InterviewSolution
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InterviewSolution
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Show that `f: N to N ` defined by ` f(n) = {[(n)/(2) if n is even ], [(n+1)/(2) if n is odd ]}` is a many -one onto function |
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Answer» We have `f(1) =((1 +1) )/(2) = (2)/(2) =1 " and " f(2) =(2)/(2) =1` Thus `f(1) =f(2) " while " 1 ne 2` `:.` f is many -one In order to show that f is onto, consider an arbitrary element `n in N` If n is odd then `(2n-1)` is odd and `f( 2n-1) =((2n-1+1))/(2) =(2n)/(2) =n` If n is even then 2n is even and ` f(2n) =(2n)/(2) =n` Thus for each `n in N` (whether even or odd ) there exists its pre-image in N. `:. ` f is onto. Hence f is many -one onto. |
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