## Relations

 Question 1

 A R1R2 = {(1, 2), (1, 4), (3, 3), (5, 4), (7, 3)} B R1R2 = {(1, 2), (1, 3), (3, 2), (5, 2), (7, 3)} C R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)} D R1R2 = {(3, 2), (3, 4), (5, 1), (5, 3), (7, 1)}
Engineering Mathematics       Relations       Gate 2004-IT
Question 1 Explanation:
From the given information,
R1 ={(1,2), (1,8), (3,6), (5,4), (7,2), (7,8)}
where x+y is divisible by 3
R2 = {(2,2), (4,4), (6,2), (6,4), (8,2)}
where x+y is not divisible by 3
Then the composition of R1 with R2 denotes R1R2, is the relation from A to C defined by property such as:
(x,z) ∈ R1R2, iff if there is a y ∈ B such that (x,y) ∈ R1 and (y,z) ∈ R2.
Thus, R1R2 = {(1,2), (3,2), (3,4), (5,4), (7,2)}
 Question 2
Let R be a symmetric and transitive relation on a set A. Then
 A R is reflexive and hence an equivalence relation B R is reflexive and hence a partial order C R is reflexive and hence not an equivalence relation D None of the above
Engineering Mathematics       RelATIONS       Gate-1995
Question 2 Explanation:
If a relation is equivalence then it must be
i) Symmetric
ii) Reflexive
iii) Transitive
If a relation is said to be symmetric and transitive then we can't say the relation is reflexive and equivalence.
 Question 3
Amongst the properties {reflexivity, symmetry, anti-symmetry, transitivity} the relation R = {(x,y) ∈ N2 | x ≠ y } satisfies __________
 A symmetry
Engineering Mathematics       Relations       Gate-1994
Question 3 Explanation:
It is not reflexive as xRx is not possible.
It is symmetric as if xRy then yRx.
It is not antisymmetric as xRy and yRx are possible and we can have x≠y.
It is not transitive as if xRy and yRz then xRz need not be true. This is violated when x=x.