Hows does this inequality work? This is a part of the proof that there is a set in $\mathbb{Q}$ that has an upper bound in $\mathbb{Q}$ but no least upper bound in $\mathbb{Q}$.

Consider the set
  $$
S=\{q\in\mathbb{Q}:q^2<2\}
$$
  Suppose that $k=\frac{a}{b}$ is an upper bound in $\mathbb{Q}$ for $S$. Suppose also that $k^2<2$. Define
  $$
\delta=2-k^2>0
$$
  Fix $N\in\mathbb{N}$ such that
  $$
N\geq \max\left(2a+1,\frac{3a}{b^2\delta}\right)
$$
  Since $N\geq 2a+1$, then $N^2\geq N(2a+1)\geq 2Na+1$ and since
  $N>\frac{3a}{b^2\delta}$, then $N^2b^2\delta>3Na\geq 2Na+1$.

My questions


*

*Since $k$ is an upper bound in $\mathbb{Q}$ for $S$, then either both $a,b$ are positive or both $a,b$ are negative; right? So can we without loss of generality consider both $a,b$ to be positive.

*The proof states $N\geq 2a+1$ and $N>\frac{3a}{b^2\delta}$. Does it assume that $2a+1>\frac{3a}{b^2\delta}$?

*$N\geq 2a+1\Rightarrow N^2\geq N(2a+1)=2Na+N\geq 2Na+1$ since $N\geq1$?

*$N>\frac{3a}{b^2\delta}\Rightarrow Nb^2\delta>3a\Rightarrow
N^2b^2\delta>3Na=2Na+Na$. On the other hand, $a>1\Rightarrow Na>N\geq1$ so $Na>1$. Combining the two, $N^2b^2\delta>3Na=2Na+Na>2Na+1$ but this is different from what is in the proof (since
$N>\frac{3a}{b^2\delta}$, then $N^2b^2\delta>3Na\geq 2Na+1$). What am I doing wrong?

 A: 

*No it does not assume that. And for everything else, you are not doing anything wrong. The proof in the text is essentially same as yours, but it skipped few details which you provided. 

A: *

*Yes.

*No. That's why they take the maximum of these two numbers — because we don't know which one is greater. (And by taking their maximum, we don't care which one is greater.)

*Yes, for the last step there.

*First of all, we only know that $a\ge1$, not that $a>1$. The latter inequality may not even be true because it's possible for $a$ to be $a=1$. Secondly, even if it can be shown that $Na>1$, then the statement $Na\ge1$ is true as well. So the author of the proof doesn't bother with the most accurate bounds: we know that $a\ge1$ and $N\ge1$, therefore $Na\ge1$, and that seems to be good enough for their purposes. Even if stronger and better inequalities are true, they just don't bother about that. After all, this is only a step towards a more important goal (proof of the overall theorem) — so if it works for the goal, it's good enough.

