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please how to prove these two properties:

$(1)\quad\forall x\in\mathbb{R},\ \forall \varepsilon>0,\ \exists r\in\mathbb{Q},\ |x-r|\leq\varepsilon$

$(2)\quad \forall x\in\mathbb{R},\ \exists (r_n)\in\mathbb{Q},\ \displaystyle\lim_{n\to\infty} r_n=x$

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    $\begingroup$ It refers to density. $\endgroup$ – Wuestenfux Jan 11 at 9:37
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    $\begingroup$ What are your definitions of $\Bbb Q$ and of $\Bbb R$? Because the details of the proof will depend on exactly how these two sets are related to one another. $\endgroup$ – Arthur Jan 11 at 9:37
  • $\begingroup$ The second property is how my professor in mathematics defined $\mathbb{R}$. And the first property is a consequence of the second one $\endgroup$ – Damien Jan 11 at 10:27
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We can show $(1)$ as follows. Let $N > \frac{1}{\varepsilon}$ be an integer. Then we observe that $$Nx - \left \lfloor{Nx}\right \rfloor < 1 \Leftrightarrow x - \frac{\left \lfloor{Nx}\right \rfloor}{N} < \frac{1}{N} < \varepsilon$$ So taking $r = \frac{\left \lfloor{Nx}\right \rfloor}{N}$, we are done.

$(2)$ follows from $(1)$ by considering a sequence $\varepsilon_1,\varepsilon_2,...$ with $\varepsilon_n > 0$ and $\lim_{n \rightarrow \infty}\varepsilon_n = 0$ (for instance $\varepsilon_n = \frac{1}{n}$). Then we can find a sequence $r_n \in \mathbb{Q}$ such that $|x - r_n| < \varepsilon_n$, from which $\lim_{n \rightarrow \infty} r_n = x$ follows since $\forall \varepsilon > 0$ there is an $N$ such that $n > N$ implies $$|x - r_n| < \varepsilon_n < \varepsilon$$ as required.

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