# What is the usefulness of classification into Transcedental and Algebraic numbers?

Classification into rationals and irrationals makes complete sense because irrational numbers seem to be completely different from rational numbers, which are terminating or repeating.

We know that all irrational numbers have non-terminating non-repeating repeating decimal expansions, then why is there need for separating numbers which are roots of polynomial equations and those who aren't?

Is there anything special in the decimal expansion of pi which the decimal expansion of root 2 doesn't have?

Could this be true that all non-transcedental irrationals can be expressed by a terminating formula in terms of radicals, rationals and arithmetic operations while transcedentals can't be? If this is true, then classification into non-transcedental irrationals and transcedentals will make sense.

EDIT- But alphacapture points out in the comments that that isn't true because polynomial equations of degree greater than 4 have no solution formula in terms of radicals. So, to me, the solution of a degree 8 polynomial equation seems to be as weird as pi, but according to maths, pi is supposed to be weirder because it's 'transcedental' and so somehow inherently differs from roots of polynomial equations.

• I can answer the question at the end right away: No. See en.wikipedia.org/wiki/Solvable_by_radicals – alphacapture Aug 6 '18 at 4:58
• – Luiz Cordeiro Aug 6 '18 at 5:05
• Algebraic numbers are "rare" in the sense that there are only countably many, so "most" numbers are transcendental. – Kusma Aug 6 '18 at 5:25
• Those numbers that are roots of an equation formed using algebra alone and rational numbers only are, in a sense, closer to being rational than numbers which are not. If I give you a number and tell you it's irrational but is the root of a simple equation formed using rational numbers, while another number is irrational and is not the root of any such equation, would you not say one of these two numbers is at least a little bit closer to being rational? – Ittay Weiss Aug 6 '18 at 10:13
• You seem to be focusing too much on decimal expansions, both in the question (viewing periodicity of the expansion as the crucial distinction) and in your answer (about approximation algorithms). In connection with the latter, note that an algebraic number is determined with complete precision by the information that it's the so-and-so many-th root of such-and-such polynomial with integer coefficients. – Andreas Blass Aug 6 '18 at 13:02

Transcendental numbers are numbers that cannot be defined in the language of algebra. Their existence shows that the basic concepts of arithmetic are not enough to fully describe all of the phenomena that occur in the real numbers.

Polynomials are precisely the formulae in one variable that can be written down using only addition, subtraction, and multiplication. Yes, they can be written as a sum of monomials, and this is a useful canonical form, but that makes for a poor definition, despite the fact that it's repeated as such endlessly by high school teachers and even most university-level sources. Thus, a polynomial equation with rational coefficients is just any equation that can be written down using the rational numbers and the $+$ and $\cdot$ signs. Of course by using negative coefficients, this also lets us use $-$ if we like. Furthermore, an equation that also uses division can always be reduced to the form $\frac {P(x)} {Q(x)}=0$, where $P$ and $Q$ are polynomials, and from there to $P(x)=0$, so if $x$ solves an equation involving division, it also solves an equation without division. And finally, any rational number can always be written using the four arithmetic operations and the numbers $0$ and $1$.

Thus, one definition for an algebraic number is a number which satisfies a formula written using only $+, -,\cdot,\div, 0, 1$, and $=$. If we consider this alphabet (and associated grammar) to be the "language of algebra", then such a formula can be taken as a definition of that number written in that language. $\sqrt 2$ can be given such a definition. $\pi$ is usually defined by making reference to geometry, and what its transcendence means is that we need geometry (or at least something bigger than algebra) to define it. A transcendental number is one such that the only predicates written in the language of algebra that the number verifies are the trivial predicates verified by all numbers, like $x+x=2x$.

You might object that something like $x^2=2$ doesn't really define $\sqrt 2$, since after all that equation is also true for $-\sqrt 2$. This is true, and in fact this insight eventually leads to Galois theory. The numbers $\sqrt 2$ and $-\sqrt 2$ cannot be distinguished using algebra and the rational numbers, in much the same way that $\pi$ cannot be defined using algebra and the rational numbers. In Galois theory we have the notion of conjugate numbers over a given field $F$, which are numbers which cannot be distinguished "from the point of view of $F$". This means that any sentence written in the "language of $F$" is either true for both elements, or true for neither. In turns out that there's always a fundamental "minimal sentence" - the minimal polynomial - such that the conjugate numbers of $a$ are precisely all of the numbers making that sentence true. Thus we cannot do better than the minimal polynomial as a definition for $a$ in the language of $F$ - it is the sentence true for $a$ which is true for the fewest other elements.

I remember reading somewhere that there's a general notion in logic called a "transcendental element" over a language or a formal system, or something like that, which is basically exactly what I outlined above: an element which verifies no sentences in the language other than the tautologies. Someone who knows more might leave a comment or an answer.

• The relevant subject is model theory, incidentally. One precise statement along the lines above is: the numbers definable in the structure $\mathcal{R}=(\mathbb{R}; +,\times)$ are exactly the algebraic numbers. In particular, we can distinguish in reals between $\sqrt{2}$ and $-\sqrt{2}$, since the ordering of real numbers is definable in $\mathcal{R}$: $a\le b$ iff for some $c$ we have $c^2=b-a$ (note that this does not work for the rationals). – Noah Schweber Aug 6 '18 at 23:01
• @NoahSchweber I think you mean the structure $\mathcal Q = (\mathbb Q, +, \times)$, no?. – Jack M Sep 14 '18 at 9:21
• No, I do not - I mean exactly what I wrote. Note that $\sqrt{2}$ (say) isn't even in $\mathbb{Q}$, so it doesn't make sense to say that it's definable in $\mathbb{Q}$. – Noah Schweber Sep 14 '18 at 12:37

I'd like to give three examples which highlight the usefulness of classifying a number as transcendental. We start with a problem stated thousands of years ago.

• Squaring the circle

This is a famous problem dating back to the ancient Greeks. Given a circle with radius one, find a construction in a finite number of steps with compass and straightedge only to obtain a square with the same area.

We know the area of the circle with radius one is $\pi$. On the one hand it can be shown that each length started from unit length, which is constructed in a finite number of steps with compass and straightedge only, is algebraic. On the other hand it was Ferdinand von Lindemann who could show that $\pi$ is transcendental (1882).

It is the transcendence of $\pi$ which clarified that this ancient problem is unsolvable.

The next example demonstrates the relevance of transcendental numbers in the 20th century.

• Hilbert's seventh problem

It was one of the great moments of the development of mathematics, when David Hilbert presented 23 problems to the mathematical community at the Paris conference of the International Congress of Mathematicians in 1900.

These problems had an enourmous influence to the development of mathematics, they even paved the way for new mathematical disciplines and inspired many of the most creative mathematicians in the 20th century.

The seventh problem, titled by Hilbert Irrationality and Transcendence of Certain Numbers addresses transcendent numbers, indicating how important a classification of them has been considered by the leading mathematicians. The problem can be stated as:

• Is ${\displaystyle a^{b}}$ always transcendental, for algebraic ${\displaystyle a\not \in \{0,1\}}$ and irrational algebraic ${\displaystyle b}$?

It needed more than three decades until an affirmative answer could be given by Aleksandr Gelfond (April 1, 1934) and independently by Theodor Schneider (May 28, 1934).

With the last example we jump right into the 21st century.

• Periods

represent another fascinating class of numbers introduced by D. Zagier and M. Kontsevich in 2000. They form a countable class of numbers lying between $\overline{\mathbb{Q}}$, the set of algebraic numbers and $\mathbb{C}$.

The properties of these numbers are an inspiring source for mathematicians and transcendent periods are highly attractive as shown in the survey Transcendence of Periods by Michel Waldschmidt (2005).

In the abstract of the paper he indicates the relevance of transcendental numbers.

Abstract: The set of real numbers and the set of complex numbers have the power of continuum. Among these numbers, those which are “interesting”, which appear “naturally”, which deserve our attention, form a countable set. Starting from this point of view we are interested in the periods as defined by M. Kontsevich and D. Zagier. We give the state of the art on the question of the arithmetic nature of these numbers: to decide whether a period is a rational number, an irrational algebraic number or else a transcendental number is the object of a few theorems and of many conjectures. We also consider the approximation of such numbers by rational or algebraic numbers.

The notion of algebraic number is a starting point for areas of mathematics such as Galois theory and algebraic number theory. Sets of algebraic numbers can he used to form finite extension fields of $\mathbf Q$, which behave in many ways like covering spaces in topology and Riemann surfaces (which are closely related to finite extensions of the field $\mathbf C(z)$).

One possible explanation I thought of myself is that there is a common iterative algorithm (like Newton's method) for all algebraic numbers (i.e. Roots of polynomials), to determine them with as much accuracy as one wants. But there are separate algorithms for e, pi, etc. And then there are non-computable transcendentals for which no algorithm exists. I guess this makes them weirder than non-transcendental irrationals.

• This should be a comment, or included in the body of the question itself, rather than an answer. – Noah Schweber Aug 6 '18 at 22:57