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Let f: N->N be defined by: f(n)={n+1,\ i...

Let `f: N->N` be defined by: `f(n)={n+1,\ if\ n\ i s\ od d n-1,\ if\ n\ i s\ e v e n` Show that `f` is a bijection.

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The correct Answer is:
`f=f^(-1)`
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