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R
Raj Pandey
Contributor-Level 9
gogog (3n+1)=gog (3n+2)=g (3n+3)=3n+1. So gogog=I.
If fog=f, then f must map range of g to values consistent with f.
There exists a one-one function f: N→N such that fog=f. e.g. f (x)=x.
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