Dedekind Cut Roots 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).
 A: 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 $r_2^2 < 2$:
$$
r_2^2 = r^2 \frac{v^2}{u^2} = r^2  \frac{(u+1)^2}{u^2}
 \frac{v^2}{(u+1)^2} \\
 \le r^2  \frac{(u+1)^2}{u^2} \frac 12 \quad\text{ because of $(4)$.} \\
 < r^2 \frac{4}{r^2}\frac 12  \quad\text{ because of $(5)$.} \\
 = 2 \, .
$$

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 $r_1, r_2 \in L_\sqrt a$.
A: 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 $\frac sq,q\in L_{\sqrt x}$.

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.

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 $\frac{190}{141},\frac{141}{100}\in L_{\sqrt 2}$.
