InterviewSolution
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InterviewSolution
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Show thatthe functionf : N to Ndefinedby f(x) = {[x-1 \ if \x \ is \ even], [x +1 \ if \ x is \ odd]} is one-oneand onto. |
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Answer» SOLUTION :Suppose `f(x_(1))= f(x_(2))` Case 1when`x_(1) ` isodd and `x_(2)`is even In THISCASE `f(x_(1)) =f(x_(2)) rArr x_(1) +1 =x_(2)-1` `rArr x_(2) - x_(1)=2` Thisis acodtradicationsince thatdifferencebetween anodd interger and anevenintegercan neverbe 2. Thusin this case`f(x_(1))NE f(x_(2))` Similarlywhen `x_(1)`is evenand `x_(2)`is oddthen `f(x_(1)) ne f(x_(2))` Case 2when `x_(1)" and " x_(2)`are bothodd In thiscase`f(x_(1))=f(x_(2)) rArr x_(1)+ 1 =x_(2) -1` `rArr x_(1)=x_(2)` `:.` f isone-one Case 3When `x_(1) " and " x_(2)`are both even In thiscase `f(x_(1))=f(x_(2))rArr x_(1) -1 =x_(2) -1` `rArr x_(1)= x_(2)` `:.`f is one-one In orderto showthat FIS ontolet `y in N`(the codomain) Case 1 wheny is odd In thiscase `(y +1)` is even `:. f( y+1) =(y+1)-1 =y` Case 2wheny is even In thiscase `(y-1) ` is odd `:. f (y-1)=y-1 +1=y` Thuseach `y in N`(codomain of f)has itspre-imagein dom (f) `:. ` f is onto. Hence f is one-oneonto. |
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