# Proving irrationals are dense in (0, 1)

The definition of dense I'm using is the following: a subset $S$ of $(0, 1)$ is dense in $(0, 1)$ if for every $x \in (0, 1)$ and every $\epsilon > 0$ there exists $s \in S$ with $|s - x| < \epsilon$.

I'm trying to show that the set of irrationals in $(0, 1)$ (i.e. the set $\{x \in \mathbb{R} \setminus \mathbb{Q} : 0 < x< 1\}$ is dense in $(0, 1)$.

My idea is this: given $x \in (0, 1)$ and $N \in \mathbb{N}$, try to find an irrational number $\alpha \in (0, 1)$ that depends on $N$ and $x$ such that $|\alpha - x| < 1/N$. I'm not sure how to construct such an $\alpha$ though. I'm thinking I need to to take some irrational multiple of $\lfloor{Nx\rfloor}$ or something but I can't quite work it out. Can anyone help?

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If it wasn't the case there would exist a whole open interval $(a,b)$ without irrationals. – Martino Oct 10 '12 at 18:57

Take any irrational number $\alpha$; now given $x$ then for every $\varepsilon$ take a rational number $q$ such that $\alpha\cdot q<\varepsilon$ and consider $s=x+\alpha\cdot q$.
(Such $q$ can be always taken $\frac1n$ for a large enough $n$)