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Transcendental number

From Wikipedia, the free encyclopedia
In mathematics, a non-algebraic number
"U-number" redirects here. For thermal conductivity, seeR-value (insulation) § U-value.

Inmathematics, atranscendental number is areal orcomplex number that is notalgebraic: that is, not theroot of a non-zeropolynomial withinteger (or, equivalently,rational)coefficients. The best-known transcendental numbers areπ ande.[1][2] The quality of a number being transcendental is calledtranscendence.

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 acountable set, while theset ofreal numbers⁠R{\displaystyle \mathbb {R} }⁠ and the set ofcomplex numbers⁠C{\displaystyle \mathbb {C} }⁠ are bothuncountable sets, and therefore larger than any countable set.

Alltranscendental real numbers (also known asreal transcendental numbers ortranscendental irrational numbers) areirrational numbers, since allrational numbers are algebraic.[3][4][5][6] Theconverse 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.[3] For example, thesquare root of 2 is an irrational number, but it is not a transcendental number as it is aroot of the polynomial equationx2 − 2 = 0. Thegolden ratio (denotedφ{\displaystyle \varphi } orϕ{\displaystyle \phi }) is another irrational number that is not transcendental, as it is a root of the polynomial equationx2 −x − 1 = 0.

History

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The name "transcendental" comes from Latin trānscendere 'to climb over or beyond, surmount',[7] and was first used for the mathematical concept inLeibniz's 1682 paper in which he proved thatsinx is not analgebraic function ofx.[8]Euler, in the eighteenth century, was probably the first person to define transcendentalnumbers in the modern sense.[9]

Johann Heinrich Lambert conjectured thate andπ were both transcendental numbers in his 1768 paper proving the numberπ isirrational, and proposed a tentative sketch proof thatπ is transcendental.[10]

Joseph Liouville first proved the existence of transcendental numbers in 1844,[11] and in 1851 gave the first decimal examples such as theLiouville constant

Lb=∑n=1∞10−n!=10−1+10−2+10−6+10−24+10−120+10−720+10−5040+10−40320+…=0.11000100000000000000000100000000000000000000000000000000000000000000000000000 …{\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 thenth digit after the decimal point is1 ifn =k! (kfactorial) for somek and0 otherwise.[12] In other words, thenth digit of this number is 1 only ifn is one of1! = 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 byrational numbers than can any irrational algebraic number, and this class of numbers is called theLiouville numbers. Liouville showed that all Liouville numbers are transcendental.[13]

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

In 1874Georg Cantor proved that the algebraic numbers are countable and the real numbers are uncountable. He also gavea new method for constructing transcendental numbers.[14] 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.[a]Cantor's work established the ubiquity of transcendental numbers.

In 1882Ferdinand von Lindemann published the first complete proof thatπ is transcendental. He first proved thatea is transcendental ifa is a non-zero algebraic number. Then, sinceeiπ = −1 is algebraic (seeEuler's identity),iπ must be transcendental. But sincei is algebraic,π must therefore be transcendental. This approach was generalized byKarl Weierstrass to what is now known as theLindemann–Weierstrass theorem. The transcendence ofπ implies that geometric constructions involvingcompass and straightedge only cannot produce certain results, for examplesquaring the circle.

In 1900David Hilbert posed a question about transcendental numbers,Hilbert's seventh problem: Ifa is analgebraic number that is not 0 or 1, andb is an irrational algebraic number, isab necessarily transcendental? The affirmative answer was provided in 1934 by theGelfond–Schneider theorem. This work was extended byAlan Baker in the 1960s in his work on lower bounds for linear forms in any number of logarithms (of algebraic numbers).[16]

Properties

[edit]

A transcendental number is a (possibly complex) number that is not the root of any integer polynomial. Every real transcendental number must also beirrational, since everyrational number is the root of some integer polynomial ofdegree one.[17] The set of transcendental numbers isuncountably infinite. Since the polynomials with rational coefficients arecountable, and since each such polynomial has a finite number ofzeroes, thealgebraic numbers must also be countable. However,Cantor's diagonal argument proves that the real numbers (and therefore also thecomplex numbers) are uncountable. Since the real numbers are the union of algebraic and transcendental numbers, it is impossible for bothsubsets to be countable. This makes the transcendental numbers uncountable.

Norational number is transcendental and all real transcendental numbers are irrational. Theirrational numbers contain all the real transcendental numbers and a subset of the algebraic numbers, including thequadratic irrationals and other forms of algebraic irrationals.

Applying any non-constant single-variablealgebraic function to a transcendental argument yields a transcendental value. For example, from knowing thatπ is transcendental, it can be immediately deduced that numbers such as5π{\displaystyle 5\pi },π−32{\displaystyle {\tfrac {\pi -3}{\sqrt {2}}}},(π−3)8{\displaystyle ({\sqrt {\pi }}-{\sqrt {3}})^{8}}, andπ5+74{\displaystyle {\sqrt[{4}]{\pi ^{5}+7}}} are transcendental as well.

However, analgebraic function of several variables may yield an algebraic number when applied to transcendental numbers if these numbers are notalgebraically independent. For example,π and(1 −π) are both transcendental, butπ + (1 −π) = 1 is not. It is unknown whethere +π, for example, is transcendental, though at least one ofe +π andeπ must be transcendental. More generally, for any two transcendental numbersa andb, at least one ofa +b andab must be transcendental. To see this, consider the polynomial(x −a)(x −b) =x2 − (a +b)x +a b . If (a +b) anda b were both algebraic, then this would be a polynomial with algebraic coefficients. Because algebraic numbers form analgebraically closed field, this would imply that the roots of the polynomial,a andb, must be algebraic. But this is a contradiction, and thus it must be the case that at least one of the coefficients is transcendental.

Thenon-computable numbers are a strict subset of the transcendental numbers.

AllLiouville numbers are transcendental, but not vice versa. Any Liouville number must have unbounded partial quotients in itssimple continued fraction expansion. Using acounting 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 ofe, one can show thate is not a Liouville number (although the partial quotients in its continued fraction expansion are unbounded).Kurt Mahler showed in 1953 thatπ 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[18] (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, seeHermite's problem).

Numbers proven to be transcendental

[edit]

Numbers proven to be transcendental:

Conjectured transcendental numbers

[edit]

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

Proofs for specific numbers

[edit]

A proof thate is transcendental

[edit]

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

Assume, for purpose offinding a contradiction, thate is algebraic. Then there exists a finite set of integer coefficientsc0,c1, ...,cn satisfying the equation:c0+c1e+c2e2+⋯+cnen=0,c0,cn≠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 irrationale, but we can absorb those powers into an integral which “mostly” will assume integer values. For a positive integerk, define the polynomialfk(x)=xk[(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∞fk(x)e−xdx ,{\displaystyle \int _{0}^{\infty }f_{k}(x)\,e^{-x}\,\mathrm {d} x\ ,}to arrive at the equation:c0(∫0∞fk(x)e−xdx)+c1e(∫0∞fk(x)e−xdx)+⋯+cnen(∫0∞fk(x)e−xdx)=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 formP+Q=0{\displaystyle P+Q=0}whereP=c0(∫0∞fk(x)e−xdx)+c1e(∫1∞fk(x)e−xdx)+c2e2(∫2∞fk(x)e−xdx)+⋯+cnen(∫n∞fk(x)e−xdx)Q=c1e(∫01fk(x)e−xdx)+c2e2(∫02fk(x)e−xdx)+⋯+cnen(∫0nfk(x)e−xdx){\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}}}HereP will turn out to be an integer, but more importantly it grows quickly withk.

Lemma 1

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There are arbitrarily largek such that Pk! {\displaystyle \ {\tfrac {P}{k!}}\ } is a non-zero integer.

Proof. Recall the standard integral (case of theGamma function)∫0∞tje−tdt=j!{\displaystyle \int _{0}^{\infty }t^{j}e^{-t}\,\mathrm {d} t=j!}valid for anynatural numberj{\displaystyle j}. More generally,

ifg(t)=∑j=0mbjtj{\displaystyle g(t)=\sum _{j=0}^{m}b_{j}t^{j}} then∫0∞g(t)e−tdt=∑j=0mbjj!{\displaystyle \int _{0}^{\infty }g(t)e^{-t}\,\mathrm {d} t=\sum _{j=0}^{m}b_{j}j!}.

This would allow us to computeP{\displaystyle P} exactly, because any term ofP{\displaystyle P} can be rewritten ascaea∫a∞fk(x)e−xdx=ca∫a∞fk(x)e−(x−a)dx={t=x−ax=t+adx=dt}=ca∫0∞fk(t+a)e−tdt{\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 achange of variables. HenceP=∑a=0nca∫0∞fk(t+a)e−tdt=∫0∞(∑a=0ncafk(t+a))e−tdt{\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 int{\displaystyle t} with integer coefficients, i.e., it is a linear combination of powerstj{\displaystyle t^{j}} with integer coefficients. Hence the numberP{\displaystyle P} is a linear combination (with those same integer coefficients) of factorialsj!{\displaystyle j!}; in particularP{\displaystyle P} is an integer.

Smaller factorials divide larger factorials, so the smallestj!{\displaystyle j!} occurring in that linear combination will also divide the whole ofP{\displaystyle P}. We get thatj!{\displaystyle j!} from the lowest powertj{\displaystyle t^{j}} term appearing with a nonzero coefficient in∑a=0ncafk(t+a){\displaystyle \textstyle \sum _{a=0}^{n}c_{a}f_{k}(t+a)}, but this smallest exponentj{\displaystyle j} is also themultiplicity oft=0{\displaystyle t=0} as a root of this polynomial.fk(x){\displaystyle f_{k}(x)} is chosen to have multiplicityk{\displaystyle k} of the rootx=0{\displaystyle x=0} and multiplicityk+1{\displaystyle k+1} of the rootsx=a{\displaystyle x=a} fora=1,…,n{\displaystyle a=1,\dots ,n}, so that smallest exponent istk{\displaystyle t^{k}} forfk(t){\displaystyle f_{k}(t)} andtk+1{\displaystyle t^{k+1}} forfk(t+a){\displaystyle f_{k}(t+a)} witha>0{\displaystyle a>0}. Thereforek!{\displaystyle k!} dividesP{\displaystyle P}.

To establish the last claim in the lemma, thatP{\displaystyle P} is nonzero, it is sufficient to prove thatk+1{\displaystyle k+1} does not divideP{\displaystyle P}. To that end, letk+1{\displaystyle k+1} be anyprime larger thann{\displaystyle n} and|c0|{\displaystyle |c_{0}|}. We know from the above that(k+1)!{\displaystyle (k+1)!} divides each ofca∫0∞fk(t+a)e−tdt{\displaystyle \textstyle c_{a}\int _{0}^{\infty }f_{k}(t+a)e^{-t}\,\mathrm {d} t} for1⩽a⩽n{\displaystyle 1\leqslant a\leqslant n}, so in particular all of thoseare divisible byk+1{\displaystyle k+1}. It comes down to the first termc0∫0∞fk(t)e−tdt{\displaystyle \textstyle c_{0}\int _{0}^{\infty }f_{k}(t)e^{-t}\,\mathrm {d} t}. We have (seefalling and rising factorials)fk(t)=tk[(t−1)⋯(t−n)]k+1=[(−1)n(n!)]k+1tk+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. HenceP≡c0∫0∞fk(t)e−tdt≡c0[(−1)n(n!)]k+1k!(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 primek+1{\displaystyle k+1}, therefore that product is not divisible byk+1{\displaystyle k+1}, and the same holds forP{\displaystyle P}; in particularP{\displaystyle P} cannot be zero.

Lemma 2

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For sufficiently largek,|Qk!|<1{\displaystyle \left|{\tfrac {Q}{k!}}\right|<1}.

Proof. Note that

fke−x=xk[(x−1)(x−2)⋯(x−n)]k+1e−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}}}

whereu(x),v(x) arecontinuous functions ofx for allx, so are bounded on the interval[0,n]. That is, there are constantsG,H > 0 such that

 |fke−x|≤|u(x)|k⋅|v(x)|<GkH 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 composingQ is bounded, the worst case being

|∫0nfke−x d x|≤∫0n|fke−x| d x≤∫0nGkH d x=nGkH .{\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 sumQ as well:

|Q|<Gk⋅nH(|c1|e+|c2|e2+⋯+|cn|en)=Gk⋅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\ ,}

whereM is a constant not depending onk. It follows that

 |Qk!|<M⋅Gkk!→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.

Conclusion

[edit]

Choosing a value ofk that satisfies both lemmas leads to a non-zero integer(Pk!){\displaystyle \left({\tfrac {P}{k!}}\right)} added to a vanishingly small quantity(Qk!){\displaystyle \left({\tfrac {Q}{k!}}\right)} being equal to zero: an impossibility. It follows that the original assumption, thate can satisfy a polynomial equation with integer coefficients, is also impossible; that is,e is transcendental.

The transcendence ofπ

[edit]

A similar strategy, different fromLindemann's original approach, can be used to show that thenumberπ is transcendental. Besides thegamma-function and some estimates as in the proof fore, facts aboutsymmetric polynomials play a vital role in the proof.

For detailed information concerning the proofs of the transcendence ofπ ande, see the references and external links.

See also

[edit]
Set inclusions between thenatural numbers (ℕ), theintegers (ℤ), therational numbers (ℚ), thereal numbers (ℝ), and thecomplex numbers (ℂ)


Notes

[edit]
  1. ^Cantor's construction builds aone-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.[15]

References

[edit]
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  57. ^abWaldschmidt, Michel (2022)."Transcendental Number Theory: recent results and open problems".Michel Waldschmidt.

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