7/25/2019 (1) Jack Huizenga's Answer to How Do You Prove That a Number is a Transcendental Number

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How do you prove that a number is a

transcendental number?

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Transcendental number theory is a notoriously ad hoc field that is woefully

underdeveloped--not for lack of try ing, but because it is extremely difficult. In

this post I will explain some of the main tools and methods in the field.

Part of my purpose for writing this answer is to draw attention away from the

answer by Alejandro Jenkins which is very pretty and has generated a lot of

views/upvotes but is ultimately completely flawed. Ne wton did not have a

proof that is transcendental, and a correct proof of this fact did not exist

until around 150 years after his death. This is a relatively difficult result, and

unfortunately there is no known simple argument which can explain the

transcend entality of in a few lines without complica ted equations.

For what follows, first recall that a transcendental number (or in the

complex numbers if y ou prefer) is any number which is not the root of a

nonzero one-variable polynomial

with integer (equivalently by clearing denominators, rational) coefficients.

Almost all numbers are transcendental. This observation is fairly useless

in practice--it will never let you say "this number is transcendental"--but it is

the easiest way to see that transcendental numbers exist at all. Observe that

the set of all finite lists of integers is countably infinite, since it is the countable

disjoint union

of the sets of lists of fixed length . Thu s the set of polynomials in one

variable is countable. Each such nonzero polynomial ha s finitely many roots,so the set of algebraic numbers is countable as well. On the other hand, the

real numbers are uncountable by Cantor's diagonalization argument, so the

complement of the set of algebraic numbers (which is simply the

transcendental numbers) is uncountable.

Liouville numbers. Historically, the first explicit transcendental numbers

were ones constructed by "brute force." One example of a Liouville number is

.

It is not hard to prove directly that this number is transcend ental. Essentially,

the ones in the powers are eventually too far apart from one

another for there to be cancellation when you compute for a degree

polynomial .

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7/25/2019 (1) Jack Huizenga's Answer to How Do You Prove That a Number is a Transcendental Number

http://slidepdf.com/reader/full/1-jack-huizengas-answer-to-how-do-you-prove-that-a-number-is-a-transcendental 2/3

Diophantine approximation. It is well-known that every irrational

number is a limit of rational numbers, for example by truncating the decimal

expansion. The subject of Diophantine approximation asks when a real

number has a good rational approximation. For example, the approximation

is fairly good, in the sense that the denominator 7 is fairly small relative to how

good the approximation is.

Roughly speaking, the irrationality measure of a number is the

largest number such that there are inifinitely many rational approximations

with

.

As gets bigger and bigger, the number can be more and more closely

approximated by rational numbers with "small" denominators. Liouville

numbers are, by definition, those numbers with infinite irrationality measure.

Irrationality measure can be used to show some numbers are transcendental by

the incredibly deep Thue-Siegel-Roth theorem, for which Roth won the FieldsMedal:

Theorem (Roth). Algebraic nu mbers have irrationality measure 2.

In particular, the only numbers with irrationality measure bigger than 2 are

transcendental; Liouville numbers are particular extreme examples.

Unfortunately, "almost all" numbers have irrationality measure 2. Thus, if you

have some particular transcendental number in mind, unless you got lucky it

will not be possible to show tha t it is transcendental by this method. Worse, it

is usually impossibly difficult to even calculate the irrationality measure for a

random number, and thus the theorem is not helpful.

For extremely special numbers, however, ideas related to Roth's theorem and

its generalization the Schmidt Subspace Thorem can actually be used to show

that the numbers are transcendental. Here is a bizarre theorem in this

direction for people who know what continued fractions are:

Theorem (Adamczewski-Bugeaud). Let be a real number

such that the continued fraction expansion

begins in arbitrarily long palindromes. Then either is a quadratic irrational

number (and the sequence eventually repeats) or is transcendental.

For example, the expansion of the number

begins in a pa lindrome of length 2, 6, and 15, and if the sequence is expa nded

in an appropriate way then will be transcend ental by the theorem.

Let me just remark that, as crazy as this result sounds, it is actually useful. In

my research in algebraic geometry I needed to know some numbers were

transcendental, and they actually fell into this very restricted class of numbers

to which the Adamczewski-Bugeaud theorem applied.

Special constants. When it comes to proving numbers like and are

transcendental, there are a small handful of happy accidents, but for the vast

majority of mathematical constants we are completely in the dark. Jaimal

Icharam's answer beautifully explains the Lindemann-Weierstrass theorem,

which is one of the only tools used for tasks like this. Again, the vast sea of

transcendental numbers cannot be studied with these results.

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7/25/2019 (1) Jack Huizenga's Answer to How Do You Prove That a Number is a Transcendental Number

http://slidepdf.com/reader/full/1-jack-huizengas-answer-to-how-do-you-prove-that-a-number-is-a-transcendental 3/3

Written Sat. 5,236 views.

Summary. I hope to have convinced you that there are a handful of beautiful

methods for studying transcendental nu mbers, which unfortunately only barely

scrape the surface of the most special transcendental numbers. When it comes

to a sufficiently "random" transcendental number, we currently have essentially

no hope of showing that number is transcendental. The explicit numbers

which ca n be showed to be transcendental either have remarkable number

theoretic, analytic, or geometric properties, and in the absence of these strong

properties essentially nothing can be said.

References:Transcendental number

Liouville number

Diophantine ap proximation

Thue–Siegel–Roth theorem

Subspace theorem

Lindemann–Weierstrass theorem

Adamzcewski-Bugeaud's paper

Application of the Ada mzcewski-Bugeaud result in Section 4 here

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