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Let `R={(a,b):a,binN" and "a=b^(2)},` Show that R satisfies none of reflexivity, symmetry and transitivity. |
Answer» (i) R is not reflexive, since `2ne2^(2)` and therefore (2,2)`!inR.` (ii) Since `4=2^(2),so(4,2)inR.` But, `2ne4^(2)`, So, `(2,4)!inR.` Thus, `(4,2)inR" but "(2,4)!inR.` `:." "` R is not symmetric. (iii) Since `16=4^(2)`, so (16,4)inR.` Also, `4=2^(2)`, so `(4,2)inR.` But, `16ne2^(2)implies(16,2)!inR.` Thus, `(16,4)inR" and "(4,2)inR." But "(16,2)!inR.` `:." "` R is not transitive. Hence, R satsfies none of reflexivity, symmetry and transitivity. |
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