Question

Let f:N→P(N) be defined by f(n)={d is an element N, d divides

n}. Prove or disprove: f is injective Prove or disprove: f is
surjective

Answer 1

The function f(n)={d is an element N, d divides n} is injective but not surjective.

Step-by-step explanation:

Definition: Let's first understand the given function, f(n)={d is an element N, d divides n}. This means that for a given natural number n, f(n) is a set of all the divisors of n.

Injective (one-to-one): To prove if the function f is injective, we need to show that different inputs result in different outputs. Let's assume f(a) = f(b). This means that the set of divisors of a is the same as the set of divisors of b. For this to happen, a and b must be the same number.

Surjective (onto): To prove if the function f is surjective, we need to show that every element in the codomain has a corresponding element in the domain. In this case, the codomain is the set of all power sets of natural numbers (P(N)). Since for any natural number n, f(n) will always be a subset of N, we can say that f is surjective.