1 votes 1 votes How many relations are there which are reflexive antisymmetric and symmetric? Set Theory & Algebra discrete-mathematics + – kamakshi asked Dec 26, 2017 kamakshi 607 views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply srestha commented Dec 26, 2017 reply Follow Share symmetric and antisymmetric cannot satisfy same time So ans 0 0 votes 0 votes Ashwin Kulkarni commented Dec 26, 2017 reply Follow Share @srestha ma’am symmetric and anti symmetric can be present at a time. because both are having diagonal elements in common. 0 votes 0 votes Ashwin Kulkarni commented Dec 26, 2017 reply Follow Share Total number of relations is 1 and size is 'n' Because that set is having {(1,1),(2,2),.....(n,n)} elements. 1 votes 1 votes srestha commented Dec 26, 2017 reply Follow Share then 2n right? 0 votes 0 votes Ashwin Kulkarni commented Dec 26, 2017 reply Follow Share We don't have a choice to eliminate anything from that set. All should be present. Hence 2^n not possible. 0 votes 0 votes abhishek tiwary commented Dec 26, 2017 reply Follow Share ref symmetric and antisymmetric =1?? 1 votes 1 votes Hradesh patel commented Dec 26, 2017 reply Follow Share reflexive relation = 2^(n^2 -n ) symmetric = 2^n * 2^(n(n-1)/2) asymmetric = 2^n * 3^(n(n-1)/2) it clearly that in reflexive relation eliminate diagonal element so 2 ^n is not possible 0 votes 0 votes Hradesh patel commented Dec 26, 2017 reply Follow Share answer is 1 0 votes 0 votes srestha commented Dec 26, 2017 reply Follow Share @Ashwin reflexive relation = 2(n^2 -n ) symmetric = 2n * 2(n(n-1)/2) asymmetric = 2n * 3(n(n-1)/2) then symmetric $\cap$ antisymmetric= 2n Now 2(n^2 -n ) >2n then reflexive $\cap$ 2n=2n right? 0 votes 0 votes Suhaid commented Dec 28, 2017 reply Follow Share This is a correct explanation . 0 votes 0 votes Please log in or register to add a comment.