15. The continuum

The aggregate of all real numbers, rational and irrational, is called the arithmetical continuum.

It is convenient to suppose that the straight line $\mathrm{\Lambda }$ of § 2 is composed of points corresponding to all the numbers of the arithmetical continuum, and of no others.¹ The points of the line, the aggregate of which may be said to constitute the linear continuum, then supply us with a convenient image of the arithmetical continuum.

We have considered in some detail the chief properties of a few classes of real numbers, such, for example, as rational numbers or quadratic surds. We add a few further examples to show how very special these particular classes of numbers are, and how, to put it roughly, they comprise only a minute fraction of the infinite variety of numbers which constitute the continuum.

(i) Let us consider a more complicated surd expression such as $$z=\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4–\sqrt{15}}.$$ Our argument for supposing that the expression for $z$ has a meaning might be as follows. We first show, as in § 12, that there is a number $y=\sqrt{15}$ such that $y^2=15$, and we can then, as in § 10, define the numbers $4+\sqrt{15}$, $4–\sqrt{15}$. Now consider the equation in $z_1$, $$z_1^3=4+\sqrt{15}.$$ The right-hand side of this equation is not rational: but exactly the same reasoning which leads us to suppose that there is a real number $x$ such that $x^3=2$ (or any other rational number) also leads us to the conclusion that there is a number $z_1$ such that $z_1^3=4+\sqrt{15}$. We thus define $z_1=\sqrt[3]{4+\sqrt{15}}$, and similarly we can define $z_2=\sqrt[3]{4–\sqrt{15}}$; and then, as in § 10, we define $z=z_1+z_2$.

Now it is easy to verify that $$z^3=3z+8.$$ And we might have given a direct proof of the existence of a unique number $z$ such that $z^3=3z+8$. It is easy to see that there cannot be two such numbers. For if $z_1^3=3z_1+8$ and $z_2^3=3z_2+8$, we find on subtracting and dividing by $z_1–z_2$ that $z_1^2+z_1{z}_2+z_2^2=3$. But if $z_1$ and $z_2$ are positive $z_1^3>8$, $z_2^3>8$ and therefore $z_1>2$, $z_2>2$, $z_1^2+z_1{z}_2+z_2^2>12$, and so the equation just found is impossible. And it is easy to see that neither $z_1$ nor $z_2$ can be negative. For if $z_1$ is negative and equal to $-\zeta$, $\zeta$ is positive and ${\zeta }^3–3\zeta +8=0$, or $3–{\zeta }^2=8/\zeta$. Hence $3–{\zeta }^2>0$, and so $\zeta <2$. But then $8/\zeta >4$, and so $8/\zeta$ cannot be equal to $3–{\zeta }^2$, which is less than $3$.

Hence there is at most one $z$ such that $z^3=3z+8$. And it cannot be rational. For any rational root of this equation must be integral and a factor of $8$ (Ex. II. 3), and it is easy to verify that no one of $1$, $2$, $4$, $8$ is a root.

Thus $z^3=3z+8$ has at most one root and that root, if it exists, is positive and not rational. We can now divide the positive rational numbers $x$ into two classes $L$, $R$ according as $x^3<3x+8$ or $x^3>3x+8$. It is easy to see that if $x^3>3x+8$ and $y$ is any number greater than $x$, then also $y^3>3y+8$. For suppose if possible $y^3\le 3y+8$. Then since $x^3>3x+8$ we obtain on subtracting $y^3–x^3<3(y–x)$, or $y^2+xy+x^2<3$, which is impossible; for $y$ is positive and $x>2$ (since $x^3>8$). Similarly we can show that if $x^3<3x+8$ and $y<x$ then also $y^3<3y+8$.

Finally, it is evident that the classes $L$ and $R$ both exist; and they form a section of the positive rational numbers or positive real number $z$ which satisfies the equation $z^3=3z+8$. The reader who knows how to solve cubic equations by Cardan’s method will be able to obtain the explicit expression of $z$ directly from the equation.

(ii) The direct argument applied above to the equation $x^3=3x+8$ could be applied (though the application would be a little more difficult) to the equation $$x^5=x+16,$$ and would lead us to the conclusion that a unique positive real number exists which satisfies this equation. In this case, however, it is not possible to obtain a simple explicit expression for $x$ composed of any combination of surds. It can in fact be proved (though the proof is difficult) that it is generally impossible to find such an expression for the root of an equation of higher degree than $4$. Thus, besides irrational numbers which can be expressed as pure or mixed quadratic or other surds, or combinations of such surds, there are others which are roots of algebraical equations but cannot be so expressed. It is only in very special cases that such expressions can be found.

(iii) But even when we have added to our list of irrational numbers roots of equations (such as $x^5=x+16$) which cannot be explicitly expressed as surds, we have not exhausted the different kinds of irrational numbers contained in the continuum. Let us draw a circle whose diameter is equal to $A_0{A}_1$, i.e. to unity. It is natural to suppose² that the circumference of such a circle has a length capable of numerical measurement. This length is usually denoted by $\pi$. And it has been shown³ (though the proof is unfortunately long and difficult) that this number $\pi$ is not the root of any algebraical equation with integral coefficients, such, for example, as $${\pi }^2=n,{\pi }^3=n,{\pi }^5=\pi +n,$$ where $n$ is an integer. In this way it is possible to define a number which is not rational nor yet belongs to any of the classes of irrational numbers which we have so far considered. And this number $\pi$ is no isolated or exceptional case. Any number of other examples can be constructed. In fact it is only special classes of irrational numbers which are roots of equations of this kind, just as it is only a still smaller class which can be expressed by means of surds.