Is this sufficient? Also, any good books/other suggestions regarding the subject will be very helpful.
Find min, max, inf, sup (if they exist):
$$B=\left\{\frac{m}{m+n}:m,n\in\mathbb{N}\right\}$$
Showing B has an upper bound:
Let $M=1$, we need to find $m,n$ fulfilling:$$\frac{m}{m+n}>1$$
As $n\in\mathbb{N}$ and is only in the denominator, the smaller it's value, the greater the value of n, the smaller $b$ will be. Therefore, let us choose $n=1$ (smallest possible value).$$\frac{m}{m+1}>1\,\,\,\,\,\leftrightarrow\,\,\,\,\,\,m>m+1$$
We got a contradiction, thus $M$ is an upper bound of $B$.
Showing $M=\sup B$: Let $\epsilon>0$, we need to find $b\in B$ fulfilling:$$\frac{m}{m+n}>1-\epsilon$$
Again, we'll choose $n=1$ to get the biggest $b$ possible:
$$\begin{align}
\frac{m}{m+1}&>1-\epsilon\\ m&>m+1-m\epsilon -\epsilon\\m&>\frac{1-\epsilon}{\epsilon}
\end{align}$$
Therefore for every $\epsilon$ we can choose $n=1,m>\frac{1-\epsilon}{\epsilon}$, which means $\sup B=1$.
Edit: Since $m,n \in\mathbb{N}$, $B>0$.
Showing $0=\inf B$: Let $\epsilon>0$, we need to find $b\in B$ fulfilling:
$$\frac{m}{m+n}<0+\epsilon$$
Choosing $m=1$ to make $b$ as small as possible:
$$1<\epsilon+n\epsilon\\n>\frac{1-\epsilon}{\epsilon}$$
We have shown that such $b$ exists for every $\epsilon$. Therefore, $\sup B = 0$
Best Answer
If you take $\,\Bbb N=\{1,2,3,...\}\,$, then I think you'll agree with
$$\forall\,\,m,n,\in\Bbb N\,\,\,,\,\,\frac{m}{m+n}>0\Longrightarrow 0\,\,\text{is a lower bound for}\,\,M\,...$$
I think it'd be a good idea to try to prove that zero is actually the infimum of $\,M\,$