Finding some rational numbers

Question

Let $p$ be a prime number and let $x$ be a rational number such that $x>1$. Prove that if $x^2<p$, then there exists a rational number $z$ such that $x<z$ and $z^2<p$. (Please, don't say that $x<\sqrt p$ because I want to do this on rational numbers).

For the case $p=2$. I have $z=x+\frac{2-x^2}{4}$ satisfies the conditions $x<z$ and $z^2<2$.

Answer 1

This is true for any rational number $p$.

If rational $x$ satisfies $x^2 < p$, then, since the rationals are dense, there is a rational $q$ such that $x < q < \sqrt{p}$, whether or not $p$ is a perfect squate.

Then $x^2 < q^2 < p$.

Answer 2

What you are asking is to prove there is a rational square between $x^2$ and $p$, or that the rational squares are dense. If $x=\frac ab$ in lowest terms, consider $y_i=\frac {(ai+1)^2}{(bi)^2}=x^2+\frac{2ai+1}{(bi)^2}$. This gets arbitrarily close to $x^2$, so just choose $i$ large enough.