For every real number $a$ there exists a sequence $r_n$ of rational numbers such that $r_n$ approaches $a$. How to prove that for every real number $a$ there exists a sequence $r_n$ of rational numbers such that $r_n \rightarrow a$.
 A: I'm surprised that no one's talked about decimal notation yet, but here's an informal proof. (For familiarity, we'll use the base-10 system.)
If $x$ is rational, just use the sequence $\left(r_n\right)=\left(x, x, x, x, \dots\right)\to x$. If it's irrational, we'll have to do some work.
Represent $x$ using decimal notation. It will be a non-repeating, non-terminating decimal. For example, let
$$x = \sqrt{2} = 1.414213562...$$
Then just make every term of your sequence a terminating decimal, a more refined approximation of $\sqrt{2}$.
$$\left(r_n\right)=\left(1, 1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414213, \dots\right)$$
Obviously, since every term in the sequence is a terminating decimal, it's a rational number. By the Cauchy criterion, the rational sequence converges, and it converges to the real number $\sqrt{2}$.
It's not a formal proof, but you can see how it works.
A: Here is another informal proof just for diversity.
Consider the the interval $ ]r_0,a[$ where $ a\in \mathbb R $ , $r \in  \mathbb Q$ and r$\neq a$,
Using dichotomy lets construct another interval $]r + \frac{|a-r|}{2},a[$ and just choose any rational in that interval and call it $r_1$
We continue in this fashion choosing $r_1,r_2 \cdots$ etc thus we have a sequence of rationals $r_n$ that is strictly increasing and is bounded above by $a$ which also happens to be the upper bound , thus it converges to $a$

$\underline{Proposition}$
Let $u_n$ be a stricly inreasing sequence from $\mathbb N \to \mathbb R$ and $A$ a non empty bounded set consiting of elements of $u_n$, the sequence $u_n$ converges to the upper bound of A.

Proof:
A is a non empty bounded set thus it has an upperbound denoted $supA$.
$u_n$ is strictly increaing and bounded thus it converges to a limit we denote by $l$ thus $\forall \epsilon > 0  $ $\exists N \in \mathbb N , \forall n \in \mathbb N : n \ge N$ $\implies$ $|u_n - l| \le \epsilon $
thus $l-\epsilon \le u_n \le l + \epsilon $ but $supA$ is the upperbound of $A$ thus $supA \le l+\epsilon$
$$\implies supA = l $$
A: Without loss of generality we may assume the real number $a$ is $\gt 0$. (If $a \lt 0$, we can apply the argument below to $|a|$ and then switch signs.)
We sketch a fairly formal proof, based on the fact that the reals are a complete ordered field. In one of the remarks at the end, we give an easy informal but incomplete "proof." 
Let $n$ be  a natural number.  Let $m=m(n)$ be the largest positive integer such that $\frac{m}{n}\lt a$. Then $\frac{m+1}{n}\ge a$, and therefore $|a-m/n|\lt 1/n$.
Let $r_n=m/n$. It is easy to show from the definition of limit that the sequence $(r_n)$ has limit $a$.
Remarks: $1.$ One really requires proof that there is a positive integer $m$ such that $\frac{m}{n}\ge a$. It is enough to show that there is a positive integer $k$ such that $k \ge a$, for then we can take $m=kn$. 
The fact that there is always an integer $\gt a$ is called the Archimedean property of the reals. We proceed to prove that the reals do have this property. 
Suppose to the contrary that all positive integers are $\lt a$. Then the set $\mathbb{N}$ of positive integers is bounded,  so has a least upper bound $b$. That means that for any $\epsilon \gt 0$ there is an integer $k$ such that $0\lt k\lt b$ and $b-k\lt \epsilon$. Pick $\epsilon=1/2$.  Then $k+1\gt b$, contradicting the assumption that $b$ is an upper bound for $\mathbb{N}$.
$2.$ One can also give a very quick but not fully persuasive "proof" of the approximation result.  Assume as before that $a\gt 0$. The numbers obtained by truncating the decimal expansion of $a$ at the $n$-th place are rational, and clearly have limit $a$. The problem is that we are then assuming that every real number has a decimal expansion. 
A: We can prove it easily by the density property of the rationals, where the later can be proved by Archimedean property in the case that you assume this is equivalent to the density property.
Let $a$ be a real number, for any $n \in\mathbb N$, $a - \frac 1n < a + \frac 1n$, so there is a rational $r_n$ that lies between them, i.e $a-\frac 1n < r_n < a+\frac 1n$, which implies that $|r_n - a|< \frac 1n$, so $r_n$ converges to $a$.
A: By Riemann's series theorem it follows, that for every $x \in \mathbb{R}$ there is an arrangement $\sigma:\mathbb{N} \to \mathbb{N}$ such that $\sum_{n=1}^\infty \frac{(-1)^{\sigma(n+1)}}{\sigma(n)} = x$. Note that $b_n := \sum_{k=1}^n \frac{(-1)^{\sigma(n+1)}}{\sigma(n)}$ is a rational sequence.
