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.

  • $\begingroup$ The proof looks fine, but there is one detail to nitpick. It is not sufficient to say that one can find a "better estimate". You have to show that for any $R$ and any $\varepsilon > 0$, there is some choice of $n$ and $p$ for which $|R - \frac{p}{(\sqrt{2})^{2n+1}}| < \varepsilon$. For example, the sequence $1 + 1/n$ is a "better estimate for $0$ as $n \rightarrow \infty$, but it converges to $1$. $\endgroup$
    – cxx
    Commented Apr 9, 2020 at 22:22
  • $\begingroup$ So I would need to specify that if $\ | R - \frac{p}{(\sqrt{2})^{2n+1}} | > \epsilon $, then we can find a closer irrational number of the same format until $\ | R - \frac{p}{(\sqrt{2})^{2n+1}} | < \epsilon $? And since $\ \epsilon $ is arbitrary, they're dense on the number line? . $\endgroup$
    – Edd
    Commented Apr 9, 2020 at 22:29
  • $\begingroup$ Yes. Neals answer is a good place to start. $\endgroup$
    – cxx
    Commented Apr 9, 2020 at 22:30

1 Answer 1


Good start! The idea is there. To make the proof precise, I'd do two things:

  1. Very skeptically home in on phrases like, "Choose $n, p$ such that..." and "Now introduce $p_1$ such that...". Why can you do that? What exactly are these numbers you're choosing and introducing?
  2. Make the phrase "better estimate" more precise. For example, how do you know your better estimate gets you arbitrarily close? The sequence $1.1 + 2^{-n}$ is successively better approximations of $1$, but it never actually gets arbitrarily close.

To start you off, I'll restate the proposition in question:

Let $R$ be an arbitrary real number. For every $\epsilon > 0$, there exist $n, p$ such that $$\bigg| R - \frac{p}{2^{n/2}}\bigg| < \epsilon. $$

Now your job is to address the challenge, "Here is $\epsilon$. What $p, n$ have you got for me?"

  • 1
    $\begingroup$ This is helpful. When I have a better proof, I'll attach it here instead of editing my answer. $\endgroup$
    – Edd
    Commented Apr 9, 2020 at 22:37
  • $\begingroup$ I think I've got it. Let $\ R $ be an arbitrary real number. If $\ R $ is arbitrary real number, then so is $\ R∗(\sqrt{2}) $. Let ϵ>0. Let $p_{0}$ be the integer part of $\ R∗(\sqrt{2}) $ such that $\ p_{0}≤ R∗(\sqrt{2}) \leq p_{0}+1 $. Clearly this interval has length 1. Now subdivide it in two and introduce a binary constant $\ p_{1} $ such that $\ p_{0} +\frac{p_{1}}2 \leq R*(\sqrt{2}) \leq p_{0} + \frac{p_{1}+1}2 $. Now this interval is half the size as before, and is therefore a better estimate. We can continue until our interval's size less than epsilon. $\endgroup$
    – Edd
    Commented Apr 14, 2020 at 0:15
  • $\begingroup$ Let's say that occurs on the nth subdivision. Then $\ p_{0} +\frac{p_{1}}2 + … + \frac{p_{n}}/2^{n}\leq R*(\sqrt{2}) \leq p_{0} + \frac{p_{1}}2 + … + \frac{p_{n}+1}/2^{n} $. Rearranging this, we get $\ 2^{n}p_{0} +2^{n-1}p_{1} + … + p_{n} \leq R*(\sqrt{2}^{2n+1}) \leq 2^{n}p_{0} + 2^{n-1}p_{1} + … + p_{n}+1 $. If we introduce $\ p = 2^{n}p_{0} +2^{n-1}p_{1} + … + p_{n} $ and rearrange, we obtain $\ \frac{p}{\sqrt{2}^{2n+1}} \leq R \leq \frac{p+1}{\sqrt{2}^{2n+1}} $. $\endgroup$
    – Edd
    Commented Apr 14, 2020 at 0:19
  • $\begingroup$ This interval has length $\ \frac{1}{\sqrt{2}^{2n+1}} $. Since $\ \epsilon $ is arbitrary, we choose a sufficiently large enough n. Then $\ | R - \frac{2}{\sqrt{2}^{2n+1}}| \leq \epsilon. $ Is this good? @Neal $\endgroup$
    – Edd
    Commented Apr 14, 2020 at 0:22

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