Class 12-science RD SHARMA Solutions Maths Chapter 1 - Relations

Given: R = {(1, 2)}

R is not reflexive as (1, 1), (2, 2), (3, 3) ∉ R.

Since (1, 2) ∈ R but (2, 1) ∉ R

Therefore, R is not symmetric.

Also, R is not transitive.

A relation R in A is said to be reflexive if aRa for all a∈A

R is said to be transitive if aRb and bRc ⇒ aRc

for all a, b, c ∈ A.

Hence for R to be reflexive (b, b) and (c, c) must be there in the set R.

Also for R to be transitive (a, c) must be in R because (a, b) ∈ R and (b, c) ∈ R so (a, c) must be in R.

So at least 3 ordered pairs must be added for R to be reflexive and transitive.

A relation R in A is said to be reflexive if aRa for all a∈A, R is symmetric if aRb ⇒ bRa, for all a, b ∈ A and it is said to be transitive if aRb and bRc ⇒ aRc for all a, b, c ∈ A.

• x > y,  x, y ϵ N

 (x, y) ϵ {(2, 1), (3, 1).......(3, 2), (4, 2)....}

This is not reflexive as (1, 1), (2, 2)....are absent.

This is not symmetric as (2,1) is present but (1,2) is absent.

This is transitive as (3, 2) ϵ R and (2,1) ϵ R also (3,1) ϵ R ,similarly this property satisfies all cases.

• x + y = 10,  x, y ϵ N

 (x, y)ϵ {(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3), (4, 6), (6, 4), (5, 5)}

 This is not reflexive as (1, 1),(2, 2)..... are absent.

 This only follows the condition of symmetric set as  (1, 9)ϵR also (9, 1)ϵR similarly other cases are also satisfy the condition.

This is not transitive because {(1, 9),(9, 1)}ϵR but (1, 1) is absent.

• xy is square of an integer, x, y ϵ N

 (x, y) ϵ {(1, 1), (2, 2), (4, 1), (1, 4), (3, 3), (9, 1),  (1, 9), (4, 4), (2, 8), (8, 2), (16, 1), (1, 16)...........}

 This is reflexive as (1,1),(2,2)..... are present.

 This is also symmetric because if aRb ⇒ bRa, for all a,bϵN.

 This is transitive also because if aRb and bRc ⇒ aRc for all a, b, c ϵ N.

• x + 4y = 10, x, y ϵ N

 (x, y) ϵ {(6, 1), (2, 2)}

 This is not reflexive as (1, 1), (2, 2).....are absent.

 This is not symmetric because (6,1) ϵ R but (1,6) is absent.

This is not transitive as there are only two elements in the set having no element common.

Given: R = {(a, b) : a < b}

Clearly, 1 < 2

That is, (1, 2) ∈ R

But 2 ≮ 1

So, (2, 1) ∉ R

Therefore, R is not symmetric.

Let (a, b), (b, c) ∈ R

∴ a < b and b < c

∴ a < c

 Therefore, (a, c) ∈ R

Thus, R is transitive.

Given: R = {(x, y) ∈ W × W such that x and y have at least one letter in common}

Clearly, (x, x) ∈ R as x and x have all the letters common.

Therefore, R is reflexive.

Now suppose, x and y have at least one letter in common.

So, y and x will also have at least one letter in common.

Therefore, R is symmetric.

Now take, x = Sun, y = Moon and z = Mars

Here, (x, y) ∈ R and (y, z) ∈ R as x and y have the letter "n" common whereas y and z have the letter "m" common.

But, x and z do not have any letter common.

Therefore, (x, z) ∉ R.

Thus, R is not transitive.

Given: R = {(a, b): a is divisor of b}

Any natural number "a" divides itself.

So, a is divisor of a.

Therefore, R is reflexive.

We know that 2 is divisor of 4.

So, (2, 4) ∈ R

But 4 does not divide 2, then (4, 2) ∉ R

Therefore, R is not symmetric.

As R is not symmetric.

Thus, R is not an equivalence relation.

To check reflexive:

Now,  which is real.

So, z₁ R z₁.

Therefore, R is reflexive.

To check symmetric:

Suppose z₁ R z₂

 is real

Let  where k is real.

Now,

Therefore,  is real as -k is real.

Therefore, R is symmetric.

To check transitive:

Suppose z₁ R z₂ and z₂ R z₃

 are real

Let

 … (i)

Now,  which is real.

Thus, R is transitive.

Hence, R is an equivalence relation.

Also we need to find the set of all elements related to 1.

Since the relation is given by, R={(a,b):a=b}, and 1 is an element of A,

R={(1,1):1=1}

Thus, the set of all elements related to 1 is 1.

R = {(2,8),(3, 27)}

∴ The range set of R is {8, 27}.

a, b ∈ Z and R is given by R={(a, b): 2 divides a-b}.

The equivalence classes can be taken as [0], [1].

Note that, for 0 ≤ i ≤ 1, [i] = {2n + i : n ∈ Z}.

So equivalence class [0] = {2n : n ∈ Z}

It is clear that all the elements of equivalence class [0] are even.

Hence, equivalence class [0]={0, ± 2, ± 4, ± 6,...}

A relation R in A is said to be reflexive if aRa for all a∈A, R is symmetric if aRb ⇒ bRa, for all a, b ∈ A and it is said to be transitive if aRb and bRc ⇒ aRc for all a, b, c ∈ A. Any relation which is reflexive, symmetric and transitive is called an equivalence relation.

Given R is reflexive and transitive but not symmetric hence to make symmetric (3,1) added to R because (1,3) is already there. Hence (3, 1) is the single ordered pair which needs to be added to R to make it the smallest equivalence relation.

A relation R in A is said to be reflexive if aRa for all a∈A, R is symmetric if aRb ⇒ bRa, for all a, b ∈ A and it is said to be transitive if aRb and bRc ⇒ aRc for all a, b, c ∈ A

R is reflexive because {(0, 0),(1, 1),(2, 2),(3, 3)}∈R .

R is symmetric, because {(0,1),(0,3)}∈R and {(1,0),(3,0)} ∈ R .

but R is not transitive since for (1, 0) ∈ R and(0, 3) ∈ R whereas (1, 3) ∉ R.

A = {1, 2, 3, 4, 5} and R = {(a, b) :|a² - b² | < 8}

If a = 1 then from R we get b = 1, 2

If a = 2 then from R we get b = 1, 2, 3

If a = 3 then from R we get b = 2, 3, 4

If a = 4 then from R we get b = 3, 4

If a = 5 then from R we get b = 5

Hence R={(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3),(3,4),(4,3), (4,4),(5,5)}

If a, b ∈ N then b must be an even integer so that a∈N

Hence only possible values for b are 2,4,6,8.

if b=2 , it gives a=12

if b=4 , it gives a=9

if b=6 , it gives a=6

if b=8 , it gives a=3

hence (a,b)ϵ{(3,8),(6,6),(9,4),(12,2)}

A relation R in A is said to be reflexive if aRa for all a∈A, R is symmetric if aRb ⇒ bRa, for all a, b ∈ A and it is said to be transitive if aRb and bRc ⇒ aRc for all a, b, c ∈ A. Any relation which is reflexive, symmetric and transitive is called an equivalence relation.

Hence smallest possible equivalence relation is {(1,1),(2,2),(3,3)}.

This relation satisfies all the conditions for an equivalence relation.