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A function `f` from the set of natural numbers to the set of integers defined by `f(n)={(n-1)/2,\ w h e n\ n\ i s\ od d-n/2,\ w h e n\ n\ i s\ e v e n` (a) neither one-one nor onto (b) one-one but not onto (c) onto but not one-one (d) one-one and onto both

Text Solution

Verified by Experts

Clearly f(6) = 3 and f(5) = 3
Hence `5 != 6` but f(5) = f(6)
Therefore, f is not one-one
Clearly, for every `m in N`, there exists `2m-1 in N`
such that `f(2m-1) = m`
Hence f is onto.
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