[Math] How to prove the density of irrational numbers in $\mathbb{R}$ without proving density of rationals first

Question

I am asked to prove the density of irrationals in $\mathbb{R}$. I understand how to do this by proving the density of $\mathbb{Q}$ first, namely, adding a known irrational number such as $\sqrt{2}$ to $x,y \in \mathbb{R}$ ($x<y$), then there exists $r_0 \in \mathbb{Q}$ such that $x+\sqrt{2}<r_0<y+\sqrt{2}$, then substract $\sqrt{2}$ from all sides of the inequality to yield $x<r_0-\sqrt{2}<y$, and $r_0-\sqrt{2}$ is irrational ($r_0$ is rational).
However, my professor has said that I can prove density of $\mathbb{R} \setminus \mathbb{Q}$ without even using the density of $\mathbb{Q}$ and it is a simple proof. I have puzzled over this for quite some time. I appreciate any help provided on this question in advance.

Answer 1

Fix $x\in \mathbb{R}$.

Case 1: If $x\in\mathbb{Q}$, then the irrational sequence $x+\sqrt{2}/n$ converges to $x$.

Case 2: If $x\notin\mathbb{Q}$, then the irrational sequence $x+1/n$ converges to $x$.