R = {(T1, T2) : T1 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.
A is the set of all books in a library of a college.
R = {(x,y) : x and y have same number of pages}
Since (x, x) ∈ R as x and x have the same number of pages ∀ x ∈ A.
∴ R is reflexive.
Also (x, y) ∈ R
⇒ x and y have the same number of pages ⇒ y and x have the same number of pages
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, (x, y) ∈ R and (y, z) ∈ R.
⇒ x and y have the same number of pages and y and z have the same number of pages
⇒ x and z have he same number of pages ⇒ (x, z) ∈ R ∶ R is transitive.
Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}.
Choose the correct answer.
(A) (2. 4) ∈ R (B) (3, 8) ∈ R (C) (6,8) ∈ R (D)(8,7) ∈ R
R = {(a, b) : a = b – 2, b > 6}
∴ (a, b) ∈ R ⇒ a = b – 2 where b > 6 ∴ (6, 8) ∈ R as 6 = 8 – 2 where b = 8 > 6 ∴ (C) is correct answer.
R = {(T1, T2) : T1 is similar to T2}
Since every triangle is similar to itself
∴ R is reflexive.
Also (T1 T2) ∈ R ⇒ T1 is similar to T2 ⇒ T2 is similar to T1 ∴ (T2,T1) ⇒ R
∴ (T1,T2) ∈ R ⇒ (T2,T1) ∈ R ⇒ R is symmetric.
Again (T1, T2), (T2, T3) ∈ R
⇒ T1 is similar to T2 and T2 is similar to T3
∴ T1 is similar to T3 ⇒ (T1,T3) ∈ R ∴ (T1, T2), (T2,T3) ∈ R ⇒ (T1, T3) ∈ R ∴ R is transitive.
∴ R is reflexive, symmetric and transitive ∴ R is an equivalence relation.
Now T1, T2, T3 are triangles with sides 3, 4, 5 ; 5, 12, 13 and 6, 8, 10.
Since
∴ T1 is similar to T3 i.e. T3 is similar to T1.
No two other triangles are similar.
Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(A) 1 (B) 2 (C) 3 (D) 4
A = {1, 2, 3}
R1 = {(1,2), (1,3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)} is the only relation on {1, 2, 3} which is reflexive, symmetric but not transitive and is such that (1, 2), (1, 3) ∈ R1.
∴ (A) is correct answer.