 Transcendental Number
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Transcendental Number

In mathematics, a transcendental number is a real number or complex number that is not an algebraic number--that is, not a root (i.e., solution) of a nonzero polynomial equation with integer coefficients. The best-known transcendental numbers are ? and e.

Though only a few classes of transcendental numbers are known, in part because it can be extremely difficult to show that a given number is transcendental, transcendental numbers are not rare. Indeed, almost all real and complex numbers are transcendental, since the algebraic numbers compose a countable set, while the set of real numbers and the set of complex numbers are both uncountable sets, and therefore larger than any countable set. All real transcendental numbers are irrational numbers, since all rational numbers are algebraic. The converse is not true: not all irrational numbers are transcendental. For example, the square root of 2 is an irrational number, but it is not a transcendental number as it is a root of the polynomial equation x2 - 2 = 0. The golden ratio (denoted $\varphi$ or $\phi$ ) is another irrational number that is not transcendental, as it is a root of the polynomial equation x2 - x - 1 = 0.

## History

The name "transcendental" comes from the Latin transcend?re 'to climb over or beyond, surmount', and was first used for the mathematical concept in Leibniz's 1682 paper in which he proved that sin(x) is not an algebraic function of x.Euler, in the 18th century, was probably the first person to define transcendental numbers in the modern sense.

Johann Heinrich Lambert conjectured that e and ? were both transcendental numbers in his 1768 paper proving the number ? is irrational, and proposed a tentative sketch of a proof of ?'s transcendence.

Joseph Liouville first proved the existence of transcendental numbers in 1844, and in 1851 gave the first decimal examples such as the Liouville constant

{\begin{aligned}L_{b}&=\sum _{n=1}^{\infty }10^{-n!}\\&=10^{-1}+10^{-2}+10^{-6}+10^{-24}+10^{-120}+10^{-720}+10^{-5040}+10^{-40320}+\ldots \\&=0.{\textbf {1}}{\textbf {1}}000{\textbf {1}}00000000000000000{\textbf {1}}00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000{\textbf {1}}00000000\ldots \\\end{aligned}} in which the nth digit after the decimal point is 1 if n is equal to k! (k factorial) for some k and 0 otherwise. In other words, the nth digit of this number is 1 only if n is one of the numbers 1! = 1, 2! = 2, 3! = 6, 4! = 24, etc. Liouville showed that this number belongs to a class of transcendental numbers that can be more closely approximated by rational numbers than can any irrational algebraic number, and this class of numbers are called Liouville numbers, named in honour of him. Liouville showed that all Liouville numbers are transcendental.

The first number to be proven transcendental without having been specifically constructed for the purpose of proving transcendental numbers' existence was e, by Charles Hermite in 1873.

In 1874, Georg Cantor proved that the algebraic numbers are countable and the real numbers are uncountable. He also gave a new method for constructing transcendental numbers. In 1878, Cantor published a construction that proves there are as many transcendental numbers as there are real numbers. Cantor's work established the ubiquity of transcendental numbers.

In 1882, Ferdinand von Lindemann published the first complete proof of the transcendence of ?. He first proved that ea is transcendental when a is any non-zero algebraic number. Then, since ei? = -1 is algebraic (see Euler's identity), i? must be transcendental. But since i is algebraic, ? therefore must be transcendental. This approach was generalized by Karl Weierstrass to what is now known as the Lindemann-Weierstrass theorem. The transcendence of ? allowed the proof of the impossibility of several ancient geometric constructions involving compass and straightedge, including the most famous one, squaring the circle.

In 1900, David Hilbert posed an influential question about transcendental numbers, Hilbert's seventh problem: If a is an algebraic number, that is not zero or one, and b is an irrational algebraic number, is ab necessarily transcendental? The affirmative answer was provided in 1934 by the Gelfond-Schneider theorem. This work was extended by Alan Baker in the 1960s in his work on lower bounds for linear forms in any number of logarithms (of algebraic numbers).

## Properties

The set of transcendental numbers is uncountably infinite. Since the polynomials with rational coefficients are countable, and since each such polynomial has a finite number of zeroes, the algebraic numbers must also be countable. However, Cantor's diagonal argument proves that the real numbers (and therefore also the complex numbers) are uncountable. Since the real numbers are the union of algebraic and transcendental numbers, they cannot both be countable. This makes the transcendental numbers uncountable.

No rational number is transcendental and all real transcendental numbers are irrational. The irrational numbers contain all the real transcendental numbers and a subset of the algebraic numbers, including the quadratic irrationals and other forms of algebraic irrationals.

Any non-constant algebraic function of a single variable yields a transcendental value when applied to a transcendental argument. For example, from knowing that ? is transcendental, it can be immediately deduced that numbers such as $5\pi$ , ${\frac {\pi -3}{\sqrt {2}}}$ , $\left({\sqrt {\pi }}-{\sqrt {3}}\right)^{8}$ , and ${\sqrt[{4}]{\pi ^{5}+7}}$ are transcendental as well.

However, an algebraic function of several variables may yield an algebraic number when applied to transcendental numbers if these numbers are not algebraically independent. For example, ? and (1 - ?) are both transcendental, but ? + (1 - ?) = 1 is obviously not. It is unknown whether ? + e, for example, is transcendental, though at least one of ? + e and ?e must be transcendental. More generally, for any two transcendental numbers a and b, at least one of a + b and ab must be transcendental. To see this, consider the polynomial (x - a)(x - b) = x2 - (a + b)x + ab. If (a + b) and ab were both algebraic, then this would be a polynomial with algebraic coefficients. Because algebraic numbers form an algebraically closed field, this would imply that the roots of the polynomial, a and b, must be algebraic. But this is a contradiction, and thus it must be the case that at least one of the coefficients is transcendental.

The non-computable numbers are a strict subset of the transcendental numbers.

All Liouville numbers are transcendental, but not vice versa. Any Liouville number must have unbounded partial quotients in its continued fraction expansion. Using a counting argument one can show that there exist transcendental numbers which have bounded partial quotients and hence are not Liouville numbers.

Using the explicit continued fraction expansion of e, one can show that e is not a Liouville number (although the partial quotients in its continued fraction expansion are unbounded). Kurt Mahler showed in 1953 that ? is also not a Liouville number. It is conjectured that all infinite continued fractions with bounded terms that are not eventually periodic are transcendental (eventually periodic continued fractions correspond to quadratic irrationals).

## Numbers proven to be transcendental

Numbers proven to be transcendental:

$2^{\sqrt {2}},$ the Gelfond-Schneider constant (or Hilbert number)
• sin(a), cos(a), tan(a), and their multiplicative inverses csc(a), sec(a), and cot(a), for any nonzero algebraic number a, expressed in radians (by the Lindemann-Weierstrass theorem).
• The fixed point of the cosine function (also referred to as the dottie number $d$ ) - the unique real solution to the equation $\cos(x)=x$ , where x is in radians (by the Lindemann-Weierstrass theorem).
• ln(a) if a is algebraic and not equal to 0 or 1, for any branch of the logarithm function (by the Lindemann-Weierstrass theorem).
• W(a) if a is algebraic and nonzero, for any branch of the Lambert W Function (by the Lindemann-Weierstrass theorem), in particular: $\Omega ,$ the omega constant
• ${\sqrt {x}}_{s},$ the square super-root of any natural number is either an integer or transcendental (by the Gelfond-Schneider theorem)
• ?(1/3), ?(1/4), and ?(1/6).
• 0.64341054629..., Cahen's constant.
• The Champernowne constants, the irrational numbers formed by concatenating representations of all positive integers.
• ?, Chaitin's constant (since it is a non-computable number).
• The so-called Fredholm constants, such as
$\sum _{n=0}^{\infty }10^{-2^{n}}=0.{\textbf {1}}{\textbf {1}}0{\textbf {1}}000{\textbf {1}}0000000{\textbf {1}}000000000000000{\textbf {1}}0000000000000000000000000000000{\textbf {1}}0000000000000000000000000000000000000000000000000000000000000000{\textbf {1}}\ldots$ which also holds by replacing 10 with any algebraic b > 1.
$\sum _{k=0}^{\infty }10^{-\left\lfloor \beta ^{k}\right\rfloor };$ where $\beta \mapsto \lfloor \beta \rfloor$ is the floor function.
• 3.300330000000000330033... and its reciprocal 0.30300000303..., two numbers with only two different decimal digits whose nonzero digit positions are given by the Moser-de Bruijn sequence and its double.

## Possible transcendental numbers

Numbers which have yet to be proven to be either transcendental or algebraic:

Conjectures:

## Sketch of a proof that e is transcendental

The first proof that the base of the natural logarithms, e, is transcendental dates from 1873. We will now follow the strategy of David Hilbert (1862-1943) who gave a simplification of the original proof of Charles Hermite. The idea is the following:

Assume, for purpose of finding a contradiction, that e is algebraic. Then there exists a finite set of integer coefficients c0, c1, ..., cn satisfying the equation:

$c_{0}+c_{1}e+c_{2}e^{2}+\cdots +c_{n}e^{n}=0,\qquad c_{0},c_{n}\neq 0.$ Now for a positive integer k, we define the following polynomial:

$f_{k}(x)=x^{k}\left[(x-1)\cdots (x-n)\right]^{k+1},$ and multiply both sides of the above equation by

$\int _{0}^{\infty }f_{k}e^{-x}\,dx,$ to arrive at the equation:

$c_{0}\left(\int _{0}^{\infty }f_{k}e^{-x}\,dx\right)+c_{1}e\left(\int _{0}^{\infty }f_{k}e^{-x}\,dx\right)+\cdots +c_{n}e^{n}\left(\int _{0}^{\infty }f_{k}e^{-x}\,dx\right)=0.$ This equation can be written in the form

$P+Q=0$ where

{\begin{aligned}P&=c_{0}\left(\int _{0}^{\infty }f_{k}e^{-x}\,dx\right)+c_{1}e\left(\int _{1}^{\infty }f_{k}e^{-x}\,dx\right)+c_{2}e^{2}\left(\int _{2}^{\infty }f_{k}e^{-x}\,dx\right)+\cdots +c_{n}e^{n}\left(\int _{n}^{\infty }f_{k}e^{-x}\,dx\right)\\Q&=c_{1}e\left(\int _{0}^{1}f_{k}e^{-x}\,dx\right)+c_{2}e^{2}\left(\int _{0}^{2}f_{k}e^{-x}\,dx\right)+\cdots +c_{n}e^{n}\left(\int _{0}^{n}f_{k}e^{-x}\,dx\right)\end{aligned}} Lemma 1. For an appropriate choice of k, ${\tfrac {P}{k!}}$ is a non-zero integer.

Proof. Each term in P is an integer times a sum of factorials, which results from the relation

$\int _{0}^{\infty }x^{j}e^{-x}\,dx=j!$ which is valid for any positive integer j (consider the Gamma function).

It is non-zero because for every a satisfying 0< a n, the integrand in

$c_{a}e^{a}\int _{a}^{\infty }f_{k}e^{-x}\,dx$ is e-x times a sum of terms whose lowest power of x is k+1 after substituting x for x-a in the integral. Then this becomes a sum of integrals of the form

$\int _{0}^{\infty }x^{j}e^{-x}\,dx$ with k+1 j, and it is therefore an integer divisible by (k+1)!. After dividing by k!, we get zero modulo (k+1). However, we can write:

$\int _{0}^{\infty }f_{k}e^{-x}\,dx=\int _{0}^{\infty }\left(\left[(-1)^{n}(n!)\right]^{k+1}e^{-x}x^{k}+\cdots \right)dx$ and thus

${\frac {1}{k!}}c_{0}\int _{0}^{\infty }f_{k}e^{-x}\,dx\equiv c_{0}[(-1)^{n}(n!)]^{k+1}\not \equiv 0{\pmod {k+1}}.$ So when dividing each integral in P by k!, the initial one is not divisible by k+1, but all the others are, as long as k+1 is prime and larger than n and |c0|. It follows that ${\tfrac {P}{k!}}$ itself is not divisible by the prime k+1 and therefore cannot be zero.

Lemma 2. $\left|{\tfrac {Q}{k!}}\right|<1$ for sufficiently large $k$ .

Proof. Note that

{\begin{aligned}f_{k}e^{-x}&=x^{k}[(x-1)(x-2)\cdots (x-n)]^{k+1}e^{-x}\\&=\left(x(x-1)\cdots (x-n)\right)^{k}\cdot \left((x-1)\cdots (x-n)e^{-x}\right)\\&=u(x)^{k}\cdot v(x)\end{aligned}} where $u(x)$ and $v(x)$ are continuous functions of $x$ for all $x$ , so are bounded on the interval $[0,n]$ . That is, there are constants $G,H>0$ such that

$\left|f_{k}e^{-x}\right|\leq |u(x)|^{k}\cdot |v(x)| So each of those integrals composing $Q$ is bounded, the worst case being

$\left|\int _{0}^{n}f_{k}e^{-x}\,dx\right|\leq \int _{0}^{n}\left|f_{k}e^{-x}\right|\,dx\leq \int _{0}^{n}G^{k}H\,dx=nG^{k}H.$ It is now possible to bound the sum $Q$ as well:

$|Q| where $M$ is a constant not depending on $k$ . It follows that

$\left|{\frac {Q}{k!}}\right| finishing the proof of this lemma.

Choosing a value of $k$ satisfying both lemmas leads to a non-zero integer ($P/k!$ ) added to a vanishingly small quantity ($Q/k!$ ) being equal to zero, is an impossibility. It follows that the original assumption, that $e$ can satisfy a polynomial equation with integer coefficients, is also impossible; that is, $e$ is transcendental.

### The transcendence of ?

A similar strategy, different from Lindemann's original approach, can be used to show that the number ? is transcendental. Besides the gamma-function and some estimates as in the proof for e, facts about symmetric polynomials play a vital role in the proof.

For detailed information concerning the proofs of the transcendence of ? and e, see the references and external links.

## Mahler's classification

Kurt Mahler in 1932 partitioned the transcendental numbers into 3 classes, called S, T, and U. Definition of these classes draws on an extension of the idea of a Liouville number (cited above).

### Measure of irrationality of a real number

One way to define a Liouville number is to consider how small a given real number x makes linear polynomials |qx - p| without making them exactly 0. Here p, q are integers with |p|, |q| bounded by a positive integer H.

Let m(x, 1, H) be the minimum non-zero absolute value these polynomials take and take:

$\omega (x,1,H)=-{\frac {\log m(x,1,H)}{\log H}}$ $\omega (x,1)=\limsup _{H\to \infty }\omega (x,1,H).$ ?(x, 1) is often called the measure of irrationality of a real number x. For rational numbers, ?(x, 1) = 0 and is at least 1 for irrational real numbers. A Liouville number is defined to have infinite measure of irrationality. Roth's theorem says that irrational real algebraic numbers have measure of irrationality 1.

### Measure of transcendence of a complex number

Next consider the values of polynomials at a complex number x, when these polynomials have integer coefficients, degree at most n, and height at most H, with n, H being positive integers.

Let m(x,n,H) be the minimum non-zero absolute value such polynomials take at x and take:

$\omega (x,n,H)=-{\frac {\log m(x,n,H)}{n\log H}}$ $\omega (x,n)=\limsup _{H\to \infty }\omega (x,n,H).$ Suppose this is infinite for some minimum positive integer n. A complex number x in this case is called a U number of degree n.

Now we can define

$\omega (x)=\limsup _{n\to \infty }\omega (x,n).$ ?(x) is often called the measure of transcendence of x. If the ?(x,n) are bounded, then ?(x) is finite, and x is called an S number. If the ?(x,n) are finite but unbounded, x is called a T number. x is algebraic if and only if ?(x) = 0.

Clearly the Liouville numbers are a subset of the U numbers. William LeVeque in 1953 constructed U numbers of any desired degree. The Liouville numbers and hence the U numbers are uncountable sets. They are sets of measure 0.

T numbers also comprise a set of measure 0. It took about 35 years to show their existence. Wolfgang M. Schmidt in 1968 showed that examples exist. However, almost all complex numbers are S numbers. Mahler proved that the exponential function sends all non-zero algebraic numbers to S numbers: this shows that e is an S number and gives a proof of the transcendence of ?. The most that is known about ? is that it is not a U number. Many other transcendental numbers remain unclassified.

Two numbers x, y are called algebraically dependent if there is a non-zero polynomial P in 2 indeterminates with integer coefficients such that P(xy) = 0. There is a powerful theorem that 2 complex numbers that are algebraically dependent belong to the same Mahler class. This allows construction of new transcendental numbers, such as the sum of a Liouville number with e or ?.

The symbol S probably stood for the name of Mahler's teacher Carl Ludwig Siegel, and T and U are just the next two letters.

### Koksma's equivalent classification

Jurjen Koksma in 1939 proposed another classification based on approximation by algebraic numbers.

Consider the approximation of a complex number x by algebraic numbers of degree n and height H. Let ? be an algebraic number of this finite set such that |x - ?| has the minimum positive value. Define ?*(x,H,n) and ?*(x,n) by:

$|x-\alpha |=H^{-n\omega ^{*}(x,H,n)-1}.$ $\omega ^{*}(x,n)=\limsup _{H\to \infty }\omega ^{*}(x,n,H).$ If for a smallest positive integer n, ?*(x,n) is infinite, x is called a U*-number of degree n.

If the ?*(x,n) are bounded and do not converge to 0, x is called an S*-number,

A number x is called an A*-number if the ?*(x,n) converge to 0.

If the ?*(x,n) are all finite but unbounded, x is called a T*-number,

Koksma's and Mahler's classifications are equivalent in that they divide the transcendental numbers into the same classes. The A*-numbers are the algebraic numbers.

### LeVeque's construction

Let

$\lambda ={\tfrac {1}{3}}+\sum _{k=1}^{\infty }10^{-k!}.$ It can be shown that the nth root of ? (a Liouville number) is a U-number of degree n.

This construction can be improved to create an uncountable family of U-numbers of degree n. Let Z be the set consisting of every other power of 10 in the series above for ?. The set of all subsets of Z is uncountable. Deleting any of the subsets of Z from the series for ? creates uncountably many distinct Liouville numbers, whose nth roots are U-numbers of degree n.

### Type

The supremum of the sequence {?(xn)} is called the type. Almost all real numbers are S numbers of type 1, which is minimal for real S numbers. Almost all complex numbers are S numbers of type 1/2, which is also minimal. The claims of almost all numbers were conjectured by Mahler and in 1965 proved by Vladimir Sprindzhuk.