91Ó°ÊÓ

: To check if the function \(d\) is an injection, we must prove either that for every \(x, y \in \mathbb{N}\), \(d(x) = d(y)\) implies \(x = y\), or we need to find a counterexample showing that there exist \(x, y \in \mathbb{N}\) such that \(d(x) = d(y)\) but \(x \neq y\). To find a counterexample, we can check the values of function \(d\) for some small numbers, like \(d(2) =2\) since \(2\) has two divisors: \(1\) and \(2\). And when we look at \(d(3)\), it also equals \(2\), because \(3\) has two divisors: \(1\) and \(3\). So, we get \(d(2) = d(3)\), but \(2 \neq 3\). Hence, the function \(d\) is not an injection.

: A function is surjective if every element in the codomain (here, \(\mathbb{N}\)) has at least one corresponding element in the domain (also \(\mathbb{N}\)). In other words, we need to show that for each natural number \(k\), we can find an \(n \in \mathbb{N}\) such that \(d(n) = k\). To prove this, let's consider the prime factorization of any natural number \(n\). Prime factorization is unique for each \(n\), and we can define it as: \(n = p_1^{e_1} \cdot p_2^{e_2} \cdot ... \cdot p_k^{e_k}\), where \(p_i\) are prime numbers and \(e_i\) are the corresponding exponents. Now, the total number of divisors for \(n\) can be represented as: \(d(n) = (e_1 + 1)(e_2 + 1)...(e_k + 1)\), as we can choose any combination of the exponents of the prime factors. We know that the multiplication of two or more natural numbers gives another natural number. Therefore, no matter which exponents we choose for the prime factors, the result of the multiplication, i.e., the total number of divisors, will be a natural number. So, for any natural number \(k\), we can find an \(n \in \mathbb{N}\) such that \(d(n) = k\). This proves that the function \(d\) is a surjection. In conclusion, the function \(d: \mathbb{N} \rightarrow \mathbb{N}\) is not an injection, but it is a surjection.