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Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th Relations and Functions

9. Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} ,given by:

(i) R = { (a,b) : |a - b| is a multiple of 4}

(ii) R = { (a,b) : a= b} is an equivalence relation. Find the set of all elements related to 1 in each case.

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Vishal Baghel | Contributor-Level 10

5 months ago

We have,

A= ${x \in 2, 0 \leq x \leq 1 2}$

The relation in set A is defined by

R= { $(a, b) : | a - b |$ is a multiple of 4}

For all $a \in A$ ,

$| a - a | = 0$ is a multiple of 4

So, $(a, a) \in R$ i.e., R is reflexive

For $a, b \in A & (a, b) \in R$ we have,

$| a - b |$ is multiple of 4

$| - (b - a) |$ is multiple of 4

$| b - a |$ is multiple of 4

So, $(b, a) \in R$

i.e., R is symmetric

for $a, b, c \in A$ $& (a, b) \in R & (b, c) \in R$

$| a - b |$ & $| b - c |$ is a multiple of 4

So $| a - b | + | b - c |$ is also a multiple of 4

$| a - b + b - c |$ is a multiple of 4

$| a - c |$ is a multiple of 4

So, $(a, c) \in R$

i.e., R is transitive

Hence, R is an equivalence relation.

Finding all set of elements related to 1

For $a \in A$

Then, $(a, 1) \in R$ i.e., $| a - 1 |$ is a multiple of 4

So, a can be 0 ≤ a ≤ 12

Only,

$| 1 - 1 | = 0$

$| 5 - 1 | = 4$ is a multiple of

...more

We have,

A= ${x \in 2, 0 \leq x \leq 1 2}$

The relation in set A is defined by

R= { $(a, b) : | a - b |$ is a multiple of 4}

For all $a \in A$ ,

$| a - a | = 0$ is a multiple of 4

So, $(a, a) \in R$ i.e., R is reflexive

For $a, b \in A & (a, b) \in R$ we have,

$| a - b |$ is multiple of 4

$| - (b - a) |$ is multiple of 4

$| b - a |$ is multiple of 4

So, $(b, a) \in R$

i.e., R is symmetric

for $a, b, c \in A$ $& (a, b) \in R & (b, c) \in R$

$| a - b |$ & $| b - c |$ is a multiple of 4

So $| a - b | + | b - c |$ is also a multiple of 4

$| a - b + b - c |$ is a multiple of 4

$| a - c |$ is a multiple of 4

So, $(a, c) \in R$

i.e., R is transitive

Hence, R is an equivalence relation.

Finding all set of elements related to 1

For $a \in A$

Then, $(a, 1) \in R$ i.e., $| a - 1 |$ is a multiple of 4

So, a can be 0 ≤ a ≤ 12

Only,

$| 1 - 1 | = 0$

$| 5 - 1 | = 4$ is a multiple of 4

$| 9 - 1 | = 8$

Hence, sets of elements related to A= ${1, 5, 9}$

The relation in set A is defined as R= ${(a, b) : a = b}$

For all $a \in A$ ,

a=a is true

so, $(a, a) \in R$

i.e., R is reflexive

for $a, b \in A & (a, b) \in R$ we have,

a=b

b=a i.e., $(b, a) \in R$

so, R is symmetric

for $a, b, c \in A$ , $(a, b) \in R$ and $(b, c) \in R$ . We have,

a=b and b=c

a=c i.e., $(a, c) \in R$

$∴$ R is transitive

Hence, R is an equivalence relation

Find all sets of elements related to 1

For $a \in A$ , we need $(a, 1) \in R$

i.e., a=1

so, set of elements related to 1 in A is {1}.

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9. Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} ,given by:

(i) R = { (a,b) : |a - b| is a multiple of 4}

(ii) R = { (a,b) : a= b} is an equivalence relation. Find the set of all elements related to 1 in each case.

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9. Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} ,given by: (i) R = { (a,b) : |a - b| is a multiple of 4} (ii) R = { (a,b) : a= b} is an equivalence relation. Find the set of all elements related to 1 in each case.

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9. Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} ,given by:

(i) R = { (a,b) : |a - b| is a multiple of 4}

(ii) R = { (a,b) : a= b} is an equivalence relation. Find the set of all elements related to 1 in each case.