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Prove that the relation R on the set `NxN`defined by`(a , b)R(c , d) a+d=b+c`for all `(a , b),(c , d) in NxN`is an equivalence relation. |
Answer» Here R is a relation on `NxxN,` defined by `(a,b)R(c,d)iffa+d=b+c" for all "(a,b),(c,d)inNxxN.` We shall show that R satisfies the following properties. (i) Reflexivity: We know that `a+b=b+a" for all "a,binN.` `:." "(a,b)R(a,b)" for all "(a,bin(NxxN).` So, R is reflexive. (ii) Symmetry: Let `(a,b)R(c,d).` Then, `(a,b)R(c,d)impliesa+d=b+c` `implies" "c+b=d+a` `implies" "(c,d)R(a,b).` `:." "(a,b)R(c,d)implies(c,d)R(a,b)" for all "(a,b),(c,d)inNxN.` This shows that R is symmetric. (iii) Transitivity: Let `(a,b)R(c,d)" and "(c,d)R(e,f)`. Then, `" "(a,b)R(c,d)" and "(c,d)R(e,f)` `implies" "a+d=b+c" and "c+f=d+e` `implies" "a+d+c+f=b+c+d+e` `implies" "a+f=b+e` `implies" "(a,b)R(e,f).` Thus, (a,b) R (c,d) and (c,d) R (e,f)`implies" "`(a,b) R (e,f). This shows that R is transitive. `:." "R is reflexive, symmetric and transitive. Hence, R is an equivalence relation on `NxxN.` |
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