1. **Problem statement:** Given that $a > b$ and both $a$ and $b$ are positive numbers, prove that for any natural number $n$, the inequality $\frac{n}{a} < \frac{n}{b}$ holds. 2. **Recall the rule:** For positive numbers, if $a > b > 0$, then $\frac{1}{a} < \frac{1}{b}$. This is because the reciprocal function is strictly decreasing on positive numbers. 3. **Apply the rule:** Since $a > b > 0$, we have $$\frac{1}{a} < \frac{1}{b}.$$ 4. **Multiply both sides by $n \in \mathbb{N}$:** Since $n$ is positive, multiplying preserves the inequality: $$n \times \frac{1}{a} < n \times \frac{1}{b}$$ which simplifies to $$\frac{n}{a} < \frac{n}{b}.$$ 5. **Conclusion:** Therefore, for any natural number $n$, if $a > b > 0$, then $$\frac{n}{a} < \frac{n}{b}.$$