Proving that the supremum is irrational The same question was asked before and it was answered clearly. The question is as follows :

Let $A$ be the set of rational solutions of $0<x^2<2$. 
  Prove that $A$ has no least upper bound belongs to $Q$.

I understood the solution written here before after reading the denseness of rationals. So I am asking for another method(if it exists) since I was given this as an homework without knowing about the denseness of rationals. I tried to prove it by contradiction (supposing that $\sup(A)$ is rational), but I reached something that is hard to progress after. Any help? 
 A: The density of rationals is the prorper way to do this. Without it you have to somehow incorporate it into the proof anyway.
We asume there is a lower upper bound $l=\frac{m}{n}\in\mathbb Q$. Then we know that $l^2\neq2$. We need a way to reach contradiction in the cases $l^2<2$ and $l^2>2$. It suffices to show a special form of density.

If $x<y$ then there is a rational $r$ such that $x<r^2<y$.

Te worst case scenario is that for any $n$ there is $m=m(n)$ such that $\frac{m^2}{n^2}\leq x$ and $\frac{(m+1)^2}{n^2}\geq y$. We fix a $n$ and we have
$$(m+1)^2\geq n^2y\geq m^2\frac{y}{x} \Rightarrow$$
$$\left(1-\frac{y}{x}\right)m^2+2m+1\geq 0 \Rightarrow$$
$$\left(\frac{y-x}{x}\right)m^2-2m-1\leq 0$$
Notice that the last inequality does not depend on $n$. Solving this iniquality shows that $m$ can only take values in a finite interval of the real number line. In other words $m$ is bounded.
However we also have $(m+1)^2\geq n^2y$ for any $n$ (remember $m(n)$ depends on $n$). This means that the sequence of $m$'s must be non bounded. This is a contradiction so we prooved the special form of density of rationals.
A: Let such a least upper bound exists, namely $l$. If $l<\sqrt 2$ there are rational numbers $l<y<\sqrt 2$ for which $$l^2<y^2<2$$which is contradiction. If $l>\sqrt2$ there is another upper bound like $d$ that $$\sqrt 2<d<l$$which contradicts with the assumption of $l$ being least upper bound. Also $l\ne\sqrt 2$ because $\sqrt 2$ is not a rational number, therefore such an $l$ does not exist and the proof is complete.
