Proof Define set $A = \{m\sqrt{k+1} \mid m \in \mathbb{N}^*, k \in P\}$, where $\mathbb{N}^*$ denotes the set of all positive integers. It is easy to check that for any $k_1, k_2 \in P, k_1 \neq k_2$, $\frac{\sqrt{k_1+1}}{\sqrt{k_2+1}}$ is an irrational number. Therefore, for any $k_1, k_2 \in P$ and positive integers $m_1, m_2$, $m_1\sqrt{k_1+1} = m_2\sqrt{k_2+1}$ implies $m_1 = m_2$ and $k_1 = k_2$.

Note that $A$ is an infinite set. We arrange the elements in $A$ in ascending order. Then we have an infinite sequence. For any positive integer $n$, suppose the $n$th term of the sequence is $m\sqrt{k+1}$. Any term before the $n$th can be written as $m_i\sqrt{i+1}$, and
$$m_i \sqrt{i+1} \le m \sqrt{k+1}.$$
Or equivalently, $m_i \le m \frac{\sqrt{k+1}}{\sqrt{i+1}}$. It is easy to see that there are $\lfloor m \frac{\sqrt{k+1}}{\sqrt{i+1}} \rfloor$ such $m_i$ for $i = 1, 2, 3, 4, 5$. Therefore,
$$n = \sum_{i=1}^{5} \lfloor m \frac{\sqrt{k+1}}{\sqrt{i+1}} \rfloor = f(m, k).$$