1 1 vote Does the given solution is correct: When S is symmetric and transitive, if S contain (3,1),(1,3) then (3,3) should also be present form transitivity. Please verify the solution. Set Theory & Algebra engineering-mathematics ace-test-series set-theory + – Overflow04 1.2k views answer comment Share Follow Print See all 5 Comments 5 5 Comments reply Show 2 previous comments [ Jiren ] commented Aug 23, 2022 reply Follow flag @GateOverflow04 Bro the Relation S is wrong because in the question they said S is symmetric and transitive but S doesnt have (3,3) pair as ( 3 S 1 ),(1 S 3 ) so according to transtive property ( 3 S 3 ) should be present 2 2 replyShare ankitgupta.1729 commented Aug 23, 2022 reply Follow flag would (1,1) not be in S ? 3 3 replyShare [ Jiren ] commented Aug 23, 2022 reply Follow flag @ankitgupta.1729 hi i saw some one calling you "sir” in one of the answers so im also sticking to this convention 😅 yes sir (1,1) should also be present in S because if we see the other way (1,3) (3,1) so from transitivity (1,1) should be present in S 3 3 replyShare Please log in or register to add a comment.
4 4 votes The best way to solve this question is by taking counter example EDIT : I forgot to write (a,a) in set S it's from transitive property (a,c),(c,a) so (a,a) should be present in S [ Jiren ] answered Aug 23, 2022 • edited Aug 23, 2022 by [ Jiren ] [ Jiren ] comment Share Follow See 1 comment 1 1 comment reply Overflow04 commented Aug 23, 2022 reply Follow flag thanks you !! 0 0 replyShare Please log in or register to add a comment.