Prove that a cut is not produced by rational number

Question

Notice: I have asked a similar question on this topic before here, but it partook an alternative approach which didn't lead me anywhere, so now I am going with the classic one, suggested by the author; these, however, are two different questions, on two different problems, and I would appreciate if the thread wasn't closed on the basis of supposed repetition. Thank you.
Problem
Let $D$ be a positive integer, but not a square of an integer, then there exists a positive integer $\lambda$ such that $$\lambda^2<D<(\lambda+1)^2$$ If we assign to the second class $A_2$, every positive rational number $a_2$ whose square is $>D$, to the first class $A_1$ all other rational numbers $a_1$, this separation forms a cut $(A_1,A_2)$ ...But this cut is produced by no rational number.
Proof

1. If there exists a rational number whose square $=D$, then there exist two positive integers $t,u$ that satisfy equation $$t^2-Du^2=0$$
2. We may assume that $u$ is the least positive integer possessing the property that its square, by multiplication by $D$, may be converted into the square of an integer $t$. Since evidently $$\lambda u<t<(\lambda+1)u$$
3. The number ${u}'=t-\lambda u$ is a positive integer certainly less than $u$. If further we put $${t}'=Du-\lambda t$$ ${t}'$ is likewise a positive integer, and we have

$${t}'^2-D{u}'^2=(\lambda^2-D)({t}^2-D{u}^2)$$

4. Which is contrary to the assumption respecting $u$ [step 2].
   
   

My question
For the life of me I do not understand how ${u}'^2$ "by multiplication by $D$, may be converted into the square of an integer $t$" when it is clearly 'converted' to ${t}'^2$, which in turn, clearly $\neq t^2$.
Can anyone explain this part to me?

Answer 1

The property isn't about that $t$ but about any possible $t$. $t'$ is a possible "$t$".

Perhaps we should put it in this overly fussy way.

Of all the $(\mu, \tau)$ integer pairs where $\tau^2 - D\mu^2 = 0$ we may assume that $u$ is the least possible $\mu$.

But as $(u', t')$ is a pair where $t'^2 - Du'^2 = 0$ but $u' < u$ that is a contradiction.

That's overly stilted but that is what they are trying to express.

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Okay. I'd translate this proof.

As $D$ is not a perfect square there is an integer $n$ (the text used $\lambda$) so that $n^2 < D < (n+1)^2$.

Book claims: There is no rational $q$ so that $D = q^2$.

The book outline goes like this:

If $(\frac tu)^2 = D$ and $\frac tu$ is in lowest terms;

Then

The book will claim If $u' = t-nu$ and $t' = Du - tn$ then $(\frac {t'}{u'})^2 = D$ and that $0< u' < u$.

So either we have two rational numbers both squaring to $D$ or that $\frac tu$ was not in lowest terms. Either one is a contradiction.

So is the claim true?

And the proof will use extensively that

$(\frac ab)^2 = D\iff \frac {a^2}{b^2} = D \iff a^2 - Db^2 = 0$.

So

First claim: $0 < u' = t-nu < u$. Is that true?

Well $\frac {t^2}{u^2} = D$ so

$t^2 = Du^2$

And $n^2 < D < (n+1)^2$ so

$n^2u^2 < Du^2 < t^2 < (n+1)^2u^2$ and

$nu < t < (n+1)u$

$0 < t-nu =u' < (n+1)u -nu = u$

So yes it is.

Second claim: $(\frac {t'}{u'})^2= D$. Is that true?

Well $(\frac {t'}{u'})^2= D \iff \frac {t'^2}{u'^2} = D\iff$

$t'^2 - u'^2D = 0$.

Well $t'^2 - u'^2D = (Du-tn)^2 -(t-nu)^2D =$

$(D^2u^2 - 2Dutn + t^2n^2) - (Dt^2 -2Dutn + Dn^2u^2)=$

$D^2u^2 + t^2n^2 - Dt^2 + Dn^2u^2 =$

$n^2(t^2 -Du^2)- D(t^2-Du^2) =$

$(n^2 -D)(t^2 - Du^2)$

But rememember $\frac {t^2}{u^2} = D$ so $t^2 - Du^2 = 0$.

So $t'^2 - u'^2D = (n^2-D)*0 = 0$.

So $\frac {t'^2}{u'^2} = D$.

And that's that.

If there is a rational number, $q$, so that $q^2 = D$ that books claims would be true which would be a contradiction

Answer 2

This is a convoluted way of saying that $u$ is the smallest integer that satisfies $\frac{t^2}{u^2} = D$ for some integer $t$. Then for such $D,u,t$, this implies $Du^2 = t^2$ [so $u^2$ times $D$ is another square $t^2$]. As $t$ is not a multiple of $u$ by assumption, implies a $\lambda$ satsifying $\lambda u < t < (\lambda+1)u$.

But I think the proof you are trying to decode was written in a very convoluted fashion. I would try to prove this by first observing the following: If $u$ does not dividing $t$, then $u^2$ does not divide $t^2$ [make sure you see why this is]. Then this implies that there is no way $\frac{t}{u}$; $u$ not dividing $t$, could square to an integer $D$; if $(\frac{t}{u})^2$ were an integer with $\frac{t}{u}$ not, this would imply $u^2$ dividing $t^2$ despite $u$ not dividing $t$.