8. Usual expansions of real numbers; the power of r

1. Approximations to a real number

Let $A$ be a dense subset of $\mathbf{R}$. For each $x \in \mathbf{R}$ and each $\varepsilon > 0$ there is an approximation by defect (resp. excess) to $x$ to within $\varepsilon$ belonging to $A$, since the interval $]x - \varepsilon, x[$ (resp. $]x, x + \varepsilon[$) contains at least one point of $A$. If we now consider a given strictly decreasing sequence $(\varepsilon_n)$ of numbers $> 0$, tending to 0, and if $r_n$ is an approximation to $x$ to within $\varepsilon_n$, then the sequence $(r_n)$ has $x$ as limit as $n$ tends to infinity.

In the case where $A$ is a subgroup of the additive group $\mathbf{R}$, and we restrict the $\varepsilon_n$ to belong to $A$, we can define canonically for each $x \in \mathbf{R}$ a sequence $(r_n)$ of approximations to $x$ by defect, belonging to $A$.

For by Archimedes' axiom (§ 2, no. 1, Theorem 1), the set of integers $p$ such that $p \varepsilon_n \leq x$ has a greatest element $p_n$; in other words, there is a unique integer $p_n$ such that

Since $|x - p_n \varepsilon_n| \leq \varepsilon_n$, it follows that $p_n \varepsilon_n$ is an approximation to $x$ to within $\varepsilon_n$ by defect, and belongs to $A$ by hypothesis; similarly $(p_n + 1) \varepsilon_n$ is an approximation to $x$ to within $\varepsilon_n$ by excess, belonging to $A$, and the two sequences $(p_n \varepsilon_n)$ and $((p_n + 1) \varepsilon_n)$ have $x$ as their limit.

2. Expansions of real numbers relative to a base sequence

We shall limit ourselves to studying the case where $\varepsilon_n = 1/d_n$, where $(d_n)$ is a strictly increasing sequence of integers such that $d_0 = 1$ and $d_n$ is a multiple of $d_{n-1}$ for $n \geq 1$. Let $a_n = d_n / d_{n-1}$ ($n \geq 1$): $a_n$ is an integer $> 1$. In this case, the sequence of approximations by defect $r_n = p_n / d_n$ is increasing, for $p_n$ is the largest integer such that $p_n / d_n \leq x$; but we have

so that $a_n p_{n-1} \leq p_n < a_n p_{n-1} + a_n$, and therefore $r_{n-1} \leq r_n \leq x$. Let

3. Definition of a real number by means of its expansion

Conversely, suppose we are given an integer $q_0$ and a sequence $(v_n)$ ($n \geq 1$) of integers such that $0 \leq v_n \leq a_n - 1$; we ask whether there is a number $x$ whose expansion (4) is such that $p_0 = q_0$ and $u_n = v_n$ for all $n$. If such a number exists it is unique, because it is equal to

Conversely, let $x$ be a rational number which can be written in the form of a fraction with denominator $d_n$ for some value of $n$. Let $m$ be the smallest integer such that $x$ is of the form $k/d_m$ ($k$ an integer); we have $r_n < x$ for $n < m$, and $r_m = x$, and therefore the expansion of $x$ is of the form (5), and $x$ has an improper expansion given by (6); moreover this improper expansion is unique.

A rational number, written in its irreducible form $p/q$, is equal to a fraction with denominator $d_n$ if and only if $q$ divides $d_n$ (the number $m$ will therefore be the smallest integer $n$ such that $q$ divides $d_n$). It can happen that every rational number has this property (for a suitably chosen $n$) : this will be the case if and only if every integer $> 0$ divides some $d_n$, e.g., if $d_n = n!$ If the $d_n$ have this property, then a number is rational if and only if its expansion, relative to the sequence $(d_n)$, terminates.

To summarize: to every sequence $s$ whose initial term $q_0$ is an arbitrary integer, and whose term $v_n (n \geq 1)$ is such that $0 \leq v_n \leq a_n - 1$, corresponds a real number equal to $q_0 + \sum_{n=1}^{\infty} \frac{v_n}{d_n}$; if $I_n$ denotes the interval $[0, a_n - 1]$ of $\mathbf{N}$, we thus define a mapping $\varphi$ of $X = \mathbf{Z} \times \prod_{n=1}^{\infty} I_n$ onto the real line $\mathbf{R}$; moreover the equation $\varphi(s) = x$, where $x \in \mathbf{R}$ is given, has one solution if $x$ is not a fraction with denominator $d_n$ (for some $n$), and two solutions otherwise.

4. Comparison of expansions

If we know the expansions of two distinct real numbers $x$ and $y$, we can determine whether $x < y$ or $x > y$.

Let $x = p_0 + \sum_{n=1}^{\infty} u_n / d_n$, $y = q_0 + \sum_{n=1}^{\infty} v_n / d_n$ be the expansions of $x$ and $y$. If $p_0 < q_0$ then $x < y$, since

More generally, suppose that $p_0 = q_0$ and $u_n = v_n$ for $1 \leq n \leq m$, but that $u_m < v_m$; if

then $r_n = s_n$ for $n < m$, and since $u_m + 1 \leq v_m$, $r_m + \frac{1}{d_m} \leq s_m$; but $r_m \leq x < r_m + \frac{1}{d_m} \leq s_m \leq y$, hence again we have $x < y$. In other words, the order of $x$ and $y$ is the same as the order of the first two distinct terms of their respective expansions.

It follows that, if $p_0 = q_0$ and $u_n = v_n$ for $n < m$, then the first $m$ terms of the expansion of every number $z$ belonging to the closed interval with end-points $x$ and $y$ are the same as those of the expansions of $x$ and $y$.

We remark also that, in this case, we have $|y - x| \leq \frac{1}{d_{m-1}}$. If we endow $\mathbf{Z}$ and the intervals $I_n$ with the discrete topology, it follows that the mapping $\varphi$ defined above is continuous on the product space $X$.

5. Expansions to base a

The most important base sequences are those for which $d_n = a^n$, where $a$ is an integer $> 1$; $a$ is then said to be the base number (or simply the base) of the corresponding expansions. For numerical calculations, expansions to base 10, which are called decimal expansions, are used; also expansions to base 2 (dyadic expansions) and base 3 (triadic expansions) are often used.

To represent the approximations by defect $r_n$ to a number $x \geq 0$, in its expansion to base $a$, the following symbolism is employed: each integer $u$ such that $0 \leq u \leq a - 1$ is denoted by a particular sign; if

we first write down, with the aid of these signs, the representation to base $a$ of the integer $p_0 = [x] \geq 0$ (Set Theory, Chapter III, § 5, no. 7), then we put a point ("decimal point") and we write successively the signs representing the numbers $u_1, u_2, \ldots, u_n$. If S is the symbol thus obtained, it is customary to write $x = S...$, by abuse of language. It should be understood once and for all that such a relation is only an abbreviated method of indicating that the right-hand side is the approximation to $x$ to within $1/a^n$ by defect.

For negative numbers the established usage is different: we write an approximation to $x' = -x > 0$ in the symbolism described above, and precede it by the sign “—”; it is thus an approximation to $x$ to within $1/a^n$ by excess that is so denoted.

This usage has its inconveniences for numerical calculations. In the notation for negative logarithms the same symbolism is adopted as for positive numbers, by putting a bar over the integral part, to indicate that it is equal to the negative of the number written.

6. The power of r

We have $\mathbf{R} = \bigcup_{n \in \mathbf{Z}} [n, n+1[$, and all the intervals $[n, n+1[$ are equipotent to $[0, 1[$. Since $[0, 1[$ is an infinite set it follows that $\mathbf{R}$ is equipotent to the interval $[0, 1[$. By considering the dyadic expansion of the numbers of the interval $[0, 1[$ we shall show that this interval is equipotent to the set S of all sequences $(u_n)$ in which each term is equal to 0 or 1.

First, it is equipotent to the subset $S'$ of S consisting of sequences $(u_n)$ such that $u_n = 0$ for an infinity of values of $n$. On the other hand, the complement $S''$ of $S'$ in S is equipotent to the set of all improper expansions of rational numbers which are equal to a fraction with denominator $2^n$; these numbers form a subset of $\mathbf{Q}$, hence the set of them is countable and therefore so is $S''$. Since $S'$ is infinite, it is equipotent to S, hence the result.

Now S is equipotent to $\mathfrak{P}(\mathbf{N})$; for we can define a bijection of $\mathfrak{P}(\mathbf{N})$ onto S by mapping each subset A of $\mathbf{N}$ to the sequence $(u_n)$ such that $u_n = 0$ if $n \in A$ and $u_n = 1$ if $n \notin CA$. We have thus proved:

Exercises

See the exercises for § 8.