IIT JEE , AIEEE Forum - Mathematics , Algebra

flowers_rsss

Let S be the set of integers .For a,bS , relation R is defined by aRb iff |a-b|<1 then R is

1)only reflexive

2)only symmetric

3)only transitive

4)equivalence

pls explain!!

joyfrancis

only reflexive!! since for a set of integers |a-b|<1 iff a=b

sachit

hey ppl at da same tym can u explain me wat is reflexive and wat is transitive and wat is symetric?plz.

flowers_rsss

yup i posted this qn to know that!!

pls any one do explain the properties!

allamraju

I think the ans is reflexive ,bcoz

1)For it to be reflexive,(a,a) shud be a member for all a $\epsilon$ S and Ia-aI=0<1.Hence reflexive.

2)For it to be symmetric,if (a,b) is a member then (b,a) shud be a member.Here,If (a,b) is a member,then Ia-bI<1 $\Rightarrow$ Ib-aI<1 $\Rightarrow$ (b,a) is a member,but for a set of integers this can't be possible.

flowers_rsss

the ans is given as only reflexive!1

joyfrancis

See this, i think it would be helpful to you.

http://www.goiit.com/posts/list/algebra-relation-55101.htm#275705

allamraju

Though the condition for symmetric relation is satisfied,there exists no (a,b) such that a,b $\epsilon$ S and Ia-bI<1.Anyway,the condition for a relation to be transitive is

Ask & Discuss Questions with Community & Experts

I think the ans is reflexive ,bcoz

1)For it to be reflexive,(a,a) shud be a member for all a $\epsilon$ S and Ia-aI=0<1.Hence reflexive.

2)For it to be symmetric,if (a,b) is a member then (b,a) shud be a member.Here,If (a,b) is a member,then Ia-bI<1 $\Rightarrow$ Ib-aI<1 $\Rightarrow$ (b,a) is a member,but for a set of integers this can't be possible.

Though the condition for symmetric relation is satisfied,there exists no (a,b) such that a,b $\epsilon$ S and Ia-bI<1.Anyway,the condition for a relation to be transitive is

If (a,b) and (b,c) are members of the relation then (a,c) shud also be a member.

Ask & Discuss Questions with Community & Experts

I think the ans is reflexive ,bcoz

1)For it to be reflexive,(a,a) shud be a member for all aS and Ia-aI=0<1.Hence reflexive.

2)For it to be symmetric,if (a,b) is a member then (b,a) shud be a member.Here,If (a,b) is a member,then Ia-bI<1Ib-aI<1(b,a) is a member,but for a set of integers this can't be possible.

Though the condition for symmetric relation is satisfied,there exists no (a,b) such that a,bS and Ia-bI<1.Anyway,the condition for a relation to be transitive is

If (a,b) and (b,c) are members of the relation then (a,c) shud also be a member.

1)For it to be reflexive,(a,a) shud be a member for all a $\epsilon$ S and Ia-aI=0<1.Hence reflexive.

2)For it to be symmetric,if (a,b) is a member then (b,a) shud be a member.Here,If (a,b) is a member,then Ia-bI<1 $\Rightarrow$ Ib-aI<1 $\Rightarrow$ (b,a) is a member,but for a set of integers this can't be possible.

Though the condition for symmetric relation is satisfied,there exists no (a,b) such that a,b $\epsilon$ S and Ia-bI<1.Anyway,the condition for a relation to be transitive is