Transcendental number

In mathematics, a transcendental number is a real or complex number that is not algebraic: that is, not the root of a non-zero polynomial with integer (or, equivalently, rational) coefficients. The best-known transcendental numbers are π and e. The quality of a number being transcendental is called transcendence.
Though only a few classes of transcendental numbers are known, partly 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 form a countable set, while the set of real numbers ⁠ R {\displaystyle \mathbb {R} } ⁠ and the set of complex numbers ⁠ C {\displaystyle \mathbb {C} } ⁠ 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 sets of rational, algebraic irrational, 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 x − 2 = 0. The golden ratio (denoted φ {\displaystyle \varphi } or ϕ {\displaystyle \phi } ) is another irrational number that is not transcendental, as it is a root of the polynomial equation x − x − 1 = 0.
The name "transcendental" comes from Latin trānscendere '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 eighteenth century, was probably the first person to define transcendental numbers in the modern sense.
Johann Heinrich Lambert conjectured that e and π were both transcendental numbers in his 1768 paper proving the number π is irrational, and proposed a tentative sketch proof that π is transcendental.
Joseph Liouville first proved the existence of transcendental numbers in 1844, and in 1851 gave the first decimal examples such as the Liouville constant
L b = ∑ n = 1 ∞ 10 − n ! = 10 − 1 + 10 − 2 + 10 − 6 + 10 − 24 + 10 − 120 + 10 − 720 + 10 − 5040 + 10 − 40320 + … = 0. 1 1 000 1 00000000000000000 1 00000000000000000000000000000000000000000000000000000 … {\displaystyle {\begin{aligned}L_{b}&=\sum _{n=1}^{\infty }10^{-n!}\\[2pt]&=10^{-1}+10^{-2}+10^{-6}+10^{-24}+10^{-120}+10^{-720}+10^{-5040}+10^{-40320}+\ldots \\[4pt]&=0.{\textbf {1}}{\textbf {1}}000{\textbf {1}}00000000000000000{\textbf {1}}00000000000000000000000000000000000000000000000000000\ \ldots \end{aligned}}}
in which the nth digit after the decimal point is 1 if n = 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 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 is called the Liouville numbers. 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 1882 Ferdinand von Lindemann published the first complete proof that π is transcendental. He first proved that e is transcendental if a is a non-zero algebraic number. Then, since e = −1 is algebraic (see Euler's identity), iπ must be transcendental. But since i is algebraic, π must therefore be transcendental. This approach was generalized by Karl Weierstrass to what is now known as the Lindemann–Weierstrass theorem. The transcendence of π implies that geometric constructions involving compass and straightedge only cannot produce certain results, for example squaring the circle.
In 1900 David Hilbert posed a question about transcendental numbers, Hilbert's seventh problem: If a is an algebraic number that is not 0 or 1, and b is an irrational algebraic number, is a 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. Every real transcendental number must also be irrational, since every rational number is the root of some integer polynomial 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, it is impossible for both subsets to 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.
Applying any non-constant single-variable algebraic function to a transcendental argument yields a transcendental value. For example, from knowing that π is transcendental, it can be immediately deduced that numbers such as 5 π {\displaystyle 5\pi } , π − 3 2 {\displaystyle {\tfrac {\pi -3}{\sqrt {2}}}} , ( π − 3 ) 8 {\displaystyle ({\sqrt {\pi }}-{\sqrt {3}})^{8}} , and π 5 + 7 4 {\displaystyle {\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, π and (1 − π) are both transcendental, but π + (1 − π) = 1 is obviously not. It is unknown whether e + π, for example, is transcendental, though at least one of e + π and eπ 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) = x − (a + b) x + a b . If (a + b) and a b 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 simple 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 have a "simple" structure, and that are not eventually periodic are transcendental (in other words, algebraic irrational roots of at least third degree polynomials do not have apparent pattern in their continued fraction expansions, since eventually periodic continued fractions correspond to quadratic irrationals, see Hermite's problem).
Numbers proven to be transcendental:
Numbers which have yet to be proven to be either transcendental or algebraic:
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 + ⋯ + c n e n = 0 , c 0 , c n ≠ 0 . {\displaystyle c_{0}+c_{1}e+c_{2}e^{2}+\cdots +c_{n}e^{n}=0,\qquad c_{0},c_{n}\neq 0~.} It is difficult to make use of the integer status of these coefficients when multiplied by a power of the irrational e, but we can absorb those powers into an integral which “mostly” will assume integer values. For a positive integer k, define the polynomial f k ( x ) = x k [ ( x − 1 ) ⋯ ( x − n ) ] k + 1 , {\displaystyle f_{k}(x)=x^{k}\left[(x-1)\cdots (x-n)\right]^{k+1},} and multiply both sides of the above equation by ∫ 0 ∞ f k ( x ) e − x d x , {\displaystyle \int _{0}^{\infty }f_{k}(x)\,e^{-x}\,\mathrm {d} x\ ,} to arrive at the equation: c 0 ( ∫ 0 ∞ f k ( x ) e − x d x ) + c 1 e ( ∫ 0 ∞ f k ( x ) e − x d x ) + ⋯ + c n e n ( ∫ 0 ∞ f k ( x ) e − x d x ) = 0 . {\displaystyle c_{0}\left(\int _{0}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)+c_{1}e\left(\int _{0}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)+\cdots +c_{n}e^{n}\left(\int _{0}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)=0~.}
By splitting respective domains of integration, this equation can be written in the form P + Q = 0 {\displaystyle P+Q=0} where P = c 0 ( ∫ 0 ∞ f k ( x ) e − x d x ) + c 1 e ( ∫ 1 ∞ f k ( x ) e − x d x ) + c 2 e 2 ( ∫ 2 ∞ f k ( x ) e − x d x ) + ⋯ + c n e n ( ∫ n ∞ f k ( x ) e − x d x ) Q = c 1 e ( ∫ 0 1 f k ( x ) e − x d x ) + c 2 e 2 ( ∫ 0 2 f k ( x ) e − x d x ) + ⋯ + c n e n ( ∫ 0 n f k ( x ) e − x d x ) {\displaystyle {\begin{aligned}P&=c_{0}\left(\int _{0}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)+c_{1}e\left(\int _{1}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)+c_{2}e^{2}\left(\int _{2}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)+\cdots +c_{n}e^{n}\left(\int _{n}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)\\Q&=c_{1}e\left(\int _{0}^{1}f_{k}(x)e^{-x}\,\mathrm {d} x\right)+c_{2}e^{2}\left(\int _{0}^{2}f_{k}(x)e^{-x}\,\mathrm {d} x\right)+\cdots +c_{n}e^{n}\left(\int _{0}^{n}f_{k}(x)e^{-x}\,\mathrm {d} x\right)\end{aligned}}} Here P will turn out to be an integer, but more importantly it grows quickly with k.
There are arbitrarily large k such that P k ! {\displaystyle \ {\tfrac {P}{k!}}\ } is a non-zero integer.
Proof. Recall the standard integral (case of the Gamma function) ∫ 0 ∞ t j e − t d t = j ! {\displaystyle \int _{0}^{\infty }t^{j}e^{-t}\,\mathrm {d} t=j!} valid for any natural number j {\displaystyle j} . More generally,
This would allow us to compute P {\displaystyle P} exactly, because any term of P {\displaystyle P} can be rewritten as c a e a ∫ a ∞ f k ( x ) e − x d x = c a ∫ a ∞ f k ( x ) e − ( x − a ) d x = { t = x − a x = t + a d x = d t } = c a ∫ 0 ∞ f k ( t + a ) e − t d t {\displaystyle c_{a}e^{a}\int _{a}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x=c_{a}\int _{a}^{\infty }f_{k}(x)e^{-(x-a)}\,\mathrm {d} x=\left\{{\begin{aligned}t&=x-a\\x&=t+a\\\mathrm {d} x&=\mathrm {d} t\end{aligned}}\right\}=c_{a}\int _{0}^{\infty }f_{k}(t+a)e^{-t}\,\mathrm {d} t} through a change of variables. Hence P = ∑ a = 0 n c a ∫ 0 ∞ f k ( t + a ) e − t d t = ∫ 0 ∞ ( ∑ a = 0 n c a f k ( t + a ) ) e − t d t {\displaystyle P=\sum _{a=0}^{n}c_{a}\int _{0}^{\infty }f_{k}(t+a)e^{-t}\,\mathrm {d} t=\int _{0}^{\infty }{\biggl (}\sum _{a=0}^{n}c_{a}f_{k}(t+a){\biggr )}e^{-t}\,\mathrm {d} t} That latter sum is a polynomial in t {\displaystyle t} with integer coefficients, i.e., it is a linear combination of powers t j {\displaystyle t^{j}} with integer coefficients. Hence the number P {\displaystyle P} is a linear combination (with those same integer coefficients) of factorials j ! {\displaystyle j!} ; in particular P {\displaystyle P} is an integer.
Smaller factorials divide larger factorials, so the smallest j ! {\displaystyle j!} occurring in that linear combination will also divide the whole of P {\displaystyle P} . We get that j ! {\displaystyle j!} from the lowest power t j {\displaystyle t^{j}} term appearing with a nonzero coefficient in ∑ a = 0 n c a f k ( t + a ) {\displaystyle \textstyle \sum _{a=0}^{n}c_{a}f_{k}(t+a)} , but this smallest exponent j {\displaystyle j} is also the multiplicity of t = 0 {\displaystyle t=0} as a root of this polynomial. f k ( x ) {\displaystyle f_{k}(x)} is chosen to have multiplicity k {\displaystyle k} of the root x = 0 {\displaystyle x=0} and multiplicity k + 1 {\displaystyle k+1} of the roots x = a {\displaystyle x=a} for a = 1 , … , n {\displaystyle a=1,\dots ,n} , so that smallest exponent is t k {\displaystyle t^{k}} for f k ( t ) {\displaystyle f_{k}(t)} and t k + 1 {\displaystyle t^{k+1}} for f k ( t + a ) {\displaystyle f_{k}(t+a)} with a > 0 {\displaystyle a>0} . Therefore k ! {\displaystyle k!} divides P {\displaystyle P} .
To establish the last claim in the lemma, that P {\displaystyle P} is nonzero, it is sufficient to prove that k + 1 {\displaystyle k+1} does not divide P {\displaystyle P} . To that end, let k + 1 {\displaystyle k+1} be any prime larger than n {\displaystyle n} and | c 0 | {\displaystyle |c_{0}|} . We know from the above that ( k + 1 ) ! {\displaystyle (k+1)!} divides each of c a ∫ 0 ∞ f k ( t + a ) e − t d t {\displaystyle \textstyle c_{a}\int _{0}^{\infty }f_{k}(t+a)e^{-t}\,\mathrm {d} t} for 1 ⩽ a ⩽ n {\displaystyle 1\leqslant a\leqslant n} , so in particular all of those are divisible by k + 1 {\displaystyle k+1} . It comes down to the first term c 0 ∫ 0 ∞ f k ( t ) e − t d t {\displaystyle \textstyle c_{0}\int _{0}^{\infty }f_{k}(t)e^{-t}\,\mathrm {d} t} . We have (see falling and rising factorials) f k ( t ) = t k [ ( t − 1 ) ⋯ ( t − n ) ] k + 1 = [ ( − 1 ) n ( n ! ) ] k + 1 t k + higher degree terms {\displaystyle f_{k}(t)=t^{k}{\bigl [}(t-1)\cdots (t-n){\bigr ]}^{k+1}={\bigl [}(-1)^{n}(n!){\bigr ]}^{k+1}t^{k}+{\text{higher degree terms}}} and those higher degree terms all give rise to factorials ( k + 1 ) ! {\displaystyle (k+1)!} or larger. Hence P ≡ c 0 ∫ 0 ∞ f k ( t ) e − t d t ≡ c 0 [ ( − 1 ) n ( n ! ) ] k + 1 k ! ( mod ( k + 1 ) ) {\displaystyle P\equiv c_{0}\int _{0}^{\infty }f_{k}(t)e^{-t}\,\mathrm {d} t\equiv c_{0}{\bigl [}(-1)^{n}(n!){\bigr ]}^{k+1}k!{\pmod {(k+1)}}} That right hand side is a product of nonzero integer factors less than the prime k + 1 {\displaystyle k+1} , therefore that product is not divisible by k + 1 {\displaystyle k+1} , and the same holds for P {\displaystyle P} ; in particular P {\displaystyle P} cannot be zero.
For sufficiently large k, | Q k ! | < 1 {\displaystyle \left|{\tfrac {Q}{k!}}\right|<1} .
Proof. Note that
f k e − x = x k [ ( x − 1 ) ( x − 2 ) ⋯ ( x − n ) ] k + 1 e − x = ( x ( x − 1 ) ⋯ ( x − n ) ) k ⋅ ( ( x − 1 ) ⋯ ( x − n ) e − x ) = u ( x ) k ⋅ v ( x ) {\displaystyle {\begin{aligned}f_{k}e^{-x}&=x^{k}\left[(x-1)(x-2)\cdots (x-n)\right]^{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), 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
| f k e − x | ≤ | u ( x ) | k ⋅ | v ( x ) | < G k H for 0 ≤ x ≤ n . {\displaystyle \ \left|f_{k}e^{-x}\right|\leq |u(x)|^{k}\cdot |v(x)|<G^{k}H\quad {\text{ for }}0\leq x\leq n~.}
So each of those integrals composing Q is bounded, the worst case being
| ∫ 0 n f k e − x d x | ≤ ∫ 0 n | f k e − x | d x ≤ ∫ 0 n G k H d x = n G k H . {\displaystyle \left|\int _{0}^{n}f_{k}e^{-x}\ \mathrm {d} \ x\right|\leq \int _{0}^{n}\left|f_{k}e^{-x}\right|\ \mathrm {d} \ x\leq \int _{0}^{n}G^{k}H\ \mathrm {d} \ x=nG^{k}H~.}
It is now possible to bound the sum Q as well:
| Q | < G k ⋅ n H ( | c 1 | e + | c 2 | e 2 + ⋯ + | c n | e n ) = G k ⋅ M , {\displaystyle |Q|<G^{k}\cdot nH\left(|c_{1}|e+|c_{2}|e^{2}+\cdots +|c_{n}|e^{n}\right)=G^{k}\cdot M\ ,}
where M is a constant not depending on k. It follows that
| Q k ! | < M ⋅ G k k ! → 0 as k → ∞ , {\displaystyle \ \left|{\frac {Q}{k!}}\right|<M\cdot {\frac {G^{k}}{k!}}\to 0\quad {\text{ as }}k\to \infty \ ,}
finishing the proof of this lemma.
Choosing a value of k that satisfies both lemmas leads to a non-zero integer ( P k ! ) {\displaystyle \left({\tfrac {P}{k!}}\right)} added to a vanishingly small quantity ( Q k ! ) {\displaystyle \left({\tfrac {Q}{k!}}\right)} being equal to zero: 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.
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.
For detailed information concerning the proofs of the transcendence of π and e, see the references and external links.