Question

Let R be the relation on the set of all real
numbers defined by aRb iffla - bls 1.
Then Ris​

Answer 1

Explanation:

Given relation is aRb is 1+ab>0,

considering both a and b are real numbers,

We know that ab=ba,

⟹aRb=1+ab>0=1+ba>0=bRa,

∴ R is a symmetric relation,

aRa=1+a

2

as $${ a }^{ 2 }$$ is always a positive real number,

$$\therefore 1+{ a }^{ 2 }>0$$,

∴ R is a reflexive relation.

Consider aRb which is 1+ab>0,

and also bRc which is equal to 1+bc>0,

if a=0.5 and b=−0.5 and c=−4,

⟹ both aRb and bRc are satisfied,

aRc=1−2<0

∴ aRc is not a realation,

Hence R is not a equivalence relation,but is a reflexive and symmetric relation.