Prove there is a square of a rational number between any two positive real numbers

Question

I'm doing this problem and the solution is below. I don't understand why this solution proved $m^2 \le x$ firstly? Any help, thanks.

Answer 1

The idea is to prove that there is a real number $m\geqslant0$ such that $m^2\leqslant x$ which is such that for any rational number $r>m$ you have $r^2>x$. So, if you add a small number to $m$ (the $\delta$ of the proof), then $(m+\delta)^2>x$. So, what you have to do next is to prove that you can find such a $\delta$ so that

• $m+\delta\in\mathbb Q$;
• $x<(m+\delta)^2<y$.

Answer 2

This is too long for a comment, so I'm posting it as an answer.

Instead of addressing the quoted proof, I give a quite short proof of the more general result that for any positive integer $k$ there exists a rational number $r$ such that $x < r^k < y$, without assuming any knowledge of the existence of $k^\text{th}$ roots. (Taking $k = 2$ doesn't make the proof any shorter.)

For all positive integers $k, p$, $$ \frac{(p + 1)^k}{p^k} = \left(1 + \frac{1}{p}\right)^k = \sum_{j=0}^k\binom{k}{j}\frac{1}{p^j} \leqslant 1 + \sum_{j=1}^k\binom{k}{j}\frac{1}{p} = 1 + \frac{2^k - 1}{p}. $$

Given $y > x > 0$, define $$ h = \frac{(2^k - 1)x}{y - x}. $$

Then $$ \tag{1}\label{ineq:p} p > h \implies \frac{(p + 1)^k}{p^k} < \frac{y}{x}. $$

There exists a positive integer $q$ such that $$ \tag{2}\label{ineq:q} q^kx \geqslant (1 + h)^k. $$

For instance, it is enough to take any $q$ such that $$ q \geqslant (1 + h)\left(1 + \frac{\max\{1 - x, 0\}}{kx}\right). $$ The inequality just given is equivalent to the conjunction of $q \geqslant 1 + h$ with $$ 1 + k\left(\frac{q}{1 + h} - 1\right) \geqslant \frac{1}{x}, $$ and therefore it implies $$ \frac{q^k}{(1 + h)^k} = \left(\frac{q}{1 + h}\right)^k = \left(1 + \frac{q}{1 + h} - 1\right)^k \geqslant 1 + k\left(\frac{q}{1 + h} - 1\right) \geqslant \frac{1}{x}, $$ as required.

Take any value of $q$ satisfying \eqref{ineq:q}.

The inequality $$ \tag{3}\label{ineq} p^k \leqslant q^kx $$ holds for some values of $p$ (for example, $p = 1$), but not for all (for example, not for $p = q\left(1 + \lceil{x}\rceil\right)$).

Therefore, there exists a unique largest value of $p$ for which \eqref{ineq} holds.

For this value of $p$, we have $(p + 1)^k > q^kx \geqslant (1 + h)^k$, and therefore $p > h$.

Invoking \eqref{ineq:p} and \eqref{ineq}, we get $$ q^kx < (p + 1)^k < p^k\frac{y}{x} \leqslant q^ky, $$ and therefore finally $$ x < \left(\frac{p + 1}{q}\right)^k < y. $$