1 votes 1 votes Consider the set S = {a, b} and ‘L’ be a binary relation such that L = {all binary relations except reflexive relation set S}. The number of relation which are symmetric _______. Set Theory & Algebra relations discrete-mathematics + – sunaina rawat asked Oct 4, 2017 sunaina rawat 1.8k views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply Show 7 previous comments sunaina rawat commented Oct 4, 2017 reply Follow Share @manu00x empty relation, {(a,a)}, {(b,b)} are not reflexive relations how?? here (aRa) also (bRb) hence should be reflexive....correct me if I am wrong 0 votes 0 votes Manu Thakur commented Oct 4, 2017 reply Follow Share @Sunania empty relation is not reflexive because (a,a) and (b,b) are missing. {(a,a)} is not reflexive relation because (b,b) is missing {(b,b)} is not reflexive relation because (a,a) is missing in a reflexive relation, all the diagonal pairs must exist. 1 votes 1 votes sunaina rawat commented Oct 5, 2017 reply Follow Share "my bad" yes I got it....Thanks 0 votes 0 votes Please log in or register to add a comment.
Best answer 8 votes 8 votes Number of symmetric relations which are not reflexive { } {(a,a)} {(b,b)} {(a,b),(b,a)} {(a,a),(a,b),(b,a)} {(a,b),(b,a),(b,b)} Hence I think 6 should be the correct ans. shivangi5 answered Oct 5, 2017 selected Oct 5, 2017 by sunaina rawat shivangi5 comment Share Follow See 1 comment See all 1 1 comment reply nikkey123 commented Nov 17, 2017 reply Follow Share can u explain your answer please 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 votes Since it is small set,it can be done in hit and trial.But for more elements we should do it systematically. We need to find number of relations which are symmetric but not reflexive = [Total Symmetric] - [Symmetric and reflexive] =[ $2 ^ {n(n+1)/2 }$] - [1*$2 ^ {n(n-1)/2 }$ ] = 8-2 =6 rahul sharma 5 answered Nov 8, 2017 rahul sharma 5 comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments targate2018 commented Nov 18, 2017 reply Follow Share How do we compute the number of symmetric AND reflexive relations. I only have idea about the individual number of symmetric relations i.e, 2n(n+1)/2 and total number of reflexive relations i.e, 2n^2 - n 0 votes 0 votes rahul sharma 5 commented Nov 18, 2017 reply Follow Share If you consider n*n matrix. Now symmteirc and reflexive we need to find.Diagnal must be 1 for reflexive.Now remaining elements in will form n will form n(n−1)/2 pairs and for each pair we have two choices ,0 or 1, so 1*2^n(n−1)/2 1 votes 1 votes targate2018 commented Nov 20, 2017 reply Follow Share Got it. Thanks :). 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes L = {all binary relations except reflexive relation set S} => L has no xRx pair => So no diagonal element present. => Total size of L = 2^(n2 - n) In symmetric relations, xRy and yRx should both come. For each x and y (x!=y), either (x,y) or (y,x) can be present. Number of symmetric relations in L = (No diagonal element selected)*(One of the non-diagonal pairs) No of non-diagonal pairs = (n2 - n)/2 => Number of symmetric relations = 2^((n2 - n)/2) agoh answered Jan 4, 2017 agoh comment Share Follow See all 0 reply Please log in or register to add a comment.