InterviewSolution
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InterviewSolution
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Let S be the set of all real numbers and let R be relation in S, defied by `R={(a,b):aleb^(3)}.` Show that R satisfies none reflexivity , symmetry and transitivity . |
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Answer» (i) Nonreflexivity Clearly ,`(1)/(2)` is a real number and `(1)/(2)le ((1)/(2))^(3)` is not true . `therefore ((1)/(2),(1)/(2))!in R.` hence ,R is not reflexive . (ii) Nonsymmetry take the real numbers `(1)/(2)and 1.` Clearly ,`(1)/(2)le 1^(3)` is true and therefore ,`((1)/(2),1) in R` But `1 le ((1)/(2))^(3)` is not true and so `(1,(1)/(2))!in R.` hemce ,R is not symmetric . (iii) Nontransitivity COnsider the real numbers `3,(3)/(2) and (4)/(3).` Clearly ,`3le((3)/(2))^(3)and (3)/(2)le((4)/(3))^(3)"but"3le((4)/(3))^(3)` is not true thus ,`(3,(3)/(2)) in R and ((3)/(2),(4)/(3))!in R.` Hence R is not transitive. thus ,R Satisfies none of reflexivity , symmetry and transitivity . |
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