CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

CBSE CLASS XII

1.1. Types of Relation.

Q.1. Show that the relation R defined by (a, b) R (c, d) => a + d = b + c on the set N×N is an equivalence relation.

Solution :

Reflexive :
As, a + b = b + a => (a, b) R (a, b).
Hence, R is reflexive.
Symmetric :
(a, b) R (c, d) => a + d = b + c
=> b + c = a + d
=> c + b = d + a
=> (c, d) R (a, b).
Hence, R is symmetric.
Transitive :
(a, b) R (c, d) and (c, d) R (e, f) => a + d = b + c and c + f = d + e
=> a + d + c + f = b + c + d + e
=> a + f = b + e => (a, b) R (e, f).
Hence, R is transitive.
Therefore, relation R is an equivalence relation. [Proved.]

Q.2. Prove that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation.

Solution :

Reflexive :
Let a ε R,
Therefore, |a – a| = 0 is an even number => (a, a) ε R,
Hence, R is reflexive.
Symmetric :
Let a, b ε R
Let (a, b) ε R => |a – b| is even number
=> | – (b – a)| is even
=> |b – a| is even
=> (b, a) ε R,
Hence, R is symmetric.
Transitive :
Let a, b, c ε R
Let (a, b) ε R and (b, c) ε R => |a – b| is even and |b – c| is even
=> a – b is even and b – c is even
=> (a – b) + (b – c) is even
=> |a – c| is even
=> (a, c) ε R.
Hence, R is Transitive.
Therefore, R ia an equivalence relation as it is reflexive, symmetric and transitive.
[Proved.]

Paper By Mr. M.P.Keshari
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