Explanation of Shakarchi's proof of 1.3.4 in Lang's Undergraduate Analysis I'm currently working through Lang's Undergraduate Analysis, and trying to understand Rami Shakarchi's proof of the following:
Let $a$ be a positive integer such that $\sqrt a$ is irrational. Let $\alpha = \sqrt a$. Show that there exists a number $c > 0$ such that for all integers $p, q$, with $q > 0$ we have $\mid q \alpha - p \mid > c/q$.
I added a screenshot of Shakarchi's proof below:

My understanding of this proof is as follows:
The suggestion given by Lang is to rationalize $q \alpha - p$, i.e. take the product $(q\alpha - p)(-q\alpha - p)$. Doing so yields
$(q\alpha - p)(-q\alpha - p) = -q^2 a + p^2$
Recall $q, a, p \in \mathbb{Z}$, with $q > 0$ and also $\sqrt a \notin \mathbb{Q}$, in particular $a \neq 0$. Then $\mid(q\alpha - p)(-q\alpha - p)\mid \geq 1 \leftrightarrow \mid q\alpha - p\mid \geq \frac{1}{\mid -q\alpha - p\mid} = \frac{1}{\mid q\alpha + p\mid}$
Where I somewhat fall down is in the next part -- we choose $c$ such that $0 < c < \text{min}(\mid\alpha\mid, \mid\frac{1}{3\mid\alpha\mid})$. I suppose we choose $c$ this way to handle the case where $\mid \alpha \mid < 1$ so that $\frac{1}{3\mid\alpha\mid} > 1$. If that is the case, then we can really choose any positive multiple of $\mid \alpha \mid$ in the demoninator, i.e. $\frac{1}{2\mid\alpha\mid}$ or $\frac{1}{4\mid\alpha\mid}$ would work just as well.
Now, using the result obtained in $\textbf{1}$ and our hypothesis, we set up the inequality in $\textbf{2}$. I'm at a loss for how the leftmost inequality is obtained -- I know that by hypothesis $\mid \alpha - p/q \mid < \mid \alpha\mid$ and we add $\mid 2\alpha \mid$ to both sides to obtain the rightmost inequality.
Then in the final inequality, it's unclear to me how we know that $\frac{1}{3\mid \alpha \mid q} < \frac{1}{\mid q\alpha + p \mid}$.
I'm looking for an answer to these two points:

*

*An explanation for the steps I outlined above as being unclear, i.e. the choice of $c$ (why do we choose $0 < c < \text{min}(\mid\alpha\mid, \mid\frac{1}{3\mid\alpha\mid})$), the leftmost inequality in $\textbf{2}$, and the middle inequality in $\textbf{3}$.

*This proof was fairly unintuitive to me -- I didn't even consider rationalizing $q \alpha - p$ when I first was working on this problem. I imagine that's the kind of thing you start to get better at seeing with practice working problems like this. Still, is there possibly a simpler or more direct proof?

 A: It's the triangle inequality
$|\alpha + \frac pq| = |2\alpha - \alpha +\frac pq| \le |2\alpha| + |-\alpha + \frac pq|=|2\alpha| + |\alpha -\frac qp| < |2\alpha| + |\alpha|=3|\alpha|$
The reason the $3$ was choosen was because:  We need to get $|\alpha -\frac pq|$ bigger  than something.  But if $|\alpha-\frac pq|< |\alpha|$ we get can't get it directly because we only know $|\alpha-\frac pq|$ is smaller than something.  Instead we have to work with $|\alpha + \frac pq|$ being bigger than something.  But how can we convert $|\alpha + \frac pq|$ to something involving $|\alpha -\frac pq|$?  Well the way they did it was $|\alpha + \frac pq| = |2\alpha - (\alpha - \frac pq)|$.  But that tosses two extra $\alpha$s into the works.

"it's unclear to me how we know that$\frac{1}{3\mid \alpha \mid q} < \frac{1}{\mid q\alpha + p \mid}$"

Well, you have $|\alpha + \frac pq| < 3|\alpha|$
So $q|\alpha + \frac pq| < 3|\alpha|q$
$0< |q\alpha + p| < 3|\alpha|q$
$\frac 1{3|\alpha| q} < \frac 1{|q\alpha + p|}$.
A: The left-most inequality in 2 took me a bit to figure out too :)
It's the triangle inequality:
$$|a|+|b| \geq |a+b| \,.$$
The middle inequality in 3 is just the overall inequality from 2.
The choice of $c$ might be more flexible, but I think using 3 just makes all the above cancel out and work more nicely.
