1.

Let `A={x in Z : 0le le 12}.` show that `R={(a,b):|a-b|` is a multiple of `4` is (i) reflexive, (ii) symmetric and (iii) transitive. Find the set of elements related to `1`.

Answer» Clearly ,`A={0,1,2,3,4. . .,10,11,12}.`
here ,R satisfies the following properties .
(i) Reflexivity
Let a be an arbitrary element of A , then ,
a-a =0 , which is a multiple of 4 . `therefore ` a R a for all ` a in A.`
(ii) symmetry
Let a R b , then
`a R bimplies |a-b|` is a maultiple of 4
`implies |-(a-b)` is a multiple of 4 .
`implies |b-a| ` is multiple of 4
`implies bR a.`
`therefore ` R is symmetric .
(iii) tranistivity
Let a R b and b Rc . then ,
a R b, b R c
`implies |a-b|` is a multiple of 4 and |b-c| is multiple of 4 .
`Let |a-b|=4k_(1)and |b-c|=4k_(2).` the n ,
`|a-c|=|(a-b)-(b-c)|=|4k_(1)-4k_(2)|`
`=|4(k_(1)-k_(2))|=4|K_(1) -K_(2)|` which is a multiple of 4
`therefore a R b,b R c implies aRc . so ` R is transitive .
thus ,R is reflexive , symmetric and transitive .
`Now ,[1] ={x in A :x R 1}.`
`={ x in A : |x-1|` is a multiple of 4}
`={1,5,9}.`
hence , the required set is `{1,5,9}.`


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