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Number that cannot be found as a result of an algebraic equation with integer coefficients
Though only a few classes of transcendental numbers are known, in part as 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 transcendental real numbers (also known as real transcendental numbers or transcendental irrational numbers) are irrational numbers, since all rational numbers are algebraic. The converse is not true: not all irrational numbers are transcendental. Hence, the set of real numbers consists of non-overlapping rational, algebraic non-rational and transcendental real numbers. 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 or ) is another irrational number that is not transcendental, as it is a root of the polynomial equation x2 - x - 1 = 0. The quality of a number being transcendental is called transcendence.
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.
in which the nth digit after the decimal point is 1 if n is equal to k! (kfactorial) 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 his honour. 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. Although this was already implied by his proof of the countability of the algebraic numbers, Cantor also 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 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).
A transcendental number is a (possibly complex) number that is not the root of any integer polynomial, meaning that it is not an algebraic number of any degree. Every real transcendental number must also be irrational, since a rational number is, by definition, an algebraic number of degree one. 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, the latter cannot both be countable. This makes the transcendental numbers uncountable.
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?, ?-3/, (-)8, and 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.
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).
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.
Numbers which have yet to be proven to be either transcendental or algebraic:
Most sums, products, powers, etc. of the number ? and the number e, e.g. e?, e + ?, ? - e, ?/e, ??, ee, ?e, ?, e?2 are not known to be rational, algebraic, irrational or transcendental. A notable exception is e? (for any positive integer n) which has been proven transcendental.
The Euler-Mascheroni constant?: In 2010 M. Ram Murty and N. Saradha considered an infinite list of numbers also containing ?/4 and showed that all but at most one of them have to be transcendental. In 2012 it was shown that at least one of ? and the Euler-Gompertz constant? is transcendental.
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:
Now for a positive integer k, we define the following polynomial:
and multiply both sides of the above equation by
to arrive at the equation:
By splitting respective domains of integration, this equation can be written in the form
Lemma 1. For an appropriate choice of k, is a non-zero integer.
Proof. Each term in P is an integer times a sum of factorials, which results from the relation
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
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
Where Aj-k is integer.
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:
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 itself is not divisible by the prime k+1 and therefore cannot be zero.
Lemma 2. for sufficiently large .
Proof. Note that
where and are continuous functions of for all , so are bounded on the interval . That is, there are constants such that
So each of those integrals composing is bounded, the worst case being
It is now possible to bound the sum as well:
where is a constant not depending on . It follows that
finishing the proof of this lemma.
Choosing a value of satisfying both lemmas leads to a non-zero integer () added to a vanishingly small quantity () 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.
^Cantor 1878, p. 254. Cantor's construction builds a one-to-one correspondence between the set of transcendental numbers and the set of real numbers. In this article, Cantor only applies his construction to the set of irrational numbers.
^J J O'Connor and E F Robertson: Alan Baker. The MacTutor History of Mathematics archive 1998.
^Allouche & Shallit 2003, pp. 385, 403. The name 'Fredholm number' is misplaced: Kempner first proved this number is transcendental, and the note on page 403 states that Fredholm never studied this number.
Lambert, Johann Heinrich (1768). "Mémoire sur quelques propriétés remarquables des quantités transcendantes, circulaires et logarithmiques". Mémoires de l'Académie Royale des Sciences de Berlin: 265-322.