Consider the mapping f:n→N, where is the set of natural
Consider the mapping f:n→N, where N is the set of natural numbers is defined as,
fn=n2, for n=odd
fn=2n+1, for n=even
For n∈N, which of the following is true about f?
- A.
Bijective
- B.
Surjective but not Injective
- C.
Injective but not surjective
- D.
Neither surjective nor injective
Attempted by 180 students.
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Correct answer: D
Answer: The function is neither injective nor surjective.
Definition: f(n) = n^2 for odd n, and f(n) = 2n+1 for even n.
Not injective: distinct inputs can give the same output. For example, f(3)=3^2=9 and f(4)=2·4+1=9.
Not surjective: every output is congruent to 1 modulo 4. If n is even (n=2k), then f(n)=2n+1=4k+1≡1 (mod 4). If n is odd (n=2k+1), then f(n)=(2k+1)^2=4k(k+1)+1≡1 (mod 4). Thus the image is a subset of numbers congruent to 1 mod 4, and numbers such as 2, 3, and 4 have no preimage.
Conclusion: Since the function fails to be one-to-one and fails to be onto, it is neither injective nor surjective.