Real Analysis – Sequence of Rational Numbers Approaching Any Real Number

Question

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

Answer 1

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.