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How to prove that for every real number $a$ there exists a sequence $r_n$ of rational numbers such that $r_n \rightarrow a$.

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The answer depends greatly on how you define your real numbers. If it is the completion of the rationals then this is trivial. – Ragib Zaman Oct 8 '12 at 0:59
up vote 11 down vote accepted

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.

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Why is it problematic to assume that real numbers have decimal expansion? – leo Oct 8 '12 at 2:04
@leo: To prove that the real numbers have certain properties, we have to start with a formal definition of the reals. Then we can prove representability by an infinite series $a_0+\frac{a_1}{10}+\frac{a_2}{100}+\cdots$ where $a_0$ is an integer and the rest of the $a_i$ are digits, that is, integers between $0$ and $9$. But before a course in "analysis" one takes these facts for granted, and then for example $3, 3.1,3.14,3.141,3.1415, 3.14159,\dots$ is a sequence of rationals with limit $\pi$. To answer your question, one has to know at what level the question is being asked. – André Nicolas Oct 8 '12 at 2:13
You don't need to assume $a>0$ if you allow $m$ to be negative. This needs the fact that any non-empty upper-bounded set of integers has a maximum element. I used this exact argument in part of a formal proof that $|\Bbb R|\le|{}^{\Bbb N}\Bbb Q|=2^{\aleph_0}$. – Mario Carneiro 2 days ago

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.

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Certainly not a formal proof -- it's borderline circular! That any real number can be approximated by a decimal requires proof. But you're absolutely right that one can find a rational number with a fixed power denominator (i.e., a number of the form p/q^n for some fixed q) to approximate any real number. PS -- according to the answer right above yours, people did talk about decimals in October. – user54535 Jan 6 '13 at 4:20

Here is another informal proof just for diversity.

Consider the the interval $ ]a,r[$ where $ a\in \mathbb R $ , $r \in \mathbb Q$ and r$\neq a$,

Using dichotomy lets construct another interval $]a,\frac{a+r}{2}[$ and denote $\phi_1 = \frac{a+r}{2}$

Thus lets define a sequence $(\phi_n)_{n\in \mathbb N} = \frac{(2^{n}-1)a + r}{2^n} for, n\ge 1$

It is clear that this sequence converges to $a$ and $\forall n\in\mathbb N$ we have $\phi_n\in\mathbb Q$ , a sequence of rational numbers converging to a real number. Lets notice that this is also valid to show that a every real number is a limit of a sequence of irrationals.

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Hint: The rational numbers are dense in the reals. If there is a real number $a$ for which there lacks such a sequence, what can we say about the neighborhood of $a$?

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This statement is equivalent to the statement that the rationals are dense in the reals. – Qiaochu Yuan Oct 8 '12 at 0:34
The point of the exercise is to see one such sequence. Isn't? – leo Oct 8 '12 at 2:03

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