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Let f: N -> N be defined by f(n) = (n + 1)/2 if n is odd and f(n) = n/2 if n is even, for all n in N. Then f is:

1. injective but not surjective
2. surjective but not injective
3. both injective as well as surjective
4. neither injective nor surjective

Correct answer: surjective but not injective

Solution

Check f(1) = (1+1)/2 = 1 and f(2) = 2/2 = 1, so two distinct inputs map to 1 => not injective. For surjectivity, any m in N is achieved: f(2m) = m (using the even branch). So every natural number is in the range => surjective. Hence f is surjective but not injective.

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