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I'm reading Dedekind's work where he defines irrational numbers, using cuts. I'm stuck where he proves (p. 13) that

there exist infinitely many cuts not produced by rational numbers

His proof includes proof of "lemma" that there is no rational (without integer) number square of which is integer number.

Let me highlight one very imortant moment.

I'M NOT INTERESTED IN ANY OTHER PROOF OF THIS "LEMMA", I WANTED TO UNDERSTAND ONLY DEDEKIND'S PROOF AND NO OTHER ONE.

Let $\lambda \in \mathbb{Z}_+$, $\sqrt{D} \notin \mathbb{Z}$:

$$ \lambda^2 < D < (\lambda+1)^2. $$

Let's take $r \in \mathbb{R}_+$ and if $r^2 > D$ then $r$ belongs to $A_2$ and if $r^2 < D$ then it belongs to $A_1$.

Dedekind says that $\sqrt{D} \notin \mathbb{Q}$ and prooves it by contradiction. Assume that $\sqrt{D} \in \mathbb{Q}$, then there are such $t$ and $u$ in $\mathbb{Z}_+$ that:

$$ t^2-Du^2=0. $$

Note: remainder of $\frac t u$ isn't $0$, cause in that way $\sqrt{D} \in \mathbb{Z}$.

I'm not quite sure how existion of such numbers ($t$ and $u$) follows from that $\sqrt{D} \in \mathbb{Q}$.

I had tried to represent $\sqrt{D}$ as $\frac m n$ and I've gotten

$$ t^2-\frac{m^2}{n^2}u^2=0, $$

$$ t^2=\frac{m^2}{n^2}u^2, $$

$$ t^2 n^2=m^2 u^2, $$

$ t, u \in \mathbb{Z}_+ $, $n \in \mathbb{N}$, and without loosing generality we can assume that $n \in \mathbb{Z}_+$ too.

$$ tn=mu. $$

Cause $t$ is not multiple of $u$, then $n$ is, let's say $u=kn$.

$$ tn=mkn, $$

$$ t=mk. $$

So if we take $t$ that it is propotional to the numerator and $u$ that is proportional for denumerator we got proof that such $t$ and $u$ can exist, aren't we?

dragondangun
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  • How does Dedekind define a rational number? – Yathi Mar 14 '24 at 10:17
  • @YathirajSharma, if I've got it right, then Dedekind defines rational number $a$ as cut $(A_1, A_2)$ where all numbers from $A_1$ are less than $a$ and all number from $A_2$ are bigger than $a$. There is no fundamental difference between cases where $a$ belongs to $A_1$ and $a$ belongs to $A_2$. It'll be $\max$ of $A_1$ in first case an $\min$ of $A_2$ in the last one. – dragondangun Mar 14 '24 at 11:22
  • And the difference (if I've got the point) between rational and irrational is that rational can belong to $A_1$ or $A_2$ and irrational can not (in other words it belongs to neither $A_1$ nor $A_2$). – dragondangun Mar 14 '24 at 16:59
  • The part of the proof you are asking about has nothing to do with Dedekind cuts. Does this answer your question? How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number? – Ethan Bolker Mar 14 '24 at 17:24
  • @EthanBolker, nope, it's not the answer. Here's the Dedekind's work from which this question had arisen. I've linked to exact page, where he states that such numbers ($t$ and $u$) exist, but lefts it without comments as it's obvious thing. – dragondangun Mar 14 '24 at 17:33
  • You ask how you know there are such numbers $t$ and $u$. That follows as in the linked answer from the assumption that $\sqrt{D}$ is rational. The argument at this point has nothing to do with Dedekind cuts. It depends only on the properties of the rational numbers. – Ethan Bolker Mar 14 '24 at 17:41
  • @EthanBolker, do you mean vacuous truth? Because of $\sqrt{D}$ can't be rational then any consequence is true and that such numbers exist in particular, do you mean that? Ok, that's cool, but Dedekind prooves that $\sqrt{D}$ can't be rational in other way and I want to understand his logic. – dragondangun Mar 14 '24 at 18:33
  • @EthanBolker, and let me explain why your answer is unacceptable. We want to proof that $\sqrt{D}$ is irrational by contradiction, so we assume that $\sqrt{D}$ is rational (let it be statement $A$), hoping to find contradiction. So, my question was why such numbers exists, assuming $A$ is true. You say that they exists because that theorem says that $A$ is false, but if we use that theorem we don't need proove that $A$ is false by contradiction, do you get it? – dragondangun Mar 14 '24 at 19:01
  • Sorry my comments don't help you. Maybe someone else here will. – Ethan Bolker Mar 14 '24 at 19:27
  • @EthanBolker, anyway thank you for your attempt (and I'm awfully sorry if my replies seems kinda rude, I'm no way meant that). – dragondangun Mar 14 '24 at 19:36
  • For a modern presentation see here and the discussion around Theorem 2 here. – Bill Dubuque Mar 18 '24 at 21:22
  • @BillDubuque, once again: my question is not about any proof of this theorem, but only Dedekind's one, that he used in his paper (link above). I'm not interested in any other proof and so my question is not duplicate of marked ones, cause all they use another ways to proof this theorem. And what is more significant that is my question is not about proof as result, but understanding of Dedekind's proof. My question pursues other goals. I've thought it was clear after my edit, but I see it's not. I'll edit question to highlight this moment. – dragondangun Mar 19 '24 at 08:58
  • Dedekind's proof is a special case of the methods I linked, see here for further details on the relationship between Dedekind's quadratic form and the more modern linear forms. – Bill Dubuque Mar 19 '24 at 09:37
  • @BillDubuque, second link has a lot of abstract algebra stuff and so it's much more complicated answer that I need. The first one use infinite descent, and yes, this is proof by contradiction, but we don't use additional variables, we only represent $D$ as $\frac n m$ and $\lfloor \sqrt{D} \rfloor$ as $q$ and we proove theorem using fraction properties. Dedekind instead introduce equation and two varibles, and doesn't comment why such equation exists (and that is what about my question). What's important that $t$ and $u$ are not defined, we only say that they exist. – dragondangun Mar 19 '24 at 09:45

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So, I had consulted with my lecturer* and he said that I'm right, but there was no reason to write so much, I could stop at $t^2=\frac{m^2}{n^2}u^2$ cause we can find such $u\in\mathbb{Z}_+$ that $n^2|u^2$ and hence we can solve equation $t^2 = q^2$, where $q\in\mathbb{Z}_+$. If we can solve this equation then such numbers ($u$ and $t$) exist.

* that wasn't my homework, I already have master's degree**
** And I have low self-esteem either, heh...

UPD: Read comments below the question. Ethan Bolker linked equivalent theorem there, but it's not Dedekind's proof and so that wasn't useful for me, but maybe it'll be the one for you. And also there is reason why we can't proof existing of such numbers by vacuous truth.

dragondangun
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