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 of finite degree with rational coefficients. The best-known transcendental numbers are π and e.[1][2] 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 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.[3][4][5][6] 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 non-rational, and transcendental real numbers.[3] 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.
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Transcription
SIMON PAMPENA: It's mind blowing. I learned this when I was at uni, the existence of transcendental numbers. And the name was a selling point. Because I was like, transcendental. You know, it's a time when you're really interested in, like, out of body experiences and whatnot. But the idea that mathematicians gave this name to numbers, numbers, these are numbers that you're familiar with. Like pi, you can write down as a decimal expansion. You'll never get it right, but it's just a number that you're familiar with. Has this property that we just didn't know about. OK, we're going to play a game, and we're going to try and understand transcendental numbers with this game. The game is reducing down numbers to 0. That's what you want to do. So the rules are you can only use whole numbers to do it, and you can add, take, multiply, and put the whole thing to any power you like, but it has to be a whole number power. OK, so let's play the game. OK, so do you have a favorite number? BRADY HARAN: Well, I like the number 10. SIMON PAMPENA: 10? BRADY HARAN: Yeah, but that seems like quite an easy one. SIMON PAMPENA: Sure. That's fine. You mean 10 in base 10? BRADY HARAN: 10 in base 10. SIMON PAMPENA: Yeah, OK. So 10 in base 10. OK, here we go. Let's start the game. So we want to get this down to 0, so the first thing we could do is multiply it by 0. But that you can do with any number, because any number times 0 is-- BRADY HARAN: 0. SIMON PAMPENA: Bingo. You can do that, but that's not very interesting. But what is interesting is that if we try and use these rules, we can go, OK, what happens if I take away 10 from this? We're done. So there you go. So that sounds kind of trivial, but it's a really good start. So we used a whole number, and we used the take. What about something else? How about 3/4? First of all, let's multiply it by 4. OK, so these things will cancel. You get 3. Now we can take away 3, we'll get 0. Excellent. But what about something crazier? What about like a really crazy number? What about like, the square root of 2? I think you guys know about the square of 2. BRADY HARAN: Yes, we do. That's irrational, isn't it? SIMON PAMPENA: It's an irrational number, and irrational means it can't be expressed as a fraction. So the square root of 2 is kind of a very strange number, and so this little thing here, I often say this little thing here is like a little sentence. It says, what number multiplied by itself gives you this number? That's the way I think of the square root sign. So I don't know what number multiplied by itself gives me 2, but that doesn't matter. Now, what we'll do is to try and get this one down to 0, OK? First of all, we'll have to-- BRADY HARAN: OK, that, I reckon I can do that. SIMON PAMPENA: Well, tell me. BRADY HARAN: I reckon if we raise that to a power-- SIMON PAMPENA: Yep. What power? BRADY HARAN: Let's raise it to the power of 2? SIMON PAMPENA: Correct, so that's multiplying it by itself. And then what do you get in the middle? BRADY HARAN: You're going to get 2, I'd bet. SIMON PAMPENA: That's right. So now you've got 2 in there, so what are you going to do? BRADY HARAN: Subtract 2. SIMON PAMPENA: Yes! So look at that. So you've just taken an irrational number, and with this game you've brought it down to 0. How about the square root of negative 1? We've gone from numbers that you know and love to fractions, OK, to irrational. This is irrational numbers. Now we've gone into what they call complex, or some people call imaginary, which is a terrible name for it. OK, so what can you do to this one here to try and get it down to 0? BRADY HARAN: Well, I'm just going to square it and add 1. SIMON PAMPENA: There you go. No flies on you, mate. So there you go. So we've been able to play this game with three or four very different types of numbers, quite special. But what about something else? What about the square root of 2 plus the square root of 3? What can you do with that? So, let's see. The square root of 2 plus the square root of 3. Now we're gonna square it. OK, so this is a little bit of high school maths. 2 plus 2 times this by this, which is 2 the square root of 2 times square root of 3 plus this squared. So that's 3. So this reduces down to 5 plus 2 by the square root of 2 by the square root of 3. OK, so this is what we've done. We've done that there, but look what's popped out. A number that we can use, a whole number. So what we'll do is on this side, we'll go 5 plus 2 by the square root of 2 by the square root of 3, and now we'll take away 5. So we'll end up getting 2 by the square root of 2 by the square root of 3. And this is good, because there's no plus sign in the middle. What can we do next? Well, we're going to square all of that. So 2 squared is 4, and the square root of 2 squared is 2. And the square root of 3 squared is 3. OK, so that dot is another way of saying times. And so that one is 2/8, two 4's are eight, eight 3's 24, done. So if we go 24, take 24, boom, we get down to 0. What I wanted to show you, the reason why I wanted to show you this is because all these different numbers look very complicated, unrelated, but let me show you. Now, let's replace all the numbers we put in with x. x take 10 is 0, 4x take 3 is 0, x squared take 2 is 0, x squared plus 1 is 0, and this one is x squared take 5, all squared, take 24 equals 0, which if we expand out, so look. These all look like algebra problems. So what we did was in our game, we picked numbers, and we tried to get them to 0. But the opposite could have been here, let's solve for x. Now, this is the stuff that you get taught in school. This is algebra, and it so happens that the family that all these numbers belong to, even the square root of negative 1, is algebraic numbers. So we've actually found a home for some of the biggest stars of maths, the numbers that have caused huge problems and schisms, what is the square root of negative 1? Square root of 2 from the ancient times, the Pythagorean times. People died because of this number. But somehow we've found a family for these numbers, algebraic numbers. OK. So next, we're going to need another sheet of paper. We've chosen some numbers. What about a special number? What about e? Now this number here-- if you're not familiar with it-- this number is a fantastic number for maths. And what it is is that if it's a function, a function of e to the x, e to any number that you raise it to, OK? On the graph, when you graph it like so, the y value is also at the slope of the tangent at that point. So it's really, really important to natural growth. It's like a really fantastic number. It means a lot to life, really, but it's actually a super crazy number. Super, super crazy. One of the expressions I can show you for it is actually an infinite sum. So I'm going to blow you away. It's 1 plus 1 on 1 plus 1 on 2 plus 1 on 6 plus 1 on 20-- anyway, it keeps going forever and ever. But can we play the game with this number? Can we bring this number down to 0 using the rules of our game? BRADY HARAN: Can we do it with algebra? SIMON PAMPENA: Can we do it with algebra? That's right. BRADY HARAN: All right. Can we? SIMON PAMPENA: Well, for ages and ages and ages, e's been around for about 400 years. Nobody really knew. I mean, this number is really, really important, and no one knew. It so happens, we can't. BRADY HARAN: It can't be done. SIMON PAMPENA: It can't be done, and I'll show you why. Well, I'll kind of try and show you why, because it's actually really tricky. But it was a guy called Charles Hermite, and he basically showed-- right, so I'm going to use these symbols here, because I don't know what the formula will be. He basically showed if you try and play the game, right, bringing e to any power that you want, whole power and timesing it by any whole number. So if you claim that there does exist some bit of algebra that you can bring it down to 0, he showed that you'll lead to a contradiction. Basically, he showed that there was a number, a whole number that existed between 0 and 1. Obviously, there's not. Obviously, there's not. But this is what you do in maths, is that if you want to show you something is impossible, you kind of assume that it's true, and then you show that it creates a contradiction. So this is amazing. So this is what Hermite discovered, and this is really, really a fantastic-- I mean, everyone should be excited by this, because e is not algebraic. So what number is it? Well, it somehow transcends what we're capable of doing. The thing with algebra is that's how we build numbers. Like, that's our world is built with algebra. Like, any number that you kind of deal with in your everyday life has a lot to do with algebra. You're just adding, taking, dividing, things to the power, but e is not. So it somehow transcends maths. So that's what they called it. e is transcendental. It's actually [INAUDIBLE] show you other than e. You know why, is because-- well, this is the interesting thing. e wasn't the first transcendental number. They discovered a transcendental number, Liouville, I think his name is, discovered a transcendental number quite a long way before this, 30 years before this. But it was like, through construction. So he was actually trying to find a number based on the rules of the game that didn't fit. What's special is that e was already, it's already a superstar of maths, e. Like, people knew about it. So this was an extra piece of information. But then people asked this question. What about pi? BRADY HARAN: Superstar. SIMON PAMPENA: The superstar. This is the superstar of math. 2,000 years old. What is pi? Is pi algebraic or transcendental? So you've got to imagine as a mathematician, OK, you love pi. Like, it comes with the territory. You cannot not like pi. So this is the thing, is that you could actually add to the knowledge of pi. You could add something new, which is incredible. I mean, I would die a happy man if I could do that. So this question came up, what is pi? Is it algebraic or transcendental? And so it was about, probably 1880s that a guy called Lindemann actually came up with the answer. He showed, and again, this is a very tricky thing that he showed, he showed e raised to any algebraic number is transcendental. So for example, e to the 1, e. That's a good thing, because e should be transcendental, because it's already been proven. Because 1 is algebraic. Your favorite number, Brady, e to the 10. That's transcendental, right? e to the square root of 2, e to the i. Right? What about pi? So how could you use this fact here, e to the a, so a is any algebraic number, is transcendental? How can you use that fact to show that pi is transcendental? OK, so this is the thing. Again, it's a proof by contradiction. So, this is what he did. He said, let's assume pi is algebraic. So pi is algebraic. That means there's a formula for it. OK, what's that formula? Who knows? Because it doesn't exist. But as an example, if you're an engineer, you'd say, oh, yeah, pi, 22 on 7. All right? OK, cool. So that means pi times 7 take 22 equals 0. Right? As an example. That's not true, by the way. There's no way I'm claiming that to be true. Don't you dare cut it and say Simon thinks that's true. It's not true. Pi, 22 on 7. Pi, 22 on 7. I know pi to quite a few decimal places, and that's obviously not true. And an actual fact, just so I can tell you, another really nice approximation of pi is the cube root of 31. It's actually pretty close. So that could be another formula. So that means if we cube that and take away 31, that equals 0. OK, so we've got like these phony equations. This is the big kicker. This is the big kicker. We're going to use another superstar equation, OK? e to the i pi equals negative 1. So this is Euler's identity. It's a famous one, isn't it? But look at it. Look what it says. e raised to the i pi is negative 1. Now, i pi, OK, if we assume pi is algebraic, that means i pi must be algebraic. So e to an algebraic number has to be transcendental. But is negative 1 transcendental? It's not, because we can play the game, and we can get it to 0. So by using another increase piece of maths in your formula, imagine this is like you're making a film, like you're doing a maths film, and you've just got the biggest Hollywood star in the world to start in it. In your proof. Starring in your proof. So this here, e to the i pi is negative 1. If indeed this was algebraic, this would have to be transcendental, so that means i pi cannot be algebraic. And who's the culprit? Well, it's not the square root of negative 1. It's pi. So pi cannot be algebraic, which means pi must be transcendental. So there's something really tricky going on, and that's why I like it. Because the tricky stuff is where all the awesome maths is. In maths, perfection is important. But then, anyone who uses maths-- for physics or chemistry, or whatever you want to do-- then they can kind of use approximations. I'm not interested in approximations.
History
The name "transcendental" comes from Latin trānscendere 'to climb over or beyond, surmount',[7] 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 .[8] Euler, in the 18th century, was probably the first person to define transcendental numbers in the modern sense.[9]
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.[10]
Joseph Liouville first proved the existence of transcendental numbers in 1844,[11] and in 1851 gave the first decimal examples such as the Liouville constant
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.[12] 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.[13]
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.[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 1882, Ferdinand von Lindemann published the first complete proof that π is transcendental. He first proved that ea is transcendental if a is a non-zero algebraic number. Then, since eiπ = −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. That π is transcendental implies that geometric constructions involving compass and straightedge 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 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).[16]
Properties
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 a rational number is the root of an integer polynomial of degree one.[17] 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 , , , 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, 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) = x2 − (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 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[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, see Hermite's problem).
Numbers proven to be transcendental
Numbers proven to be transcendental:
- ea if a is algebraic and nonzero (by the Lindemann–Weierstrass theorem).
- π (by the Lindemann–Weierstrass theorem).
- eπ, Gelfond's constant, as well as e−π/2 = ii (by the Gelfond–Schneider theorem).
- ab where a is algebraic but not 0 or 1, and b is irrational algebraic (by the Gelfond–Schneider theorem), in particular:
- , the Gelfond–Schneider constant (or Hilbert number)
- sin a, cos a, tan a, csc a, sec a, and cot a, and their hyperbolic counterparts, 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).[19]
- 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), in particular: the universal parabolic constant.
- logb a if a and b are positive integers not both powers of the same integer, and a is not equal to 1 (by the Gelfond–Schneider theorem).
- Non-zero results of arcsin a, arccos a, arctan a, arccsc a, arcsec a, arccot a and their hyperbolic counterparts, for any algebraic number a (by the Lindemann–Weierstrass theorem).
- The Bessel function of the first kind Jν(x), its first derivative, and the quotient are transcendental when ν is rational and x is algebraic and nonzero,[20] and all nonzero roots of Jν(x) and J'ν(x) are transcendental when ν is rational.[21]
- W(a) if a is algebraic and nonzero, for any branch of the Lambert W Function (by the Lindemann–Weierstrass theorem), in particular: Ω the omega constant
- W(r,a) if both a and the order r are algebraic such that , for any branch of the generalized Lambert W function.[22]
- √xs, the square super-root of any natural number is either an integer or transcendental (by the Gelfond–Schneider theorem)
- ,[23] ,[24] and .[24] The numbers and are also known to be transcendental. The numbers and are also transcendental.[25]
- The values of Euler beta function (in which a, b and are non-integer rational numbers).[26]
- 0.64341054629 ... , Cahen's constant.[27]
- .[28] In general, all numbers of the form are transcendental, where are algebraic for all and are non-zero algebraic for all (by the Baker's theorem).
- The Champernowne constants, the irrational numbers formed by concatenating representations of all positive integers.[29]
- Ω, Chaitin's constant (since it is a non-computable number).[30]
- The supremum limit of the Specker sequences (since they are non-computable numbers).[31]
- The so-called Fredholm constants, such as[11][32][b]
- which also holds by replacing 10 with any algebraic number b > 1.[34]
- , for rational number x such that .[28]
- The values of the Rogers-Ramanujan continued fraction where is algebraic and .[35] The lemniscatic values of theta function (under the same conditions for ) are also transcendental.[36]
- j(q) where is algebraic but not imaginary quadratic (i.e, the exceptional set of this function is the number field whose degree of extension over is 2).
- The values of the infinite series with fast convergence rate as defined by Y. Gao and J. Gao, such as .[37]
- The real constant in the definition of van der Corput's constant involving the Fresnel integrals.[38]
- The real constant in the definition of Zolotarev-Schur constant involving the complete elliptic integral functions.[39]
- Gauss's constant and the related lemniscate constant.[40]
- Any number of the form (where , are polynomials in variables and , is algebraic and , is any integer greater than 1).[41]
- Artificially constructed non-periodic numbers.[42]
- The Robbins constant in three-dimensional line picking problem.[43]
- The aforementioned Liouville constant for any algebraic b ∈ (0, 1).
- The sum of reciprocals of exponential factorials.[28]
- The Prouhet–Thue–Morse constant[44] and the related rabbit constant.[45]
- The Komornik–Loreti constant.[46]
- Any number for which the digits with respect to some fixed base form a Sturmian word.[47]
- The paperfolding constant (also named as "Gaussian Liouville number").[48]
- Constructed irrational numbers which are not simply normal in any base.[49]
- For β > 1
- where is the floor function.[50]
- 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.[51]
- The number , where Yα(x) and Jα(x) are Bessel functions and γ is the Euler–Mascheroni constant.[52][53]
- Nesterenko proved in 1996 that and are algebraically independent.[25] This results in the transcendence of the Weierstrass constant[54] and the number .[55]
Possible transcendental numbers
Numbers which have yet to be proven to be either transcendental or algebraic:
- Most sums, products, powers, etc. of the number π and the number e, e.g. eπ, e + π, π − e, π/e, ππ, ee, πe, π√2, eπ2 are not known to be rational, algebraically irrational or transcendental. A notable exception is eπ√n (for any positive integer n) which has been proven transcendental.[56] At least one of the numbers ee and ee2 is transcendental, according to W. D. Brownawell (1974).[57] It has been shown that both e + π and π/e do not satisfy any polynomial equation of degree and integer coefficients of average size 109.[58]
- The Euler–Mascheroni constant γ: In 2010 M. Ram Murty and N. Saradha found an infinite list of numbers containing γ/4 such that all but at most one of them are transcendental.[59][60] In 2012 it was shown that at least one of γ and the Euler–Gompertz constant δ is transcendental.[61]
- Apéry's constant ζ(3) (whose irrationality was proved by Apéry).
- The reciprocal Fibonacci constant and reciprocal Lucas constant[62] (both of which have been proved to be irrational).
- Catalan's constant, and the values of Dirichlet beta function at other even integers, β(4), β(6), ... (not even proven to be irrational).[63]
- Khinchin's constant, also not proven to be irrational.
- The Riemann zeta function at other odd positive integers, ζ(5), ζ(7), ... (not proven to be irrational).
- The Feigenbaum constants δ and α, also not proven to be irrational.
- Mills' constant and twin prime constant (also not proven to be irrational).
- The second and later eigenvalues of the Gauss-Kuzmin-Wirsing operator, also not proven to be irrational.
- The Copeland–Erdős constant, formed by concatenating the decimal representations of the prime numbers.
- The relative density of regular prime numbers: in 1964, Siegel conjectured that its value is .
- has not been proven to be irrational.[25]
- Various constants whose value is not known with high precision, such as the Landau's constant and the Grothendieck constant.
Related conjectures:
Proofs for specific numbers
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:
By splitting respective domains of integration, this equation can be written in the form
Lemma 1
There are arbitrarily large k such that is a non-zero integer.
Proof. Recall the standard integral (case of the Gamma function)
- if then .
This would allow us to compute exactly, because any term of can be rewritten as
Smaller factorials divide larger factorials, so the smallest occurring in that linear combination will also divide the whole of . We get that from the lowest power term appearing with a nonzero coefficient in , but this smallest exponent is also the multiplicity of as a root of this polynomial. is chosen to have multiplicity of the root and multiplicity of the roots for , so that smallest exponent is for and for with . Therefore divides .
To establish the last claim in the lemma, that is nonzero, it is sufficient to prove that does not divide . To that end, let be any prime larger than and . We know from the above that divides each of for , so in particular all of those are divisible by . It comes down to the first term . We have (see falling and rising factorials)
Lemma 2
For sufficiently large k, .
Proof. Note that
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
So each of those integrals composing Q is bounded, the worst case being
It is now possible to bound the sum Q as well:
where M is a constant not depending on k. It follows that
finishing the proof of this lemma.
Conclusion
Choosing a value of k 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.
The transcendence of π
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.
See also
- Transcendental number theory, the study of questions related to transcendental numbers
- Transcendental element, generalization of transcendental numbers in abstract algebra
- Gelfond–Schneider theorem
- Diophantine approximation
- Periods, a set of numbers (including both transcendental and algebraic numbers) which may be defined by integral equations.
Notes
- ^ 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.[15]
- ^ 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.[33]
References
- ^ Pickover, Cliff. "The 15 most famous transcendental numbers". sprott.physics.wisc.edu. Retrieved 2020-01-23.
- ^ Shidlovskii, Andrei B. (June 2011). Transcendental Numbers. Walter de Gruyter. p. 1. ISBN 9783110889055.
- ^ a b Bunday, B. D.; Mulholland, H. (20 May 2014). Pure Mathematics for Advanced Level. Butterworth-Heinemann. ISBN 978-1-4831-0613-7. Retrieved 21 March 2021.
- ^ Baker, A. (1964). "On Mahler's classification of transcendental numbers". Acta Mathematica. 111: 97–120. doi:10.1007/bf02391010. S2CID 122023355.
- ^ Heuer, Nicolaus; Loeh, Clara (1 November 2019). "Transcendental simplicial volumes". arXiv:1911.06386 [math.GT].
- ^ "Real number". Encyclopædia Britannica. mathematics. Retrieved 2020-08-11.
- ^ "transcendental". Oxford English Dictionary. s.v.
- ^ Leibniz, Gerhardt & Pertz 1858, pp. 97–98; Bourbaki 1994, p. 74
- ^ Erdős & Dudley 1983
- ^ Lambert 1768
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External links
- Weisstein, Eric W. "Transcendental Number". MathWorld.
- Weisstein, Eric W. "Liouville Number". MathWorld.
- Weisstein, Eric W. "Liouville's Constant". MathWorld.
- "Proof that e is transcendental". planetmath.org.
- "Proof that the Liouville constant is transcendental". deanlmoore.com. Retrieved 2018-11-12.
- Fritsch, R. (29 March 1988). Transzendenz von e im Leistungskurs? [Transcendence of e in advanced courses?] (PDF). Rahmen der 79. Hauptversammlung des Deutschen Vereins zur Förderung des mathematischen und naturwissenschaftlichen Unterrichts [79th Annual, General Meeting of the German Association for the Promotion of Mathematics and Science Education]. Der mathematische und naturwissenschaftliche Unterricht (in German). Vol. 42. Kiel, DE (published 1989). pp. 75–80 (presentation), 375–376 (responses). Archived from the original (PDF) on 2011-07-16 – via University of Munich (mathematik.uni-muenchen.de ). — Proof that e is transcendental, in German.
- Fritsch, R. (2003). "Hilberts Beweis der Transzendenz der Ludolphschen Zahl π" (PDF). Дифференциальная геометрия многообразий фигур (in German). 34: 144–148. Archived from the original (PDF) on 2011-07-16 – via University of Munich (mathematik.uni-muenchen.de/~fritsch).