I'm self studying through Courant's Introduction to Calculus and Analysis and I came across a Solutions manual with additional exercises by Arthur Blank. I'm working through these problems alone and I'd like some feedback on this proof. I've only taken one proof focused class two years ago so I'm not completely experienced.
Statement to be proved: For $n,p$ ranging over all positive integers, the irrational numbers
$\ x = \frac{p}{(\sqrt{2})^{2n+1}} $
are dense on the real number line.
Proof: Take any point on the real number line and call it $\ R$.
Choose $\ n, p $ such that $\ R \in \left[\frac{p}{(\sqrt{2})^{2n+1}},\frac{p+1}{(\sqrt{2})^{2n+1}}\right]$.
So then $\ \frac{p}{(\sqrt{2})^{2n+1}} \leq R \leq \frac{p+1}{(\sqrt{2})^{2n+1}} $.
Rearranging this, $\ p \leq R {(\sqrt{2})^{2n+1}} \leq p+1 $.
Now separate this interval into two equal parts.
Now introduce $\ p_{1}$ such that $\ R (\sqrt{2})^{2n+1} \in \left[p + \frac{p_1}2,p+ \frac{p_1+1}2\right]$.
So then $\ p + \frac{p_{1}}2 \leq R(\sqrt{2})^{2n+1} \leq p+ \frac{p_1+1}2 $.
We can multiply by a factor to get $\ 2p + p_{1} \leq R(\sqrt{2})^{2n+1}(2) \leq 2p+ p_{1}+1 $.
But $\ 2 = \sqrt{2}^2$, so we can arrange the inequality and introduce the constant $\ p_\text{new} = 2p+ p_{1} $ to obtain
$\ p_\text{new} \leq R{(\sqrt{2})^{2(n+1)+1}} \leq p_\text{new} $.
If we let $\ n_\text{new} = (n+1) $, then we have just obtained a number $ x = \frac{p_\text{new}}{(\sqrt{2})^{2n_\text{new}+1}}$ that is a better estimate for $\ R $ than $ \frac{p}{(\sqrt{2})^{2n+1}}$.
Since we can always obtain a better estimate by dividing our interval into two parts, the irrational numbers $\ x = \frac{p}{(\sqrt{2})^{2n+1}} $
are dense on the real number line.
Best Answer
Good start! The idea is there. To make the proof precise, I'd do two things:
To start you off, I'll restate the proposition in question:
Now your job is to address the challenge, "Here is $\epsilon$. What $p, n$ have you got for me?"