The question is as follows:
Suppose $\beta \in \mathbb{R}$ is irrational and the set $S$ is defined by $S \equiv \left \{ m+n\beta : m,n \in \mathbb{Z} \right \}.$
$i)$ Show that $ra+sb \in S$ for all $a,b \in S$ and all $r,z \in \mathbb{Z}.$
$ii)$ Show that there exists a strictly monotone increasing sequence of natural numbers $\left \{ n_{k} \right \}$ such that the sequence $\left \{ n_{k}\beta-\left \lfloor n_{k}\beta \right \rfloor \right \}$ converges.
$iii)$ Show that $S$ is dense in $\mathbb{R}$.
For part $i$, I know that since $a,b \in S$, then $ra+sb$ can be written as $$r(m+n\beta) + s(m'+n'\beta')$$
$$= rm + sm' + rn\beta + sn'\beta'.$$
Then I can say that $x = rm+sm' \in \mathbb{Z}$, but I need to write this in the form $x+y\beta$ where $x,y \in \mathbb{Z}$ and $\beta \in \mathbb{R}$ is irrational. However, I cannot seem to simply this any further to show this expression is indeed in $S$.
For part $ii$, I am pretty sure that I just have to show that the sequence $\left \{ n_{k}\beta-\left \lfloor n_{k}\beta \right \rfloor \right \}$ is strictly monotone increasing and that is bounded. By the Monotone Convergence Theorem, the sequence will converge.
For part $iii$, I know that $$\left \lfloor \frac{k}{\tau } \right \rfloor \leq \frac{k}{\tau } \leq \left \lfloor \frac{k}{\tau } \right \rfloor + 1,$$ but I am not sure how to continue.
I would much appreciate if someone could give me a starting point for each of the problems or help me solve them. Studying even introductory real analysis is very different and tricky for me.
Thanks for reading this long question!
For your (i), why do you have a $\beta'$ and what is $\alpha$? If $a,b \in S$, there exist $m_a, n_a, m_b, n_b \in \mathbb{Z}$ such that $a = m_a + n_a \beta$ and $b = m_b + n_b \beta$. To show $ra+sb \in S$, you need to find $x,y \in \mathbb{Z}$ such that $ra+sb = x + y \beta$.
(Edited to follow edits to the Question) For your (ii): Generally, you need to say what $n_k$ is. With your edit, $\left( n \sqrt{2} - \lfloor n \sqrt{2}\rfloor\right)_{n \in \mathbb{N}}$ is bounded (by $0$ and $1$), but is not monotonic.
However, think about $\frac{m}{n}$ being close to $\beta$. Then $m$ is close to $n\beta$. For each $n$, it is automatic that you can pick $m$ such that $\left|\frac{m}{n} - \beta \right| \leq \frac{1}{2n}$. (The fractions with denominator $n$ partition the line into segments of width $\frac{1}{n}$ and $\beta$ is in one of those segments. It can't be more than halfway away from the closest endpoint.) Rather than think of incrementing $n$, perhaps think of choosing an $n$ that exactly subdivides the interval containing $\beta$ more finely.
For (iii): Let $\varepsilon > 0$ and $r \in \mathbb{R}$. Now pretend to solve for $\beta$: $m + n\beta = r$, so $\beta = \frac{r-m}{n}$. Can you find an $n$ large enough (and a matching $m$) so that the gap between $\beta$ and $\frac{r-m}{n}$ is less than $\varepsilon$?