1. The problem asks to find the supremum (least upper bound) and infimum (greatest lower bound) of the set $$S=\left\{\frac{n-m}{n+m}: n,m \in \mathbb{N}\right\}$$ where $n$ and $m$ are natural numbers. 2. Consider the expression $$\frac{n-m}{n+m}$$ with $n,m \geq 1$. 3. Since $n$ and $m$ are both positive, the denominator $n+m > 0$. 4. The value of the fraction depends on the difference between $n$ and $m$: - If $n \gg m$, then $$\frac{n-m}{n+m} \to 1$$ from below, because $n-m \approx n$ and $n+m \approx n$. - If $m \gg n$, then $$\frac{n-m}{n+m} \to -1$$ from above, because $n-m \approx -m$ and $n+m \approx m$. 5. However, the fraction can never be exactly $1$ or $-1$ because that would require $m=0$ or $n=0$, which are not in $\mathbb{N}$ if natural numbers start at 1. 6. Therefore, the supremum of $S$ is $$\sup S = 1$$ and the infimum of $S$ is $$\inf S = -1$$. 7. To summarize: - The set $S$ is bounded above by 1 (supremum). - The set $S$ is bounded below by -1 (infimum). Final answer: $$\sup S = 1, \quad \inf S = -1$$