Question

Q1:Let R={(x y):x+y< 0 and x y belong to set A } be a relation defined on A where A={1 2 3 4} then R is

Answer 1

Step-by-step explanation:

R={(1,4),(4,1),(2,3)(3,2)

so R is not reflexive as (1,1)∈ / R

R is symmetric by defination and R is not transitive

as (1,4)∈R, (4,1)∈R but (1,1)∈

Answer 2

Now Vq € Q1, (f(6f(q, ex,)), f(6; (q., ex. ))) ... Let the result be true for all y e X such that |y| = n – 1, n > 0 and a = ya, where a ... Let (Q, R ...