Prove that there is no rational lowest upper bound for $\sqrt{3}$. I am trying to prove this and have looked at similar questions to gauge how to approach this. I have:
Suppose that there exists a smallest rational number greater than $\sqrt{3}$.
We shall call that number $n$, which, as it is rational, can be expressed as $\frac{p}{q}$
$\frac{\sqrt{3}+n}{\sqrt{3}}$ is a number greater than $\sqrt{3}$ but less than $n$, but this number would no longer be rational now, would it?
 A: if you have positive integers $x,y$ with
$$  \frac{x}{y} > \sqrt 3 $$
we also have
$$ x^2 - 3 y^2 > 0  $$
is an integer
so that there is some positive $T$ with
$$ x^2 - 3 y^2 = T.  $$
Well we calculate that
$$  u = 2x + 3y \; \; , \hspace{9mm}  v = x+2y $$
satisfy
$$ u^2 - 3 v^2 = T > 0 $$
also.
$$ u^2 > 3 v^2 $$
$$\frac{u^2}{v^2} > 3 $$
$$ \left(  \frac{u}{v}  \right)^2 > 3 $$
Well
$$ x^2 - 3 y^2 > 0  $$
$$ x^2 > 3 y^2  $$
$$ x^2 + 2 x y > 2xy + 3 y^2 $$
$$ x (x+2y) > y(2x+3y) $$
$$ \frac{x}{y}  > \frac{2x+3y}{x+2y}  $$ so
$$ \frac{x}{y}  > \frac{2x+3y}{x+2y} > \sqrt 3 $$
A: Let $S=\{q\in \mathbb Q:q^2>3\}$ and suppose $q\in S.$ Set $p=\frac{3q+3}{q+3}$.
Then,  $p<q$ and $p\in S$ because  $\left(\frac{3q+3}{q+3}\right)^2-3=\frac{6(q^2-3)}{(q+3)^2}>0.$
A: Let $q$ be least integer greater than $\sqrt{3}$, then $q^2>3$.If we can get a rational $q-\frac1 n >\sqrt{3}$ then we'll get a contradiction.
Observe $\left(q-\frac1 n\right)^2\geq q^2-\frac{2q}{n}$.
By Archimedean principle,$\exists N $such that $\frac 1 N < \frac{q^2-3}{2q}$. $$\therefore q^2-3>\frac{2q}{n}\implies q^2-\frac{2q}{n}>3 \implies .\left(q-\frac1 n\right)^2>3 \implies \left(q-\frac1 n\right) \geq \sqrt{3}$$
$\therefore$ we found a rational $q-\frac 1 N$ which is less than $q$ but greater than $\sqrt{3}$.Which is a contradiction.
A: Suppose $q \in \mathbb{Q} > \sqrt{3}$; then $q^2 > 3$, or $q^2= 3 + \delta$ with $\delta \in\mathbb{Q} > 0$.  Then we want to choose some rational $\varepsilon>0$ such that $$(q-\varepsilon)^2=q^2-2q\varepsilon+\varepsilon^2=3+\delta-2q\varepsilon+\varepsilon^2 > 3+\delta-2q\varepsilon\ge 3,$$
so that $q-\varepsilon\in\mathbb{Q}$ and $q >q-\varepsilon > \sqrt{3}$.  The inequality holds provided that $2q\varepsilon \le \delta.$  In particular, we can just choose $\varepsilon=\delta/(2q)$, noting that this is rational whenever $\delta$ and $q$ are.
We conclude that, for any rational $q > \sqrt{3}$, the number $q - (q^2-3)/(2q)=\frac{1}{2}q+\frac{3}{2q}$ is a rational smaller than $q$ but still larger than $\sqrt{3}$.
