🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

Suppose for the sake of contradiction that $\sqrt{n}$ was rational so that $\sqrt{n}=\frac{p}{q}$ therefore $q^{2}n=p^2$, since $n$ is not a perfect square then we know that $n={\prod }_{i\in {\mathbb{N}}_1}{p}_i^{{\alpha }_i}$ where there is some $j\in {\mathbb{N}}_1$ such that ${\alpha }_j$ is odd, now we also know that $q^2$ is equal to ${\prod }_{i\in {\mathbb{N}}_1}{p}_i^{2{\beta }_i}$ since odd plus even is still odd then we know that in the prime factorization of $q^{2}n$ the power on the $j$-th prime is still odd, thus $q^{2}n$ is not a perfect square, but on the other hand $p^2$ is, which is a contradiction, therefore $\sqrt{n}$ must be irrational.

Since $x<y$ we have $0<y-x$ therefore by the archimediean property we have an $n\in {\mathbb{N}}_1$ such that $n(y-x)>1$

Consider the set $X=\{z\in \mathbb{Z}:z>nx\}$, this set is not empty because $nx+1\in X$, additionally it is bounded below, and so as a subset of $\mathbb{Z}$ it has a least element say $m$ such that $m>nx$ . Since it is the least element we know that $m-1\le nx$, which implies that $m\le nx+1$.

Recall that our $n$ satisfies $n(y-x)>1$ therefore $m\le nx+1\le nx+\left(n(y-x)\right)=ny$, thus we have $nx<m<ny$ and since $n\ne 0$ we have $x<\frac{m}{n}<y$ as needed.

If $x<y$ then $\sqrt{2}+x<\sqrt{2}+y$, and thus there is a rational number $r\in \mathbb{Q}$ such that $${\displaystyle \sqrt{2}+x<r<\sqrt{2}+y}$$ and therefore $x<r-\sqrt{2}<y$, but $r-\sqrt{2}$ is irrational, because the sum of a rational and irrational is irrational, then we are done.