Dedekind Cut Roots

Question

I was studying Dedekind Cuts in my history of math class and was looking at the following formulas for the multiplication of Dedekind Cuts:

$L_\sqrt2 = \{ r \in \mathbb{Q} : r^2 < 2 $ and $ r>0 \} $

$L_2 = \{ r \in \mathbb{Q} : 0 < r < 2 \} $

$L_\sqrt2 \cdot L_\sqrt2 \subset L_2$ where $\cdot$ is set multiplication.

It was easy for me to prove that this is a subset. Later on the book claims that in fact $L_\sqrt2 \cdot L_\sqrt2 = L_2$ and suggests the reader test out a few examples to see that this is true and then prove it.

Something I tried is assuming that because $\frac{19}{10}$ is in $L_2$ that I should be able to find two positive rational values both less than $\sqrt2$ that multiply to $\frac{19}{10}$ but I've been unable to find such numbers.

I'm looking for help not in just finding an example, but for in general finding a way to show that if I have a prime number in the numerator of a rational number on the right hand side of this equation, how I can find two rational factors of such a number on the left hand side of the equation less than the sqrt of the number the equation is made for (so for in general, $L_\sqrt{x} \cdot L_\sqrt{x} = L_x$ since I have heard this is also true).

Answer 1

Let $r \in L_2$, i.e. $0 < r < 2$. We want to show that there exist $r_1, r_2 \in L_\sqrt2$ such that $r = r_1 \cdot r_2$.

Since $\dfrac 2r > 1$ we can choose $v \in \mathbb N$ such that $$ 1 + \frac 1v < \frac{2}{r} \, . \tag 1 $$ Then define $$ u = \max \{ x \in \mathbb N \mid \frac{x^2}{v^2} < 2 \} \tag 2 $$ so that $$ \frac{u^2}{v^2} < 2 \tag 3 $$ and $$ \frac{(u+1)^2}{v^2} \ge 2 \, \tag 4 $$ Note that $u \ge v$ follows from $(2)$, and together with $(1)$ implies $$ 1 + \frac 1u < \frac{2}{r} \, . \tag 5 $$

With $r_1 := \dfrac uv$, $r_2 := r \dfrac vu$ it is clear that $r = r_1 \cdot r_2$, and $r_1^2 < 2$ follows from $(3)$.

It remains to show that $r_2^2 < 2$: $$ r_2^2 = r^2 \frac{v^2}{u^2} = r^2 \frac{(u+1)^2}{u^2} \frac{v^2}{(u+1)^2} \\ \le r^2 \frac{(u+1)^2}{u^2} \frac 12 \quad\text{ because of $(4)$.} \\ < r^2 \frac{4}{r^2}\frac 12 \quad\text{ because of $(5)$.} \\ = 2 \, . $$

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This can easily be generalized to the case $r \in L_a$ where $a$ is any positive rational number.

First choose $v \in \mathbb N$ such that $$ 1 + \frac 1v < \frac{a}{r} $$ and then define $$ u = \max \{ x \in \mathbb N \mid \frac{x^2}{v^2} < a \} $$

Then $r_1 := \dfrac uv$, $r_2 := r \dfrac vu$ satisfy $r = r_1 \cdot r_2$ and $r_1, r_2 \in L_\sqrt a$.

Answer 2

Suppose we have already proven that given $s\in L_x$ it is always possible to find $q\in\mathbb Q$ such that $$ s<q^2<x $$ Then we can factor $s$ as $s=\frac sq\cdot q$ and check that $\frac sq,q\in L_{\sqrt x}$.

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Let us show how to find $q\in\mathbb Q$ with $s<q^2<x$. Fix $t^2>x$ and consider some $q^2<x$. Then $q<t$ and for $0<\varepsilon<1$ we have $$ (q+\varepsilon)^2-q^2=2q\varepsilon+\varepsilon^2<(2t+1)\varepsilon $$ Now we can choose a rational $0<\varepsilon<1$ small enough to have $(2t+1)\varepsilon<x-s$ and define $q$ as the largest multiple of $\varepsilon$ satisfying $q^2<x$. With this setup we have $$ s<q^2<x $$ as desired.

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As alluded to in the comments to the OP, one way to choose $q$ is to take the decimal approximation to $\sqrt x$ to a precision high enough to match the $s<q^2<x$ criterion. For $x=2$ and $s=\frac{19}{10}$ we have $$ \sqrt x=\sqrt 2\approx 1.414213562373 $$ and we see that truncating as $q=1.4$ already works since $1.4^2=1.96>\frac{19}{10}$. This corresponds to setting $\varepsilon=\frac1{10}=0.1$ in the above setup.

If we should have followed the entire construction in the proof, we should have chosen for instance $t=2$ with $t^2>x$ and then $(2t+1)\varepsilon<x-s$ would become $5\varepsilon<\frac{1}{10}$ so that $\varepsilon<\frac{1}{50}$. A natural choice could then be $\varepsilon=\frac1{100}=0.01$ so that $q=1.41$ and with that we have $$ 1.9<q^2=1.9881<2 $$ and thus $\frac{19}{10}=\frac{190}{141}\cdot\frac{141}{100}$ with $\frac{190}{141},\frac{141}{100}\in L_{\sqrt 2}$.