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Question
Symmetric but neither reflexive nor transitive
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Solution
(1,1)∉R, hence not reflexive (1,2)∈R
⇒(2,1)∈R, also (2,1)∈R
⇒(1,2)∈R.
(a,b)∈R=9(b,a)∈R, hence symmetric
(1,2)∈R(2,1)∈R but (1,1)∈R hence not transitive.
⇒(2,1)∈R, also (2,1)∈R
⇒(1,2)∈R.
(a,b)∈R=9(b,a)∈R, hence symmetric
(1,2)∈R(2,1)∈R but (1,1)∈R hence not transitive.
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