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Consider the mapping f from the natural numbers to the integers given by f(n) = (n - 1)/2 for odd n, and f(n) = -n/2 for even n. Which of the following best describes f?

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

Correct answer: both injective and surjective

Solution

Odd n give f=(n-1)/2 = 0,1,2,... (all non-negative integers); even n give f=-n/2 = -1,-2,... (all negative integers). Together they hit every integer with no repeats, so f is a bijection (both injective and surjective).

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