• NCERT Chapter
Question 9

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

A = { x ∈ Z : 0 ≤ x ≤ 12} = {0,1,2,3,4,5,6,7,8,9,10,11,12}

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

For any element a ∈A, we have (aa) ∈ R as |a - a = 0|is a multiple of 4.

∴R is reflexive.

Now, let (ab) ∈ R ⇒ |a - b| is a multiple of 4.

⇒ |-(a - b)| = ⇒ |b - a| is a multiple of 4.

⇒ (ba) ∈ R

∴R is symmetric.

Now, let (ab), (bc) ∈ R.

⇒ |(a - b)| is a multiple of 4 and |(b - c)| is a multiple of 4.

⇒ (a - b) is a multiple of 4 and (b - c) is a multiple of 4.

⇒ (a - c) = (a – b) + (b – c) is a multiple of 4.

⇒ |a - c| is a multiple of 4.

⇒ (ac) ∈R

∴ R is transitive.

Hence, R is an equivalence relation.

The set of elements related to 1 is {1, 5, 9} since

|1 - 1| = 0 is a multiple of 4,

|5 - 1| = 4 is a multiple of 4, and

|9 - 1| = 8 is a multiple of 4.

(ii) R = {(ab): a = b}

For any element a ∈A, we have (aa) ∈ R, since a = a.

∴R is reflexive.

Now, let (ab) ∈ R.

⇒ a = b

⇒ b = a

⇒ (ba) ∈ R

∴R is symmetric.

Now, let (ab) ∈ R and (bc) ∈ R.

⇒ a = b and b = c

⇒ a = c

⇒ (ac) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.

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Let A = R – {3} and B = R – {1}. Consider the function  f : A → B defined by

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