÂ ZÂ as follows:
(a,Â b) âˆˆÂ RÂ â‡”Â aÂ âˆ’Â bÂ is divisible byÂ n.
Show thatÂ RÂ is an equivalence relation onÂ Z.

Solution:

Given (a,Â b) âˆˆÂ RÂ â‡”Â aÂ âˆ’Â bÂ is divisible byÂ n is a relation R defined on Z.

To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

Let us check these properties on R.

Reflexivity:

LetÂ aÂ âˆˆÂ N

Here, aÂ âˆ’Â aÂ =Â 0Â =Â 0Â Ã—Â n

â‡’Â a âˆ’ aÂ isÂ divisibleÂ byÂ n

â‡’Â (a,Â a)Â âˆˆÂ R

â‡’Â (a,Â a)Â âˆˆÂ RÂ forÂ allÂ aÂ âˆˆÂ Z

So,Â RÂ isÂ reflexiveÂ onÂ Z.

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R

Here, a âˆ’ bÂ isÂ divisibleÂ byÂ n

â‡’Â a âˆ’ bÂ =Â n pÂ forÂ someÂ pÂ âˆˆÂ Z

â‡’Â b âˆ’ aÂ =Â nÂ (âˆ’p)

â‡’Â b âˆ’ aÂ isÂ divisibleÂ byÂ nÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â [Â pÂ âˆˆÂ Zâ‡’Â âˆ’Â pÂ âˆˆÂ Z]

â‡’Â (b,Â a)Â âˆˆÂ R

So,Â RÂ isÂ symmetricÂ onÂ Z.

Transitivity:

LetÂ (a,Â b)Â andÂ (b,Â c)Â âˆˆÂ R

Here,Â a âˆ’ bÂ isÂ divisibleÂ byÂ nÂ andÂ b âˆ’ cÂ isÂ divisibleÂ byÂ n.

â‡’Â a âˆ’ b=Â n pÂ forÂ someÂ pÂ âˆˆÂ Z

AndÂ bâˆ’cÂ =Â n qÂ forÂ someÂ qÂ âˆˆÂ Z

a â€“ b +Â b âˆ’ cÂ =Â n pÂ +Â n q

â‡’Â a âˆ’ cÂ =Â nÂ (p + q)

â‡’Â (a,Â c) âˆˆÂ RÂ forÂ allÂ a,Â cÂ âˆˆÂ Z

So,Â RÂ isÂ transitiveÂ onÂ Z.

Therefore R is reflexive, symmetric and transitive.

Hence,Â R is an equivalence relation on Z.

5. LetÂ ZÂ be the set of integers. Show that the relation RÂ = {(a,Â b):Â a,Â bÂ âˆˆÂ ZÂ andÂ aÂ +Â bÂ is even} is an equivalence relation onÂ Z.

Solution:

Given RÂ = {(a,Â b):Â a,Â bÂ âˆˆÂ ZÂ andÂ aÂ +Â bÂ is even} is a relation defined on R.

Also given that ZÂ be the set of integers

To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

Let us check these properties on R.

Reflexivity:

LetÂ aÂ beÂ anÂ arbitraryÂ elementÂ ofÂ Z.

Then, aÂ âˆˆÂ R

Clearly,Â a + aÂ =Â 2aÂ isÂ evenÂ forÂ allÂ aÂ âˆˆÂ Z.

â‡’Â (a,Â a)Â âˆˆÂ RÂ forÂ allÂ aÂ âˆˆÂ Z

So,Â RÂ isÂ reflexiveÂ onÂ Z.

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R

â‡’Â a + bÂ isÂ even

â‡’Â b + aÂ isÂ even

â‡’Â (b,Â a)Â âˆˆÂ RÂ forÂ allÂ a,Â bÂ âˆˆÂ Z

So,Â RÂ isÂ symmetricÂ onÂ Z.

Transitivity:

LetÂ (a,Â b)Â andÂ (b,Â c)Â âˆˆÂ R

â‡’Â a + bÂ andÂ b + cÂ areÂ even

Now,Â letÂ a + bÂ =Â 2xÂ forÂ someÂ xÂ âˆˆÂ Z

AndÂ b + cÂ =Â 2yÂ forÂ someÂ yÂ âˆˆÂ Z

AddingÂ theÂ aboveÂ two equations,Â weÂ get

A + 2bÂ + cÂ =Â 2xÂ +Â 2y

â‡’Â a + cÂ =Â 2Â (x + y âˆ’ b),Â whichÂ isÂ evenÂ forÂ allÂ x,Â y,Â bÂ âˆˆÂ Z

Thus,Â (a,Â c)Â âˆˆÂ R

So,Â RÂ isÂ transitiveÂ onÂ Z.

Therefore R is reflexive, symmetric and transitive.

Hence,Â RÂ is an equivalence relation onÂ Z

6. mÂ is said to be related toÂ nÂ ifÂ mÂ andÂ nÂ are integers andÂ mÂ âˆ’Â nÂ is divisible by 13. Does this define an equivalence relation?

Solution:

Given that mÂ is said to be related toÂ nÂ ifÂ mÂ andÂ nÂ are integers andÂ mÂ âˆ’Â nÂ is divisible by 13

Now we have to check whether the given relation is equivalence or not.

To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

LetÂ R = {(m,Â n):Â m,Â nÂ âˆˆ ZÂ :Â m âˆ’ nÂ isÂ divisibleÂ byÂ 13}

Let us check these properties on R.

Reflexivity:

LetÂ mÂ beÂ anÂ arbitraryÂ elementÂ ofÂ Z.

Then, mÂ âˆˆÂ R

â‡’Â m âˆ’ mÂ =Â 0Â =Â 0Â Ã—Â 13

â‡’Â m âˆ’ mÂ isÂ divisibleÂ byÂ 13

â‡’Â (m,Â m)Â isÂ reflexiveÂ onÂ Z.

Symmetry:

LetÂ (m,Â n)Â âˆˆÂ R.

Then, m âˆ’ nÂ isÂ divisibleÂ byÂ 13

â‡’Â m âˆ’ nÂ =Â 13p

Here,Â pÂ âˆˆÂ Z

â‡’Â n â€“ m = 13Â (âˆ’p)

Here,Â âˆ’pÂ âˆˆÂ Z

â‡’Â n âˆ’ mÂ isÂ divisibleÂ byÂ 13

â‡’Â (n,Â m) âˆˆÂ RÂ forÂ allÂ m,Â nÂ âˆˆÂ Z

So,Â RÂ isÂ symmetricÂ onÂ Z.

Transitivity:

LetÂ (m,Â n)Â andÂ (n,Â o) âˆˆR

â‡’Â m âˆ’ nÂ andÂ n âˆ’ oÂ areÂ divisibleÂ byÂ 13

â‡’Â m â€“ n = 13pÂ andÂ nÂ âˆ’Â oÂ = 13qÂ forÂ someÂ p,Â qÂ âˆˆÂ Z

AddingÂ theÂ aboveÂ two equations,Â weÂ get

m â€“Â n + n âˆ’ oÂ = 13pÂ +Â 13q

â‡’Â mâˆ’oÂ = 13Â (p + q)

Here,Â p + qÂ âˆˆÂ Z

â‡’Â m âˆ’ oÂ isÂ divisibleÂ byÂ 13

â‡’ (m,Â o)Â âˆˆÂ RÂ forÂ allÂ m,Â oÂ âˆˆÂ Z

So,Â RÂ isÂ transitiveÂ onÂ Z.

Therefore R is reflexive, symmetric and transitive.

Hence,Â RÂ is an equivalence relation onÂ Z.

7. LetÂ RÂ be a relation on the setÂ AÂ of ordered pair of integers defined by (x,Â y)Â RÂ (u,Â v) ifÂ xvÂ =Â y u. Show thatÂ RÂ is an equivalence relation.

Solution:

First let R be a relation on A

It is given that setÂ AÂ of ordered pair of integers defined by (x,Â y)Â RÂ (u,Â v) ifÂ xvÂ =Â y u

Now we have to check whether the given relation is equivalence or not.

To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

Reflexivity:

LetÂ (a,Â b)Â beÂ anÂ arbitraryÂ elementÂ ofÂ theÂ setÂ A.

Then, (a,Â b)Â âˆˆÂ A

â‡’Â a bÂ =Â b a

â‡’Â (a,Â b)Â RÂ (a,Â b)

Thus,Â RÂ isÂ reflexiveÂ onÂ A.

Symmetry:

LetÂ (x,Â y)Â andÂ (u,Â v) âˆˆAÂ suchÂ thatÂ (x,Â y)Â RÂ (u,Â v).Â Then,

x v = y u

â‡’Â v x = u y

â‡’Â u y = v x

â‡’Â (u,Â v)Â RÂ (x,Â y)

So,Â RÂ isÂ symmetricÂ onÂ A.

Transitivity:

LetÂ (x,Â y),Â (u,Â v)Â andÂ (p,Â q) âˆˆRÂ suchÂ thatÂ (x,Â y)Â RÂ (u,Â v)Â andÂ (u,Â v)Â RÂ (p,Â q)

â‡’Â x vÂ =Â y uÂ andÂ u qÂ =Â v p

MultiplyingÂ theÂ correspondingÂ sides,Â weÂ get

x vÂ Ã—Â u qÂ =Â y uÂ Ã—Â v p

â‡’Â x qÂ =Â y p

â‡’Â (x,Â y)Â RÂ (p,Â q)

So,Â RÂ isÂ transitiveÂ onÂ A.

Therefore R is reflexive, symmetric and transitive.

Hence,Â RÂ is an equivalence relation onÂ A.

8. Show that the relationÂ RÂ on the setÂ AÂ = {xÂ âˆˆÂ Z; 0 â‰¤Â xÂ â‰¤ 12}, given byÂ RÂ = {(a,Â b):Â aÂ =Â b}, is an equivalence relation. Find the set of all elements related to 1.

Solution:

Given setÂ AÂ = {xÂ âˆˆÂ Z; 0 â‰¤Â xÂ â‰¤ 12}

Also given that relation RÂ = {(a,Â b):Â aÂ =Â b} is defined on set A

Now we have to check whether the given relation is equivalence or not.

To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

Reflexivity:

LetÂ aÂ be an arbitrary element ofÂ A.

Then,Â aÂ âˆˆÂ R

â‡’Â aÂ =Â aÂ  Â  Â  Â  Â Â [Since,Â everyÂ elementÂ isÂ equalÂ toÂ itself]

â‡’Â (a,Â a)Â âˆˆÂ RÂ forÂ allÂ aÂ âˆˆÂ A

So,Â RÂ isÂ reflexiveÂ onÂ A.

Symmetry:

LetÂ (a,Â b)Â âˆˆÂ R

â‡’Â aÂ b

â‡’Â bÂ =Â a

â‡’Â (b,Â a)Â âˆˆÂ RÂ forÂ allÂ a,Â bÂ âˆˆÂ A

So,Â RÂ isÂ symmetricÂ onÂ A.

Transitivity:

LetÂ (a,Â b)Â andÂ (b,Â c)Â âˆˆÂ R

â‡’Â aÂ =bÂ andÂ bÂ =Â c

â‡’Â aÂ =Â bÂ c

â‡’Â aÂ =Â c

â‡’Â (a,Â c)Â âˆˆÂ R

So,Â RÂ isÂ transitiveÂ onÂ A.

Hence,Â RÂ is an equivalence relation onÂ A.

Therefore R is reflexive, symmetric and transitive.

The set of all elements related to 1 is {1}.