For every $x\in\mathbb R$ and $\varepsilon$ > 0 , there exist $\,q,q'\in\mathbb Q$, such that $q<x<q'$ and $\left |q-q' \right |< \varepsilon$

Question

I'm asked to prove that for every $\varepsilon$ > 0 , there exists two rational numbers $q$ and $q'$ such that $q<x<q'$ and $\left |q-q' \right |<\varepsilon$ where $x$ is a real number.

I have to confess that I've quite a hard time grasping the "concept of $\varepsilon$", some proofs where it is used are still a bit obscur to me. Anyway, here is my try :

Let $w,x,y \in \mathbb R$ such that $w<x<y$ and we define $\varepsilon := y-w$.

I proved earlier in class that it always exists a rational number between two real numbers, i.e. we have $w<q<x<q'<y$ with $q,q' \in \mathbb Q$

Hence, for all $x \in \mathbb R$ we have $q<x<q'$ with $\left |q-q' \right |<\varepsilon$.

This proof seems a bit too "easy", and I guess there is something wrong with it. Can I actually define $\varepsilon$ as I wish ?

Thanks for your help

Answer 1

Let $m$ be a positive integer such that $$ \frac{2}{\varepsilon}<m. $$ Then $$ \frac{\lfloor mx\rfloor}{m} \leqslant x<\frac{\lfloor mx\rfloor+1}{m} $$ hence $$ \frac{\lfloor mx\rfloor-1}{m} < x<\frac{\lfloor mx\rfloor+1}{m} $$ and $$ 0<\frac{\lfloor mx\rfloor+1}{m}-\frac{\lfloor mx\rfloor-1}{m}=\frac{2}{m}<\varepsilon. $$