3. The field of real numbers

1. Multiplication in $\mathbf{R}$

The topology of the rational line $\mathbf{Q}$ is compatible not only with the additive group structure, but also with the field structure of $\mathbf{Q}$. For the function $xy$ is continuous at $(0, 0) \in \mathbf{Q} \times \mathbf{Q}$, since for each integer $n > 0$ the relations $|x| \leq 1/n$ and $|y| \leq 1/n$ together imply that $|xy| \leq 1/n^2 \leq 1/n$; on the other hand, if $a$ is any non-zero rational number, the function $ax$ is continuous at $x = 0$, since for each integer $n > 0$ the relation $|x| \leq 1/|n|a|$ implies that $|ax| \leq 1/n$. This shows that $xy$ is continuous at every point of $\mathbf{Q} \times \mathbf{Q}$ (Chapter III, § 6, no. 3).

To show that $1/x$ is continuous on $\mathbf{Q}^*$ we shall establish more precisely that $1/x$ is uniformly continuous (with respect to the additive structure) in the complement of any neighbourhood V of 0. Namely we have

The image, under the function $1/x$, of any Cauchy filter on $\mathbf{Q}^*$ (with respect to the additive uniformity) which does not have 0 as a cluster point, is a Cauchy filter (with respect to the additive uniformity). Hence (Chapter III, , no. 8, Proposition 7)

2. The multiplicative group $\mathbf{R}^*$

We know from Chapter III, § 6, no. 7, that the topology induced on $\mathbf{R}^*$ by the topology of the real line is compatible with the multiplicative group structure of $\mathbf{R}^*$; since $\mathbf{R}^*$ is an open subset of the locally compact space $\mathbf{R}$, it follows that $\mathbf{R}^*$ is a locally compact topological group (Chapter I, § 9, no. 7, Proposition 13) and is therefore complete (Chapter III, § 3, no. 3, Corollary 1 to Proposition 4; this follows also from Chapter III, § 6, no. 8, Proposition 8); of course, this latter property relates to the multiplicative uniformity on $\mathbf{R}^*$ and not to the uniformity induced on $\mathbf{R}^*$ by the additive uniformity of $\mathbf{R}$.

The function $xy$ maps the set $\mathbf{Q}_+ \times \mathbf{Q}_+$ into $\mathbf{Q}_+$, and therefore it maps $\mathbf{R}_+ \times \mathbf{R}_+$ into $\mathbf{R}_+$ (Chapter I, § 2, no. 1, Theorem 1); in other words, the product of two real numbers $\geqslant 0$ is $\geqslant 0$. The formulae $(-x)y = -xy$ and $(-x)(-y) = xy$ then show that the product of a number $\geqslant 0$ and a number $\leqslant 0$ is $\leqslant 0$, and that the product of two numbers $\leqslant 0$ is $\geqslant 0$; from this it follows that

(which might also have been obtained by extension of the corresponding relation on $\mathbf{Q} \times \mathbf{Q}$).

If $x > 0$ and $y > 0$ we have $xy \neq 0$, and therefore $xy > 0$; likewise, if $x < 0$ and $y > 0$, then $xy < 0$; and if $x < 0$ and $y < 0$, then $xy > 0$. In particular, if $x \neq 0$ we have $x^2 > 0$, so that a sum of squares of real numbers cannot be zero unless each of the numbers is zero.

If $x > 0$ and $y \leqslant z$ (resp. $y < z$) we have $xy \leqslant xz$ (resp. $xy < xz$) in other words, a homothety of ratio $> 0$ preserves order on $\mathbf{R}$. Since $(-x)y = -xy$, a homothety of ratio $< 0$ changes the ordering on $\mathbf{R}$ into the opposite ordering.

If $x > 0$ we have $1/x > 0$, since $x.(1/x) = 1 > 0$. If $0 < x < y$ we have $xy > 0$, hence $x.(1/xy) < y.(1/xy)$, that is $1/y < 1/x$. Hence the mapping $x \to 1/x$ of the set $\mathbf{R}_+^*$ of real numbers $> 0$ onto itself is strictly decreasing.

We see in the same way that the function $1/x$ is strictly decreasing in ]$\leftarrow, 0[$, and therefore the function $\frac{1}{x-a}$ is strictly decreasing in each of the intervals ]$\leftarrow, a[$ and $]a, \to[$.

It follows from what precedes that $\mathbf{R}_+^*$ is a subgroup of the multiplicative group $\mathbf{R}^*$; moreover, the order relation $x \leq y$ is compatible with the multiplicative, group structure of $\mathbf{R}_+^*$; in other words, $\mathbf{R}_+^*$ is a linearly ordered group.

The fact that the product of two real numbers $\geq 0$ is $\geq 0$ can be expressed by saying that $\mathbf{R}$ is an ordered field: all the above properties are common to all ordered fields.

3. Nth roots

Let $n$ be any integer $> 0$. From the relation $0 < x < y$ we deduce, by induction on $n$, that $0 < x^n < y^n$. In other words, the function $x \to x^n$ is strictly increasing for $x \geq 0$; it is clearly continuous at every point and therefore (§ 2, no. 6, Theorem 5) it is a homeomorphism of $\mathbf{R}_+$ onto an interval $I$. On the other hand, since $x \geqslant 1$ implies $x^{n-1} \geqslant 1$ and therefore $x^n \geqslant x$, it follows that $I$ is not bounded and hence $I = \mathbf{R}_+$. The value, for $x \geqslant 0$, of the inverse of the mapping $x \to x^n$ is denoted by $x^{1/n}$ or $\sqrt[n]{x}$ and is called $x$ to the power $1/n$ or the nth root of $x$ (for $n = 2, 3$ we say square root, cube root; for $n = 2$ we write $\sqrt{x}$ in place of $\sqrt[2]{x}$). The positive number $x^{1/n}$ is thus defined as the unique positive solution of the equation

In particular we see that there is a real number $x$ such that $x^2 = 2$, whereas no rational number has this property; thus we recover the fact that the rational line $\mathbf{Q}$ is not a complete space.

The mapping $x \to x^{1/n}$ of $\mathbf{R}_+$ onto itself is strictly increasing and continuous. By (2) we have $0^{1/n} = 0, 1^{1/n} = 1$, also

hence $x \to x^{1/n}$ is an automorphism of the topological group $\mathbf{R}_+^*$.

In Chapter V, § 4, no. 1, we shall generalize this result by finding all the automorphisms of the multiplicative group $\mathbf{R}_+^*$.

Exercises

See the exercises for § 3.