(1) Suppose $\frac{p_1}{q_1}$ and $\frac{p_2}{q_2}$ are rationals with $0<\frac{p_1}{q_1} <\frac{p_2}{q_2}$. We want to find a rational $\frac{a}{b}$ such that $\frac{p_1}{q_1}<\frac{a^2}{b^2} <\frac{p_2}{q_2}$.
I know that if we choose any rational in the open interval between $\sqrt{\frac{p_1}{q_1}}$ and $\sqrt{\frac{p_2}{q_2}}$, then this rational will have the desired property. However, the issue is that I am trying to use this proof as a part of a larger proof to prove the existence of the square root of any positive real number. So I cannot use any argument that directly references square roots.
Here is some work I have done so far. I went down the rabbit hole a bit and this may be a dead end, I'm not sure.
Since $\frac{p_1}{q_1} = \frac{p_1q_1q_2^2}{q_1^2q_2^2}$ and $\frac{p_2}{q_2} = \frac{p_2q_1^2q_2}{q_1^2q_2^2}$, it is sufficient to prove that:
(2) there exists a perfect square, $n^2$, (possibly 1) such that the interval between $n^2p_1q_1q_2^2$ and $n^2p_2q_1^2q_2$ contains a perfect square.
To prove (2) it is sufficient to prove that:
(3) for all positive integers $n$ there exist perfect squares, $k^2$ and $m^2$, such that $k^2n<m^2<k^2(n+1)$
So my question is either for assistance proving any of (1), (2), or (3) without using the existence of square roots of positive real numbers,
OR if this is something that MUST be proven using the existence of square roots, then just state that. I would NOT like help in proving the existence of square roots another way. If it MUST be done another way, I would like to figure out the other way myself.
That being said, I am sure that proving the existence of square roots of positive reals CAN be proven another way (likely much simpler), but I am hoping that the path I am taking may work too. I just need help with this last part.
Best Answer
The following proof does not assume any knowledge of real numbers and square roots and uses only the properties of rational numbers. The argument below is typical of many analytical arguments and is the essence of definition of real numbers as envisaged first by Dedekind.
Let $x, y$ be two positive rationals with $x<y$. We are supposed to find a rational $a$ such that $x<a^2<y$. The problem is easily handled if $y$ itself is a perfect square. Thus if $y=t^2$ then we can choose $a<t$ such that $$y-a^2=t^2-a^2<y-x$$ ie $$(t-a) (t+a) <y-x$$ and this is easily done if we choose $$t-a<\frac{y-x} {2t}$$
We are thus left with the difficult case when $y$ is not a square. Let's consider the set $$A=\{a\mid a\in\mathbb {Q}, a>0,a^2<y\}$$ Clearly this set is non empty and bounded above. We will prove that there is no greatest member in this set. Suppose on the contrary that there is a greatest member, say $m$, in $A$. Then we have $m^2<y$ but for any rational $n>m$ we have $n^2>y$. Let's further assume $m<n<m+1$ and then we have $$n^2-m^2=(n-m)(n+m)<(n-m)(2m+1)$$ And we have an obvious contradiction if $$0<n-m<\min\left(1,\frac{y-m^2}{2m+1}\right)$$ because then we have $n^2-m^2<y-m^2$ or $n^2<y$.
In a similar manner we can show that the set $$B=\{b\mid b\in\mathbb {Q}, b>0,b^2>y\}$$ is non-empty, bounded below and has no least member. Now we can choose some specific members $a\in A, b\in B$ and an arbitrary positive integer $n$ and consider the sequence of numbers $$x_i=a+i\cdot\frac{b-a} {n} $$ so that $$a=x_0<x_1<x_2<\dots<x_n=b$$ In the above sequence there is a last, say $x_r$, which lies in $A$ and the next one $x_{r+1}\in B$ and we have $x_{r+1}-x_r=1/n$. Since $n$ is arbitrary it follows that give any positive rational number $\epsilon$ we can find members $a\in A, b\in B$ such that $b-a<\epsilon$.
Next we set $\epsilon=y-x>0$ and consider a specific number $k\in B$. Now we can choose $a\in A, b\in B$ with $b-a<\epsilon/(2k)$ and also ensure $b<k$. This implies that $a<b<k$ and thus $a+b<2k$ and $$b^2-a^2=(b+a)(b-a)<2k\cdot\frac{\epsilon}{2k}=\epsilon$$ or $$(b^2-y)+(y-a^2)<\epsilon$$ Each term in parentheses in positive and hence individually less than $\epsilon $. We thus have $$y-a^2<\epsilon=y-x$$ or $x<a^2<y$. Thus we have found a rational number $a$ as desired.
The above proof can be worked out (to aid in understanding) for specific example. Let's say we wish to find a square between $x=1.9$ and $y=2$. We take $a=1\in A, b=2\in B$ and set $n=100$ ($n$ is choosen such that $1/n< (y-x) / 2b$) and consider sequence of numbers $x_i=1+(i/100)$. Clearly $x_{41}=1.41\in A, x_{42}=1.42\in B$ and $1.9<1.41^2<2$.