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Is my solution sufficient to solve the problem?

I am attempting to provide a solution for SECTION 1.1a, page 2 Q1.(b) of Introduction to Calculus and Analysis I by Richard Courant. The question: "Show that between any two rational numbers there exists at least one irrational number and, consequently, infinitely many."

This post is a follow up to Show that between any two rational numbers there exists at least one irrational number and, consequently, infinitely many..

With help from users in the comments to my earlier post, I am satisfied that between any two rational numbers there exists at least one irrational number. As the justification is relevant to this question I will restate it:

Let $p,q\in\mathbb{Q}$ where $p<q$. Their midpoint, $M=(p+q)/2$ is also rational, because addition and division between rational numbers output rational numbers - for division this is excepting denominators of 0.

The book gave a proof by contradiction that $\sqrt{2}$ is irrational already, so we will assume that. Our goal is to find a positive integer, $n>0\in\mathbb{Z}$, that we can divide $\sqrt{2}$ by, so its image is less than half of $q-M$. $\sqrt{2}<2$, so it would suffice to find such an $n$ for 2.

$$(2/n)<(q-M)\iff2/(q-M)<n\implies n=\lceil2/(q-M)\rceil$$

Therefore, if $k=M+\frac{\sqrt{2}}{n}$, then $k$ is both irrational as well as between $q$ and $p$.

Aside from the comments in the earlier linked post, https://math.stackexchange.com/a/46824/1499599 is also worthy of credit.

Now, based on this conclusion, I am to show that there exists infinitely many irrational numbers between any two rational numbers.

I was thinking of showing this using the following logic:

Let $j=M+\frac{2}{n}$.

We know that $p<k<j<q$.

$p,j,q\in\mathbb{Q}$, and $k$ is irrational. By the same logic we used in our proof, there exists an irrational number, $k^\prime$ between $j$ and $q$.

In calculating $k^\prime$, there will be a corresponding $j^\prime\in\mathbb{Q}$.

$$p<k<j<k^\prime<j^\prime<q$$

We could iterate this further by taking an irrational point between $j^\prime$ and $q$ and so on to obtain infinitely many irrational numbers between $p$ and $q$.

Therefore, there are infinitely many irrational numbers between any two rational numbers.

Alternatively, we could elaborate on the comment to How many Irrational numbers? to say there exists an irrational number, $l=\frac{\sqrt{2}}{b}$ between $p$ and $q$ for every $b$ in

$$\{\lceil2/(q-M)\rceil,\lceil2/(q-M)\rceil+1,\lceil2/(q-M)\rceil+2,...\}$$

P123
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    You wrote $k=M+n$. Should this have been $k=M+\frac{\sqrt{2}}{n}$? Also, I can't make sense of your inequality $p<k<n<q$. If the previous suggested modification is correct, then $p<M<k<q$ would make sense to me. – Will Orrick Dec 05 '24 at 03:40
  • @WillOrrick That modification is correct! Thank you for pointing that out, I will edit the post. Would it suffice to make $n$ large enough, so that $M<k<(M+q)/2=g$? Because then, the interval $[g,q]$ does not contain $k$, and is bounded by rational numbers. – P123 Dec 05 '24 at 03:44
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    Here is a shorter way: if $p<q$ are rational then $\alpha =p+{1 \over \sqrt{2}} (q-p)$ is irrational and lies in $(p,q)$. Then, for any rational $t>0$, note that $p+t(\alpha-p)$ is irrational. It is straightforward to produce an infinite number of rationals in $(0,1]$. – copper.hat Dec 05 '24 at 03:48
  • @copper.hat Thank you, I think I will use that instead of my proposed solution. I am still curious if it is right though. – P123 Dec 05 '24 at 03:54
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    Yes, that should work. But it might be easier to work with the interval $[p,M]$ which, so far, contains no identified irrational number. Use your construction to find an irrational number in the second half of this interval. Keep repeating this, always finding the next irrational number in the second half of the current interval and reserving the first half for future irrational numbers. – Will Orrick Dec 05 '24 at 03:54
  • @WillOrrick Ahh, the interval can be divided into two rational numbers infinitely many times, and each subdivision contains an irrational number. We don't even need all of these irrational numbers for the solution to be correct. – P123 Dec 05 '24 at 03:58
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    @P123 It is straightforward to prove. Adding a rational to an irrational is always irrational (yet I still do it), and multiplying an irrational by a non zero rational is also always irrational. Producing a supply of rationals in $(0,1]$ is prescriptive. – copper.hat Dec 05 '24 at 03:58
  • @copper.hat Thank you, please pardon my inexperience in the subject. – P123 Dec 05 '24 at 03:59
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    I am very experienced with irrationality in all aspects of life. – copper.hat Dec 05 '24 at 03:59
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    If for irrational $r$ and rationals $a, b$ such that $a < r < b$, knowing that for any rational $f$ such that $0 < f < 1$, we would have $fr$ as irrational, we have $a + fr$ irrational, where $a < a + fr < b$. I suppose the constraint would be that $fr < b - a$ – Hudjefa Dec 05 '24 at 06:54

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