Short Answer TypeR = {(T1, T2) : T1Â is congruent to T2}
(i) Since every triangle is congruent to itself
∴ R is reflexive.
(ii) Also (T1, T2) ∈ R ⇒ T1 is congruent to T2 ⇒ T2 is congruent to T1 (T2,T1) ∈ R
(T1,T2) ∈ R ⇒ (T2,T1) ∈ R ⇒ R is symmetric.
(iii) Again (T1, T2), (T2, T3) ∈ R ⇒ T1 i is congruent to T2 and T2 is congruent to T3 ∴ T1 is congruent to T3 ∴ (T1,T3) ∈ R (T1,T2), (T2,T3) ∈ R ⇒ (T1,T3) ∈ R ∴ R is transitive.
From (i), (ii), (iii), it is clear that R is reflexive, symmetric and transitive
∴ R is an equivalence relation.
Let R be the relation defined on the set of natural numbers N as R = {(x, y) : x ∈ N, y ∈ N, 2 x + y = 41 }
Find the domain and range of this relation R. Also verify whether R is (i) reflexive (ii) symmetric (iii) transitive.
The following three relations are defined on the set of natural numbers :
R = {(x, y) : x < y, x ∈ N, y ∈ N}
S = { (x,y) : x + y = 10, x ∈ N, y ∈ N}
T = { (x, y) : x = y or  x – y = 1, x ∈ N, y ∈ N } Explain clearly which of the above relations are (i) Reflexive (ii) Symmetric (iii) Transitive.
Long Answer TypeShow 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}Â is an equivalence relation. Find the set of all elements related to 1 in each case.
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