<s>[INST] Proving that $x,y\in\mathbb{R}, x>0,$ and $x^2Suppose $x,y\in \mathbb{R}$ and $x>0$. We also know that ${{x}^{2}}<y$. I'm struggling to find a formal proof that there's always an $n\in \mathbb{N}$ such that ${{\left(x+\frac{1}{n} \right)}^{2}}<y$. A hint would be much appreciated.
Edit: Basically, any $n\in \mathbb{N}$ does the job as long as $n>\left\lfloor \frac{1}{\sqrt{y}-x} \right\rfloor $? That got me thinking... Would this still hold if both $x,y$ were rational numbers $x,y\in \mathbb{Q}$? I mean, we wouldn't be to use square roots.
 [/INST] $\left(x+\frac{1}{n} \right)^{2}
=x^2+\frac{2x}{n}+\frac1{n^2}
$
so we want
$x^2+\frac{2x}{n}+\frac1{n^2}
\lt y
$
or
$\frac{2x}{n}+\frac1{n^2}
\lt y-x^2
$.
Let $d = y-x^2$.
We are done if
we can find an $n$
such that
$\frac{2x}{n}
\lt \frac{d}{2}
$
and
$\frac1{n^2}
\lt \frac{d}{2}
$.
The first is equivalent to
$n 
\gt \frac{4x}{d}
$.
The second is equivalent to
$n^2 > \frac{2}{d}
$.
Since $n^2\ge n$,
this will be true if
$n > \frac{2}{d}
$.
Since $d > 0$
and
$x > 0$,
$ \frac{4x}{d}
\gt 0
$
and
$ \frac{2}{d}
\gt 0
$.
Therefore
if 
$n 
\ge \frac{4x}{d}+\frac{2}{d}
=\frac{4x+2}{d}
$,
then
${{\left(x+\frac{1}{n} \right)}^{2}}<y$.
</s>