I am trouble getting this proof and idea behind it. I have taken example of a and b as (3,4). I have problem with understanding the part where author writes " first step is to choose n large enough so that increments of sizes $\frac {1}{n}$ are too close to step over (3,4). Now this will happen if increments are smaller than size of interval, which he writes in next line using archimedian property.
After that he says now choose m to be smallest natural number greater than na. I don't get this.. Hope someone explains more clearly
Let $a<b$ (you may ignore the condition that $a\geq 0$). Then $\frac{1}{b-a}>0$ and by the Archimedian Property, there is $n\in\mathbb{N}$ such that $n>\frac{1}{b-a}$, which implies that $nb-na>1$. It remains to show that there is $m\in\mathbb{Z}$ such that $na<m<nb$, that is the interval $(na,nb)$ whose size is greater than 1 contains at least an integer $m$. Now take the largest integer $M\in \mathbb{Z}$ such that $M\leq na$ (see Every non-empty subset of the integers which is bounded above has a largest element.) then $$na<M+1\leq na+1<nb$$ and we can take $m=M+1$.
In other words, we "expand" the real line by a positive integer factor $n$, so that the interval $(a,b)$ becomes $(na,nb)$ with $nb-na>1$. Then some integer $m$ will get trapped in $(na,nb)$.