Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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
add comment

3 Answers 3

up vote 4 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.

share|improve this answer
    
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
add comment

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 $a$?

share|improve this answer
1  
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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.