Elusive Transcendentals

The numbers are usually studied in layers of increasing subtlety and intricacy. We start with the natural, or counting, numbers ${\mathbb{N} = \{ 1, 2, 3, \dots \}}$ . Then come the whole numbers or integers, ${\mathbb{Z} = \{ \dots, -2, -1, 0, 1, 2, \dots \}}$ . All the ratios of these (avoiding division by 0) yield the rationals ${\mathbb{Q} = \{ \frac{p}{q} : p, q \in \mathbb{Z} \}}$ . Then come all the real numbers ${\mathbb{R}}$ .

Algebraic and Transcendental Numbers

The real numbers fall into two categories. The algebraic numbers ${\mathbb{A}}$ are all those numbers that are roots of polynomial equations with integer coefficients. For example, ${\sqrt{2}}$ , the square root of 2, is algebraic, since it is a root of the polynomial equation ${x^2-2=0}$ . Finally, the real numbers that are not algebraic are called transcendental numbers. They are members of the set ${\mathbb{R}\setminus\mathbb{A}}$ , that is, in ${\mathbb{R}}$ but not in ${\mathbb{A}}$ .

We thus have an ascending chain of sets

$\displaystyle \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{A} \subset \mathbb{R}$

The transcendental numbers are those in ${\mathbb{R}}$ but in none of the other sets, that is, in the outermost layer in the figure above, but not in any of the inner layers.

Existence of Transcendentals

Although there are infinitely more transcendental numbers than algebraic numbers, they are not so easy to find. The first proof that transcendental numbers exist was given by Joseph Liouville. He found a large class of numbers that are transcendental, and showed how to construct such numbers. An example of a Liouville number is ${\tau = \sum_{j=1}^\infty 1/2^{j!}}$ . This gives us an explicit way of constructing a transcendental. Expanded in binary form, this is

$\displaystyle \tau = 0.110001000000000000000001000 \dots$

where the 1’s occur at positions ${1!, 2!, 3!, \dots}$ .

Later, Georg Cantor gave an existence proof for transcendentals by showing that the real algebraic numbers are countable, whereas the set of all real numbers is uncountable, so there must exist real transcendental numbers, indeed infinitely more that algebraic numbers. But Cantor’s proof was non-constructive: it provided no means of finding particular examples of transcendentals.

A Non-constructive Proof

The set ${\mathbb{Q}}$ of rational numbers forms a field, closed under addition and multiplication. Thus, for any two rationals, ${a\in\mathbb{Q}}$ and ${b\in\mathbb{Q}}$ , the sum ${a+b}$ and the product ${a\cdot b}$ are also rational. It is trivial to show this.

Can combinations of irrational numbers be rational? Obviously, if ${a = \sqrt{2}}$ and ${b = -\sqrt{2}}$ then ${a+b = 0}$ and ${a\cdot b=-2}$ are both rational. But is it possible for ${a^b}$ to be rational if both ${a}$ and ${b}$ irrational?

The answer is “Yes”. Let ${a = b = \sqrt{2}}$ and ${x = a^b = \sqrt{2}^{\sqrt{2}}}$ . If ${x}$ is rational, we have a solution. But if ${x}$ is irrational, we define ${a =x}$ and ${b = \sqrt{2}}$ . Then

$\displaystyle a^b = \left( \sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \left(\sqrt{2}\right)^{\sqrt{2}\cdot\sqrt{2}} = \left(\sqrt{2}\right)^{2} = 2$

We thus have the result that ${a^b}$ is rational for both ${a}$ and ${b}$ irrational, with

$\displaystyle \text{Either }\quad a = b = \sqrt{2} \qquad\text{or}\qquad a = \sqrt{2}^{\sqrt{2}} \text{ and } b = \sqrt{2} \,.$

The proof of this result is non-constructive. It does not tell us which of the alternatives is true, and it gives us no inkling of how to answer this. The answer is known, but it cannot be proved in the simple manner we have used here.

The Gelfond-Schneider Theorem

Aleksandr Gelfond (left) and Theodor Schneider (right).

In 1900, David Hilbert gave a talk at the International Congress of Mathematicians in which he described some major unsolved problems. The full list of 23 problems appeared in the Proceedings of the conference. The Gelfond-Schneider theorem solved Problem 7 of Hilbert’s list.

The Gelfond-Schneider Theorem states that all members of a large set of numbers are transcendental. The result was proved independently in 1934 by Soviet mathematician Aleksandr Gelfond and German mathematician Theodor Schneider.

The theorem may be stated thus:

If ${\alpha}$ is an algebraic number (not equal to 0 or 1), and if ${\beta}$ is an irrational algebraic number, then ${\alpha^\beta}$ is a transcendental number.

Before this theorem was proved, numbers like ${e}$ and ${\pi}$ were known to be transcendental, but the character of numerous numbers like ${2^{\sqrt{2}}}$ was unknown. After the proof, the transcendental nature of a huge set of numbers was known. Some of these were given names, for example, the Gelfond-Schneider constant ${2^{\sqrt{2}}}$ .

Since ${\sqrt{2}}$ is algebraic, we can apply the Gelfond-Schneider theorem using ${a = b = \sqrt{2}}$ to conclude that ${\sqrt{2}^{\sqrt{2}}}$ is transcendental. But we cannot use the number ${\sqrt{2}^{\sqrt{2}}}$ , which is not algebraic, in the theorem. Indeed, as we saw,

$\displaystyle \left( \sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \left(\sqrt{2}\right)^{\sqrt{2}\cdot\sqrt{2}} = \left(\sqrt{2}\right)^{2} = 2 \,.$

The Gelfond-Schneider theorem extends to complex numbers, and can be used to show that Gelfond’s constant ${e^\pi}$ is transcendental. The status of the number ${\pi^e}$ is still unknown.