For each of the following, state whether it is possible to have a function meeting the criteria given, explain why orwhy not, and if it is possible, give two examples.(a) A function f : N≥0 → N≥0 which is not surjective.(b) A function f : N≥0 → Z which is injective.(c) A function f : Q → Q which is bijective.
Question
For each of the following, state whether it is possible to have a function meeting the criteria given, explain why orwhy not, and if it is possible, give two examples.(a) A function f : N≥0 → N≥0 which is not surjective.(b) A function f : N≥0 → Z which is injective.(c) A function f : Q → Q which is bijective.
Solution
(a) Yes, it is possible to have a function f : N≥0 → N≥0 which is not surjective. A surjective function is one where every element in the codomain is mapped to by at least one element in the domain. If there exists at least one element in the codomain that is not mapped to by any element in the domain, then the function is not surjective.
Examples:
- f(n) = n + 1. This function is not surjective because there is no natural number that maps to 0 in the codomain.
- f(n) = 2n. This function is not surjective because no natural number maps to any odd number in the codomain.
(b) Yes, it is possible to have a function f : N≥0 → Z which is injective. An injective function is one where every element in the domain maps to a unique element in the codomain. Since the set of integers (Z) is larger than the set of non-negative integers (N≥0), it is possible to have an injective function from N≥0 to Z.
Examples:
- f(n) = n. This function is injective because every non-negative integer maps to a unique integer.
- f(n) = -n. This function is also injective because every non-negative integer maps to a unique integer.
(c) Yes, it is possible to have a function f : Q → Q which is bijective. A bijective function is one that is both injective and surjective. This means that every element in the domain maps to a unique element in the codomain, and every element in the codomain is mapped to by at least one element in the domain. Since the set of rational numbers (Q) is infinite, it is possible to have a bijective function from Q to Q.
Examples:
- f(q) = q + 1. This function is bijective because every rational number maps to a unique rational number, and every rational number is mapped to by at least one rational number.
- f(q) = 2q. This function is also bijective because every rational number maps to a unique rational number, and every rational number is mapped to by at least one rational number.
Similar Questions
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