The most prominent examples of transcendental numbers are π and e. Only a few classes of transcendental numbers are known, indicating that it can be extremely difficult to show that a given number is transcendental.
However, transcendental numbers are not rare: indeed, almost all real and complex numbers are transcendental, since the algebraic numbers are countable, but the sets of real and complex numbers are uncountable. All transcendental numbers are irrational, since all rational numbers are algebraic. (The converse is not true: not all irrational numbers are transcendental.)
in which the nth digit after the decimal point is 1 if n is a factorial (i.e., 1, 2, 6, 24, 120, 720, ...., etc.) and 0 otherwise. Liouville showed that this number is what we now call a Liouville number; this essentially means that it can be particularly well approximated by rational numbers. Liouville showed that all Liouville numbers are transcendental.
Johann Heinrich Lambert conjectured that e and π were both transcendental numbers in his 1761 paper proving the number π is irrational. The first number to be proven transcendental without having been specifically constructed for the purpose was e, by Charles Hermite in 1873. In 1874, Georg Cantor found the argument mentioned above establishing the ubiquity of transcendental numbers.
In 1882, Ferdinand von Lindemann published a proof that the number π is transcendental. He first showed that e to any nonzero algebraic power is transcendental, and since eiπ = −1 is algebraic (see Euler's identity), iπ and therefore π must be transcendental. This approach was generalized by Karl Weierstrass to the Lindemann–Weierstrass theorem. The transcendence of π 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).
Transcendental numbers are never rational, but some irrational numbers are not transcendental. For example, the square root of 2 is irrational, but it is a solution of the polynomial x2 − 2 = 0, so it is algebraic, not transcendental.
Any non-constant algebraic function of a single variable yields a transcendental value when applied to a transcendental argument. So, for example, from knowing that π is transcendental, we can immediately deduce that numbers such as 5π, (π − 3)/√2, (√π − √3)8 and (π5 + 7)1/7 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, π and 1 − π are both transcendental, but π + (1 − π) = 1 is obviously not. It is unknown whether π + e, for example, is transcendental, though at least one of π + e and πe 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; however, not all transcendental numbers are Liouville numbers. 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 π 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 for which it is unknown whether they are transcendental or not:
Assume, for purpose of finding a contradiction, that e is algebraic. Then there exists a finite set of integer coefficients satisfying the equation:
and such that and are both non-zero.
Depending on the value of n, we specify a sufficiently large positive integer k (to meet our needs later), and multiply both sides of the above equation by , where the notation will be used in this proof as shorthand for the integral:
We have arrived at the equation:
which can now be written in the form
The plan of attack now is to show that for k sufficiently large, the above relations are impossible to satisfy because
The fact that is a nonzero integer results from the relation
To show that
A similar strategy, different from Lindemann's original approach, can be used to show that the number π 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.