Proof that all real numbers have a rational Cauchy sequence? I saw in an article that for every real number, there exists a Cauchy sequence of rational numbers that converges to that real number. This was stated without proof, so I'm guessing it is a well-known theorem in analysis, but I have never seen this proof. So could someone give a proof (preferably with a source) for this fact? It can use other theorems from analysis, as long as they aren't too obscure.
 A: Let $\alpha$ be any real number. 
Define 
$$a_n = \frac{\lfloor n \alpha \rfloor}{n}$$
where $\lfloor , \rfloor $ denotes the floor function.
Then $a_n \in \mathbb Q$. Moreover we have 
$$\frac{n \alpha -1}{n} \leq a_n \leq \alpha$$
and hence $a_n \to \alpha$.
A: Based on the context gathered from your question and your comments it is clear that you are concerned about the equivalence of construction of reals as Dedekind cuts of rationals and as equivalence classes of Cauchy sequences of rationals.
We can do this in the following manner. First we show that corresponding to a given Dedekind cut of rationals there exists a Cauchy sequence of rationals. To do so in proper manner it is best to explicitly state the definition of a Dedekind cut of rationals.
A Dedekind cut of rationals is a set $A\subseteq\mathbb{Q} $ with the following properties:


*

*$\emptyset\neq A\neq \mathbb{Q} $. 

*If $p\in A$ and $q\in\mathbb{Q} $ with $q<p$ then $q\in A$. 

*If $p\in A$ then there exists a $q\in A$ with $p<q$.


The above definition is one provided by Rudin in his Principles of Mathematical Analysis. But the true picture of a Dedekind cut is revealed when one considers not only the set $A$ but also its complement $B=\mathbb{Q} - A$ as done by Dedekind himself.
It is easy to prove that given any positive rational number $\epsilon$ one can find rationals $p, q$ such that $p\in A, q\in B$ such that $0<q-p<\epsilon $. Thus for every positive integer $n$ we can find rationals $a_n, b_n$ such that $a_n\in A, b_n\in B, 0<b_n-a_n<1/n$ and further we can ensure that $b_{n} \geq b_{n+1},a_{n}<a_{n+1}$. Now we can show that $\{a_n\} $ is a Cauchy sequence of rationals. This is how one can find a Cauchy sequence of rationals corresponding to a given Dedekind cut of rationals. Note that the sequence $b_n$ can also be considered here instead of $a_n$. 
To do the reverse let us consider a Cauchy sequence of rationals $\{a_n\} $ and we then need to find a Dedekind cut corresponding to it. Consider a rational number $r$. If there is a positive integer $m$ such that $a_n>r$ for all $n\geq m$ then we say that that $r$ is inferior to the sequence $\{a_n\} $. Consider the set $$A=\{p\mid p\in\mathbb {Q}, \exists q\in\mathbb{Q}, q>p, q\text{ is inferior to sequence }\{a_n\} \} $$ Since a Cauchy sequence is bounded it is clear that the set $A$ and its complement $B=\mathbb{Q} - A$ are both non-empty. Also if $p\in A$ then there is some $r\in\mathbb{Q} $ such that $r>p$ and $r$ is inferior to $\{a_n\} $. And if we take a rational $q$ with $q<p$ then $q<r$ and hence $q\in A$. Moreover we also have $(p+r) /2<r$ so that $p<(p+r) /2\in A$. Thus $A$ satisfies all the properties of a Dedekind cut. This is how one finds a Dedekind cut of rationals corresponding to a given Cauchy sequence of rationals.
We still need to show that if $\{b_n\} $ is another Cauchy sequence of rationals which is equivalent to $\{a_n\} $ (ie $a_n-b_n\to 0$) and if we apply the above mentioned procedure to find a Dedekind cut $A'$ corresponding to the sequence $\{b_n\} $ then $A'=A$. This is not difficult to prove and you should try it yourself.

The full equivalence of the construction of reals via Dedekind cuts and via Cauchy sequences of rationals can be done by proving that the correspondence between these approaches described above is an isomorphism which preserves arithmetic as well as order relationship. But carrying out such a proof is boring and lengthy. 
A: Note that  the set of real numbers is a complete set that is every Cauchy sequence of real numbers is convergent also every convergent sequence is a Cauchy sequence. Therefore once you have a rational sequence converging to a real number the same sequence is a  rational Cauchy sequence converging to the same real number.   
A: One way to define the real numbers is as equivalence classes of Cauchy sequences (see exercise 3.24 in Princples of Mathematical Analysis by Walter Rudin for the details of the (actually, a more general) construction). Under this construction, it is obvious that every real number has such a Cauchy sequence, since every real number literally is a (set of) Cauchy sequence(s). Once you've done this, it is an easy exercise to prove that $\mathbb{R}$ is the only complete ordered field up to isomorphism, so it doesn't matter which construction you use for $\mathbb{R}$, the Cauchy construction guarantees it.
A: Let $x \in \mathbb R$.  
Let $q_i \in \mathbb Q$ so that $x - \frac 1i < q_i < x$.
Claim 1: Such a sequence $\{q_i\}$ exists (because the rationals are dense in the reals)
Claim 2: $q_i \to x$ (by the very definition of limits)
Claim 3: $\{q_i\}$ is cauchy.  (For $n > \frac 2\epsilon; i,j > n$ then $|q_i - x| < \frac 1n < \frac \epsilon 2$ and $|q_i - q_j| < |q_i - x| + |q_j - x| < \epsilon$.)
Basically this is the FUNDAMENTAL principal of the meaning of real numbers.
Real numbers are such that every bounded above sequence has a least upper bound.  That is the fundamental claim of the nature of real numbers.  The means that any sequence in which all the terms get "close together" must have an existent real limit.
A Cauchy sequence is just such a sequence.  To be informal, a Cauchy sequence is a sequence that "ought to" converge although the actual limit point need need be specified (or even be specifiable).
Another way, perhaps without choice is, by the Archmedian principal there exists a unique integer $k_1$ so that $k_1 \le x < k_1 + 1$ and for every integer $n$ there is a unique $k_n$ so that $k_n \le x*n < k_n + 1$ or $\frac {k_n}n \le x < \frac {k_n}n+ \frac 1n$.  Let $q_n = \frac {k_n}n$.  That is a cauchy sequence converging to $x$.
In fact, if $k_n$ is the unique integer so that $k_n \le x*10^n \le k_n +1$ then the sequence $q_n = \frac {k_n}{10^n}$ is nothing more or less then the sequence of $n$th place decimal expansions.  (If $x = \pi$ then $\{q_n\} = \{3, \frac {31}{10}, \frac {314}{100}, \frac {3141}{1000}, ....\}$ etc.)
It is precisely because all reals have cauchy sequences of rationals converging to them, that the concept of decimal numbers even works.
