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.
 A: Hint. $\sqrt{y}-x>0$, and the sequence $(1/n)$ tends to $0$. What does it mean for a sequence to tend to $0$?
A: Let $q$ and $r$ be positive rational numbers, and suppose that $r^2<q$. Motivated by the Babylonian method, define the rational number 
$$s\stackrel{\text{def}}{=}\frac{2qr}{q+r^2}\text{.}$$
Then
$$\begin{align}
s-r&=r\left(\frac{q-r^2}{q+r^2}\right)>0 \\
q-s^2&=q\left(\frac{q-r^2}{q+r^2}\right)^2 > 0
\end{align}$$
so we have found $s$ such that $r<s$ and $r^2<s^2<q$. (If you like, instead choose $n=\lceil(s-r)^{-1}\rceil$.)
In the Dedekind-cut formulation, a real number $x$ is defined to be a pair of unary predicates in the rational numbers
$$\begin{align}
(\quad)<x&:\mathbb{Q}\to\Omega & q\mapsto q<x \\
x<(\quad)&:\mathbb{Q}\to\Omega & q\mapsto x<q 
\end{align}$$
satisfying certain closure properties.
Let $x$ and $y$ be real numbers such such that $0<x$ and $0<y$. Then it is by definition that $x^2<y$ if and only if there exists rational $q$ such that $x^2<q$ and $q<y$. Likewise, it is either a definition or basic lemma that $x^2<q$ if and only if there exists a rational $r$ such that $x<r$ and $r^2<q$. Hence the real-number result follows from the corresponding statement for rational numbers.
In the Dedekind-cut formulation, it is circular reasoning to invoke the real-number $\sqrt{y}$ in proving your theorem. For $\sqrt{y}$ even to be well-defined, one must have already shown that for any rational $q$ such that $q<\sqrt{y}$ one can find a refinement $r$ such that $q<r$ and $r<\sqrt{y}$.
A: This is really just a question of continuity; in particular, we are interested in the function
$$f(x)=x^2$$
and are try to prove that if $f(x) < y$ then $f(x+1/n) < y$ for some $n$. However, this falls into place easily with the machinery of continuity. Note that saying $f(x) < y$ is the same as saying that $f(x) \in (-\infty,y)$. 
Here, we see some analytical ingredients that are worth understanding: $f$ is a continuous function. The interval $(-\infty,y)$ is open. Therefore, $f^{-1}((-\infty,y))$ - that is, the set of points $x$ so that $f(x) < y$ - must also be open. This means that for some $\varepsilon > 0$ we have that for every $x'\in (x-\varepsilon,x+\varepsilon)$ it holds that $f(x') < y$. In particular, if $1/n < \varepsilon$, it holds that $f(x+1/n) < y$ as desired.
Note that there are really only two tools being used here - but that they are really important tools for doing real analysis:


*

*The preimage of an open set under a continuous function is open. 

*For every $\varepsilon > 0$, there is some $n\in\mathbb N$ with $1/n < \varepsilon$.
These tools capture the intuition we want in an elegant form: essentially, our argument boils down to that if $x^2 < y$ then there is some distance between $x^2$ and $y$ - but since there is some wiggle room in the output and the function is continuous, there has to be some wiggle room in the input - and, no matter how small that is, we can change $x$ by some $1/n$ without changing whether its square is less than $y$.

It's worth noting - especially if you want to prove this for $\mathbb Q$ in a way that follows quickly from axioms - that you can also solve this in a really elementary way (though, doing so is essentially repeating a weaker version of the proof that $f$ is continuous that has a more confusing proof). Suppose that
$$x^2<y.$$
We are looking for some $n$ such that $(x+1/n)^2 < y$. We can split into cases; if $x$ is negative, any $n$ so that $x+1/n$ is negative suffices since $z\mapsto z^2$ is decreasing for negative inputs. Otherwise, observe that if we expand the square on the left hand side we get:
$$\left(x+1/n\right)^2 = x^2 + \frac{2}n\cdot x + \frac{1}{n^2}.$$
We need $\frac{2}n\cdot x + \frac{1}{n^2}$ to be less than $y - x^2$ (which is positive by hypothesis). First, note that if $n > \frac{1}{x}$ then $\frac{1}{n^2} < \frac{1}n\cdot x.$
Therefore, if we choose $n$ to be greater than $\frac{1}x$ as well as $\frac{3x}{y-x^2}$, we can find that $\frac{2}n\cdot x + \frac{1}{n^2} < y-x^2$ which implies that $(x+1/n)^2 < y$ by our earlier algebra. Such an $n$ clearly exists - for instance, take the ceiling of the maximum of $1/x$ and $\frac{3x}{y-x^2}$ and add one.
A: $\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
${{\left(x+\frac{1}{n} \right)}^{2}}<y$.
A: A very simple proof is the following. Let $f(z)=(x+z)^{2}$. Clearly $f$ is continuous and $f(0)=x^{2}<y$ by hypothesis. By continuity there is an interval containing 0 and so in the form $(-a,a)$  with $a>0$ such that for all $z$ in this interval $f(z)<y$. Then there is $n$ big enough such that $1/n$ is in this interval and so $f(1/n)=(x+1/n)^{2}<y$.
