Real Analysis Inequality Proof Involving Reals and Rationals $0 < |r - q| < \varepsilon$

Question

I'm having difficulties making progress in proving: $$\forall \varepsilon > 0, \ \exists q \in Q \text{ where } 0 < |r - q| < \varepsilon $$

To clarify, $r$ is a real number and $q$ is a rational number.

Is there some theorem I should be using? This exercise is presented in the same section/chapter as the Completeness Axiom (each nonempty set has a least upper bound or supremum), the Archimedean Property of Real Numbers ($ \exists n \in Z^{+}$ such that $na>b $ for positive real numbers $a$ and $b$), and a theorem stating there is a rational and irrational number between any two distinct real numbers.

I'm just not seeing the connection (if any at all). Any help in the right direction would be much appreciated. Thank you!

Answer 1

Choose some positive integer $n$ such that $n>\frac{1}{\varepsilon}$. Define $q$ as $\frac{[nr]+1}{n}$. Then, $|r-q|=\frac{1-\{nr\}}{n}\in (0;\varepsilon)$.

Answer 2

No need to any theorem. Let $r\notin\Bbb Q$ and $$x=10^k\cdot r$$for some value of $k\in \Bbb R$ then define $$n\triangleq\Big\lfloor10^k\cdot r\Big\rfloor$$which leads to $$n\le 10^k\cdot r<n+1$$or $${n\over 10^k}-{1\over 10^k}<r<{n+1\over 10^k}={n\over 10^k}+{1\over 10^k}$$finally $$0<|r-{n\over 10^k}|<{1\over 10^k}<\epsilon$$ and $q={n\over 10^k}$ when $k>-\log \epsilon$

For the case $r\in \Bbb Q$, define $q=r+{1\over 10^k}$ for large enough $k\in \Bbb N$.