TRANSCENDENTAL NUMBERS

TRANSCENDENTAL NUMBERS JEREMY BOOHER. Introduction The Greeks tried unsuccessfully to square the circle with a comass and straightedge. In the 9th century, Lindemann showed that this is imossible by demonstrating that π is not a root of any olynomial with integral coefficients. Such numbers are said to be transcendental. Although the distinction between algebraic and transcendental numbers seems similar to the distinction between rational and irrational numbers, transcendental numbers are considerably more mysterious. For examle, it is a standard fact that the algebraic numbers are countable, while the transcendentals are uncountable. Although there are more transcendental numbers, it is surrisingly difficult to exhibit any transcendental number, let alone show a naturally occurring number like e or π is transcendental. We will first discuss Liouville s theorem, which shows that certain secially constructed numbers are transcendental. We then follow Hermite in roving that e is transcendental. The idea is to aroximate the function e z by rational functions, known as Padé aroximants, and use these good aroximations to show e cannot be algebraic. The actual roof is short and self-contained, but crytic in isolation. We then illustrate how Padé aroximants give a natural derivation of the continued fraction exansion of e, following Cohn [3]. Finally, we rove the Lindemann-Weierstrass theorem and as a consequence shows that π is transcendental, following the standard method as resented in [4] or [2]. 2. Aroximation and Transcendental Numbers We first rove a classical aroximation theorem for algebraic numbers which will let us show that certain exlicit, secially constructed numbers are transcendental. Theorem (Liouville). Let β be a solution to a olynomial f(x) = i a ix i of degree d with integral coefficients. Then there exists a constant δ, deending on β, such that for any rational number q β with q >, we have > δ q d. Proof. We factor f(x) = a d (x β) m (x α ) m... (x α n ) mn over the comlex numbers. Let γ be the minimum distance between β and any of the α i. For any δ < γ, no root α i of f(x) satisfies α i β < δ q d < γ. Thus we may assume that q is not a root, so qd f( q ) is a non-zero integer. Thus q d a + q d a +... + a d d Date: 22, udated Aril 7, 26. In contrast, it is elementary to find families of irrational numbers, like n for n square-free.

2 JEREMY BOOHER and hence using the factorization that m () a d q d ( q α ) m... ( q α n). mn If > γ and δ < γ, the roof is finished. So suose that that q β γ. Under this assumtion, letting ρ be the maximum of α i β we see that α i q α i β + β q ρ + γ This gives the estimate q α m... q α n n m (ρ + γ) d m. Taking δ to be ositive but less than both γ and ( a d (ρ + γ) d m) m m δm and hence after taking mth roots that Thus no rational number satisfies q d δ q d/m δ q d. < δ q d. and using () gives This theorem can be used to exhibit an exlicit transcendental number. Consider the number β = k= 2 k!. If it were algebraic, then it would be the root of a degree d olynomial. By Liouville s theorem, there would be a δ > such that no rational number q satisfies < δ q d. This is not ossible as the non-zero binary digits of β are far aart, so aroximation. More recisely, let q = 2 K! and = = k=k+ 2 k! k=(k+)! n 2 k! will be too good an k= K 2 K! k!. Then we have k= If we ick K so K > d and 2 2 K! < δ, we get 2q q K δ q d. 2 k = 2 (K+)! 2 = 2q (K+). This contradicts Liouville s theorem, and hence β is transcendental. Liouville s theorem is a simle way to exhibit transcendental numbers. It is more work to show natural constants such as e and π are transcendental.

TRANSCENDENTAL NUMBERS 3 3. The Transcendence of e This section will establish that e is a transcendental number. We first resent the elementary argument that e is irrational, and then look at the theory of Padé aroximants to rove that e r is irrational for r Q\{}. We then simlify the roof and use its essential features to rove that e is transcendental. 3.. The irrationality of e. It is very easy to show that e is irrational using its definition as Suose e = a b e := n=. with a and b relatively rime integers. Then for n b, ( ) n N := e k! is an integer because e and k! are integers. However, estimating N shows that < N = n + + (n + )(n + 2) +... < n + + (n + ) 2 +... = n which is imossible. Thus e is irrational. 3.2. Interlude. The above argument demonstrates the general method of roving that numbers are transcendental: first use analytic roerties to aroximating the number, and then suose the number is algebraic and use the assumtion to show the error is an integer between and. Sometime it is more convenient to use the assumtion to show the estimate is a non-zero multile of a large integer but show the error cannot be that big. The analytic content used to rove that e is irrational was the ower series exansion of e z around z =. Unfortunately, this does not generalize to showing e is transcendental, or even that related numbers like e 5 are irrational. The aroximation of e z by a truncated ower series is too weak to readily give roofs of these facts. Power series determine holomorhic functions which are the basic building blocks of the world of comlex analysis. Meromorhic functions are the next simlest object, so we can hoe to better aroximate e z using rational functions, just as ratios of integers rovide better aroximations to real numbers than integers do. 3.3. Aroximations by Rational Functions. Hermite was able to use the idea of aroximating e z by rational functions to show that e is transcendental. The theory of aroximating general ower series was first studied systematically by Hermite s student Padé near the end of the nineteenth century, so these rational functions are called Padé aroximants desite the secial cases we care about being used earlier. Definition 2. For an analytic function f(z) and air (n, m) of non-negative integers, a Padé aroximant of order (n, m) at is a rational function R(z) = P (z) Q(z) with deg(p ) n and deg(q) m such when exanded as ower series around we have k= f(z) R(z) = O(z n+m+ ). Padé aroximants are analogous to rational aroximations of irrational numbers, so we hoe for similarities with the theory of aroximating real numbers by rationals. From the theory of continued fractions, we know:

4 JEREMY BOOHER Fact 3. Let α be an irrational number. For any n, there exists a rational number q with and q relatively rime integers and q > n such that q α < q 2. The quantity qα is an estimate in the error of the aroximation relative to the size of the denominator, and goes to zero as the size of the denominator becomes large. For a Padé aroximant P (z) Q(z) of f(z), the analogous term is P (z) Q(z)f(z) and being small means vanishing to a large order at z =. We are mainly interested in Padé aroximants for e z, so instead of dealing with the general theory we will show that they are unique and then construct Padé aroximants for e z directly. Proosition 4. The Padé aroximant of order (n, m) is unique u to scaling P and Q. Proof. Suose P (z) Q (z) and P 2(z) Q 2 (z) are both Padé aroximants to the same function, both of order (n, m). Then when exanding them as ower series around, we have P (z) Q (z) P 2(z) Q 2 (z) = O(zm+n+ ) Combining the fractions, we see that since Q (z) and Q 2 (z) are nonzero at z =, the order of vanishing of P (z)q 2 (z) P 2 (z)q (z) must be at least m+n+. But this is a olynomial of degree at most m + n, so it vanishes identically. Thus P = cp 2 and Q = cq 2. Since there are m + n + 2 coefficients to ick and P (x) Q(x) deends only on the olynomials u to scalars, we exect to be able to force m + n + coefficients to agree. So general existence theorems for Padé aroximants are not surrising. However, we simly need to understand the aroximants for e z. One simle way to do this is to reformulate definition of Padé aroximants in terms of comlex analysis. Clearing the denominator in the definition, we can equally well require that the olynomials P (z) and Q(z) satisfy as ower series around, or equivalently that P (z) e z Q(z) = O(z m+n+ ) P (z) e z Q(z) z m+n+ is holomorhic at z =. To find and q, the key is exerience with evaluating integrals involving the exonential function multilied by a olynomial. For examle, xe z xz dx = z xez xz + In general, if r(x) is a olynomial of degree n, then e z xz r(x)e z xz dx = P (z)ez Q(z) z n+ z dx = ez z z 2. where Q(z) and P (z) are olynomials obtained via reeated integration by arts. To find Padé aroximants, we need the olynomials P (z) and Q(z) roduced by integration by arts to have low degrees. This requires a secial choice of r(x). Proosition 5. For any olynomial r(x) of degree m + n, r(x)e z xz dx = F (z, )ez F (z, ) z m+n+

TRANSCENDENTAL NUMBERS 5 where F (z, x) is the sum (2) F (z, x) := r(x)z m+n + r (x)z m+n +... + r (n+m) (x). In articular, for any m and n, there exist a olynomial r(x) of degree m + n such that where deg(p ) m and deg(q) n. r(x)e z xz dx = P (z)ez Q(z) z m+n+ Proof. The formula for F can be found by reeated integration by arts, but we can just verifying it directly: d ( F (z, x)e z xz ) = dx n+m i= ( r (i) (x)z m+n+ i e z xz r (i+) (x)z m+n i e z xz) = r(x)z m+n+ e z xz. Selecting r(x) so that r() = r () =... = r (n) () = and r() = r () =... = r (m) () = forces the degree of P (z) = F (z, ) and Q(z) = F (z, ) to be at most m and n resectively. To do this, take r(x) = x n ( x) m. Corollary 6. For any air (n, m) there exist a Padé aroximant of order (n, m) for e z. 3.4. The Irrationality of Nonzero Rational Powers of e. These basic ideas about Padé aroximants rovide a method (following []) to show owers of e are irrational. Proosition 7. Let a be a non-zero rational number. Then e a is irrational. Proof. It suffices to rove this when the exonent is a ositive integer, since owers of rational numbers are rational. Let e = a b with a, b ositive and relatively rime. By our analogy with aroximation of rational numbers, we want to take z = and use the fact that P () Q()e is an estimate on the error. Combined with the hyothesis that e is rational, it will yield a contradiction. The error is closely related to the integrals of the revious section. If we set r(x) = x n ( x) n, the relevant integral becomes r(x)e x dx = F (, )e F (, ) 2n+. Using the assumtion that e r is rational, a nice bound is af (, ) bf (, ) = b 2n+ e x x n ( x) n dx b 2n+ e = a 2n+. However, we know that r (k) () = r (k) () = for k < n. Thus the only derivatives aearing in (??) are at least the nth derivative. Since r(x) has integral coefficients the nth and further derivatives at and must be multiles of by Taylor s formula. Thus af (, ) bf (, ) is a multile of. It is not zero since the integral is clearly ositive. If we chose n so that > a 2n+ (ossible since > e ( ) n n) e we get a contradiction. Thus e is irrational. As we attemt to generalize this, there is one simlification to kee in mind. We argued that F (, ) and F (, ) are multiles of a large number ( in this case). This haens because the first n derivatives of f(x) are zero at z = and z =. Since f has integral coefficients, Taylor s formula imlies the higher derivatives are multiles of. It would be better to divide r(x) by so that F (, ) and F (, ) are integers and show the error is an integer between and than to try to kee track of divisibility. This will be imlemented in the next section as we show e is transcendental.

6 JEREMY BOOHER 3.5. Transcendence of e. Suose e is algebraic, so there exists integers b i Z such that b + b e +... + b d e d =. We want to aroximate e, e 2,..., e d by rational functions, and then evaluate the error terms. One way to aroximate the owers of e would simly be to use Padé aroximants for each one. But they will be unrelated, and the argument becomes unnecessarily comlicated. 2 The correct aroach is a generalization of the roblem of simultaneous aroximation of irrationals. Fact 8 (Dirichlet). Given α,..., α r in R and n >, there exist,..., r, q Z with < q n such that i q α i <. The key is to aroximate the irrationals by fractions of the same denominator. As before, the quantity i qα i is an estimate of the error relative the size of the denominator that becomes small as q becomes large. By analogy, we need to aroximate e, e 2,..., e d by rational functions with the same denominator. We look at integrals of the form qn r r(x)e z xz dx = F (z, )ez F (z, ) z n+ If we simly try lugging in different values of z, the denominator changes. To get the same denominator, we need to fix z and change the uer bound of integration. This will relace the e by e m while keeing the F (z, ) term. Therefore we are led to look at m r(x)e m x dx = F (, )e m F (, m) as a measure of the error of aroximating e m by F (,m) F (,). So to set u the roof, define F (x) = r(x) + r (x) +... + r deg(r) (x) and I(m) := m Since we want the degrees of F (j) to be small for j =,, 2..., d, take r(x) = (n )! xn ((x )(x 2)... (x d)) n. r(x)e m x dx = F ()e m F (m). This is very similar to the choice of x n ( x) n from the irrationality argument, excet we look at more oints and we simlify by dividing by (n )!. If we aroximate e m by F (m) F (), the error is ( + b e F () ) ( +... + b d e d F (m) ). F () F () Multilying through by the common denominator F () gives + b I() + b 2 I(2) +... + b d I(d). as an estimate of the error. The term I(m) = F (m) F ()e m corresonds to the error term bα i a i when we aroximate the real number α i by a i b. Theorem 9. The number e is transcendental. 2 This aroach is needed to rove the Lindemann-Weierstrass theorem in section 5.. That roof uses the same ideas as in Hermite s roof but requires more work. The transcendence of e does follow immediately.

Proof. Suose e is algebraic and satisfies TRANSCENDENTAL NUMBERS 7 b + b e +... + b d e d = where the b i are integers. We have defined r(x) = (n )! xn ((x )(x 2)... (x d)) n We consider the error term F (x) = r(x) + r (x) +... + r (nd ) (x) I(m) = m e m x r(x)dx = F ()e m F (m). J : = b I() + b 2 I(2) +... + b d I(d) ( = b (F () F ()) + b (F ()e F ()) + b 2 F ()e 2 F (2) ) ) +... + b d (F ()e d F (d) = (b F () + b F () +... + b d F (d)). The first ste is to show that J is a nonzero integer. We will do this when n is a rime larger than b. First, a lemma. Lemma. Let m =,,..., d and k be a ositive integer. Then r (k) (m) is an integer. Unless k = n and m =, r (k) (m) is a multile of n. Proof. The derivative r (k) (m) is k! times the coefficient of (x m) k in the ower series exansion of r(x) around m. By the definition of r(x), it vanishes to order n at and to order n at, 2,..., d. Therefore when r (k) (m) is non-zero k n and hence the denominator is canceled out. Provided n is much larger than d, r n () is clearly the only derivative not a multile of n. This immediately shows that F (m) is an integer, and hence J is an integer also. To show J is a nonzero integer, note that all of the derivatives aearing in the sums F (), F ()... F (d) are multiles of n excet for r (n ) (). But since b < n and n is rime, the term b r (n ) () is not a multile of n. Therefore J is a non-zero integer. The second ste is to estimate the integral. If y is in the interval [, d] then r(y) (n )! (2d)n (2d) n... (2d) n = (n )! (2d)dn. As (n )! grows faster than c n, this is less than any ɛ > for sufficiently large n. But then m I(m) = r(x)e m x dx mem ɛ, so J b I() + b 2 I(2) +... + b d I(d) ɛ (b e + b 2 2e 2 +... + b d de d) which can be made less than by aroriate choice of ɛ. This contradicts the fact that J is a non-zero integer. Therefore e is transcendental. Corollary. For any non-zero rational number a, e a is transcendental. Proof. Suose β = e a = e q is algebraic where and q are integers. We may assume both are ositive. Then e satisfies the equation z β q which has algebraic coefficients. Since the algebraic numbers form an algebraically closed field, e is algebraic, a contradiction. Remark 2. The choice to use the different exonents n and n in r(x) = x n ((x )(x 2)... (x d)) n simlifies Hermite s original by making it clear that J is a non-zero integer.

8 JEREMY BOOHER 4. An Alication of Padé Aroximants to the Continued Fraction Exansion of e Continued fractions of rational numbers and of quadratic irrationals are finite and eriodic resectively. There are also surrising atterns in the continued fraction exansions of transcendental numbers like e. Padé aroximants rovide a way to find the continued fraction exansion of e. (Another way is to follow Euler and look at a differential equation called the Riccati equation.) A few minutes with a calculator will suggest that the continued fraction exansion of e is [2,, 2,,, 4,,, 6,,, 8,...]. (Sometimes this is re-written as [,,,, 2,,, 4,,,...] to extend the attern.) Using the standard notation for continued fractions, for i we set a 3i = 2i, a 3i+2 = a 3i+ =. Letting [a, a 2,..., a i ] = i q i, in general we have In this case, we can recursively calculate i+2 = a i+2 i+ + i and q i+2 = a i+2 q i+ + q i. 3n = 2n 3n + 3n 2, q 3n = 2nq 3n + q 3n 2 3n+ = 3n + 3n, q 3n+ = q 3n + q 3n 3n+2 = 3n+ + 3n, q 3n+2 = q 3n+ + q 3n. The question is whether lim n, a, a 2,..., a n ] = lim equals e. To check this, we can just n n q n check that lim q ne n =. n Using the connection with Padé aroximants, this should be given by an integral r(x)e x dx. Again, we have a choice of what to use for r(x). If we take r(x) = xn ( x) n, the integral evaluates to some of the convergents of e. Denote the integral by I(n). For examle, n = gives 3 e, n = 2 gives 7e 9, and I(3) = 93 7e. The second, fifth, and eighth convergents to e are 3, 9/7, and 93/7. Where are the other convergents coming from? Our choice of r(x) gives a good aroximation because the degree of the numerator and denominator are balanced. We would still exect a good aroximation if the degrees where n and n +. Define J(n) and K(n) to be J(n) := x n+ ( x) n e x dx and K(n) := Lemma 3. The integrals are related to the convergents of e as follows: I(n) = ( ) n (q 3n e 3n ) J(n) = ( ) n+ (q 3n e 3n ) K(n) = ( ) n (q 3n+ e 3n+ ). x n ( x) n+ e x dx. Proof. We will rove the results by showing that the integrals satisfy the same initial conditions and recurrence relations as the convergents. Now we have that I() = J() = K() = Furthermore, integration by arts gives that I(n + ) = x n+ ( x) n+ e x dx = (n + )! x( x)e x dx = 3 e x 2 ( x)e x dx = 3e 8 x( x) 2 e x dx = 4e. x n+ ( x) n + x n ( x) n+ e x dx = J(n) K(n).

In addition, J(n) = TRANSCENDENTAL NUMBERS 9 x n+ ( x) n e x dx (n + )x n ( x) n nx n+ ( x) n = e x dx (2n + )x n ( x) n nx n ( x) n = e x dx = (2n + )I(n) + J(n ) = 2nI(n) + K(n ) Finally, elementary algebra shows that K(n) = x n ( x) n+ e x dx = x n ( x) n e x dx x n+ ( x) n e x dx = I(n) J(n). In articular, these relations among the integrals are the same recurrence relations as the numerator and denominator of the convergents. It is now easy to derive the continued fraction exansion of e. Theorem 4. We have that e = [2,, 2,,, 4,,,...]. Proof. It suffices to show that lim n n q n e =. But this exression is given by I(m), J(m), or K(m) where m n/3. But as n the size of all three integrals goes to since the interval is a constant length, x j ( x) k e x is bounded, and the denominator goes to infinity. Therefore the limit is, so [2,,, 4,,, 6,...] gives a continued fraction exansion of e. 5. More General Transcendence Results It is ossible to rove that π is irrational and transcendental using arguments analogous to those for e. This is not surrising, as e πi = connects the two. However, it is also a consequence of a much more general theorem which Lindemann roved for n = and Weierstrass roved in general. Theorem 5 (Lindemann-Weierstrass). Let β,..., β n be algebraic numbers not all zero, and α,..., α n be distinct algebraic numbers. Then Corollary 6. The number π is transcendental. β e α + β 2 e α 2 +... + β n e αn. Proof. Suose π is algebraic. Then so is πi, and taking α =, α 2 = πi and β = β 2 = we have that + e πi. Corollary 7. If α is a non-zero algebraic integer, e α, sin(α), and cos(α) are transcendental. Proof. Suose e α satisfies a olynomial equation of degree n. Then letting α j = jα for j n and β j be the (algebraic) coefficient of x j, the theorem roduces a contradiction. For the other arts, write the trigonometric functions in terms of exonentials. The roof of Lindemann s theorem has two comonents. The first is an analytic argument similar to the roof that e is transcendental that establishes the theorem in the secial case that the β i are rational and that there are integers n = < n < n 2 <... < n r = n such that (3) β nt+ = β nt+2 =... = β nt+ and α nt+, α nt+2,..., α nt+ are all of the conjugates of α nt+. The second comonent is algebraic in nature, and uses ideas from the theory of symmetric olynomials to extend the first art to the general theorem.

JEREMY BOOHER 5.. The Analytic Part. With the hyothesis (3) on the α i and the β i given above, fix a b for which the bα i is an algebraic integer for each i. We need to aroximate e α iz by a quotient of rational functions, but the argument is now more comlicated because the α i are no longer just, 2,..., d. Trying a similar aroach, let (4) where is a large rime number. Define r i (x) = b n() (x α )... (x α n ) /(( )!(x α i )) I i (α) = α r i (z)e α z dz. This is a generalization of the I(m), and is well defined indeendent of the ath from to z because the integrand is holomorhic. Via integration by arts, we can evaluate this integral as (5) I i (α) = F i ()e α F i (α) where F i (z) = n k= r (k) i (z). Lemma 8. Let m =,..., n and k be a ositive integer. Then r (k) i (α m ) is an algebraic integer. Unless k = and m = i, r (k) i (m) is times an algebraic integer. This is essentially the same as Lemma. Finally define J i = β I i (α ) + β 2 I i (α 2 ) +... + β n I i (α n ). We will give bounds on J... J n using the hyothesis that β e α +... + β n e αn =. Using (5) on J i, the terms involving owers of e in J i disaear and we have J i = β F i (α ) β 2 F i (α 2 )... β n F i (α n ). By the lemma, all of the terms in the sums for F i (α j ) are times an algebraic integer excet for the term r ( ) i (α i ). Now recognize that r ( ) i (α i ) = b n (α i α k ) = (g (α i )) where g(x) = k n k i n (bx bα k ) Z[X]. k= If we now look at the roduct J J 2... J n, all but one term in the roduct will be times an algebraic integer. This term is the roduct of the terms from each of the J i that are not multiles of, and so is g (α ) g (α 2 )... g (α n ). It is a rational integer, since all of the g (α i ) are algebraic integers and it is invariant under the action of the Galois grou since the α i include comlete sets of conjugates. Since g (α )g (α 2 )... g (α n ) is indeendent of, for large it cannot be a multile of. Thus J J 2... J n is a non-zero algebraic integer. We will next show that J J 2... J n is rational by showing it is symmetric in α nt+,..., α nt+ (recall this is the grouing of the α i into sets of conjugates). So let σ be any ermutation of n t +,..., n t+ and extend it by the identity to be a ermutation on, 2,..., n. Now J J 2... J n = ( ) n n i= (β F i (α ) + β 2 F i (α 2 ) +... + β n F i (α n )).

TRANSCENDENTAL NUMBERS Unraveling the definition of F i (α j ), we see that alying σ to it gives F σ(i) (α σ(j) ). But then alying σ to the roduct and using the fact that β i Q, we obtain n ( β F σ(i) (α σ() ) +... + β n F σ(i) (β σ(n) ) ) n ( = βσ ()F i (α ) +... + β σ (n)f i (β n ) ) i= by re-indexing. By hyothesis β nt+ = β nt+2 =... = β nt+, so the roduct is invariant. Therefore it is a non-zero rational integer. However, estimating the integral as in the roof of the transcendence of e gives that n J... J n βk α k e αk r i (α k ) < i= for large enough as ( )! grows faster than c. But a non-zero rational integer cannot have absolute value less than, so our hyothesis that leads to a contradiction. i= β e α +... + β n e αn = 5.2. The Algebraic Part. We now reduce from the general case of Lindemann s theorem to the case (3) when the β i are rational and that there are integers n = < n < n 2 <... < n r = n such that β nt+ = β nt+2 =... = β nt+ and α nt+, α nt+2,..., α nt+ are a comlete set of conjugates. To reduce to the case that the β i are rational, we will inductively construct a sequence of non-zero olynomials F i with rational coefficients in 2n+ i variables such that F i (β i,..., β n, e α,..., e αn ) = β e α +... + β n e αn. To construct the F i, let F (x,..., x n, y,..., y n ) = x y +... + x n y n. In general, let F i (x i,..., x n, y,..., y n ) := F i (β, x i,..., x n, y,..., y n ) β where the roduct ranges over all β that are conjugate to β i. The definition is invariant under ermuting the conjugates of β i, so F i has rational coefficients. By induction, if F i has degree N i the coefficient of y N i is always non-zero. Next we exand out the olynomial F n+ (e α,..., e αn ) (of degree N) in the form β e α +... + β m e α m We wish to show that at least one of the coefficients β i is non-zero. Without loss of generality, we may assume that α has the largest absolute value of the α i. The exonents α i that arise are n those of the form c j α j where j c j = N. The only way to get an exonent of Nα is to ick j= c = N, c j = for j since α is the largest of all α j. Since the coefficient of y is non-zero, this imlies that the term e Nα has a non-zero coefficient. Thus we are in the case where the β i are rational, the α n are algebraic, and not all of the β i are zero. This exression is zero if the original exression was, so it suffices to rove Lindemann s theorem when all of the β i are rational. Multilying through by the common denominator, we may also assume the β i are rational integers. We now will further restrict to the case when all of the conjugates of α i aear with the same rational coefficient. Pick a olynomial with integral coefficients that has all of the α i as roots. Let α n+,..., α N be the additional roots of this olynomial. Extend the definition of β i to be zero for n < i N, and consider the roduct (β e ασ() +... + β N e ασ(n) ). σ S N

2 JEREMY BOOHER If β e α +... + β n e αn = the entire roduct is zero. If we were to exand the roduct, it would again be of the form β e α +... + β m e α m. Now if βe α arises as a term in this roduct, α = c α +...+c N α N where c +c 2 +...+c N = N! and β is a roduct of rational integers. There will also be terms with exonent c α σ() +... + c N α σ(n) for any ermutation. These terms will run through a comlete set of conjugates for α (along with other things) but have the same value of β. By grouing terms in this way, it is clear that the coefficient for each conjugate e α will be the same. Furthermore, at least one will be non-zero: in each roduct (β e α σ() +... + β N e α σ(n) ) ick the non-zero term with the largest value of α i, where the ordering is done first by real art and then by imaginary art. This term in the exansion will be non-zero, and cannot be canceled by anything else because each of the exonents is the largest ossible. This comletes the reduction to the case where (3) holds which we already roved using the analytic argument. 6. Later Results The two results resented here were known in the 9th century. Over the course of the 2th century, the integrals and use auxiliary functions r(x) were generalized considerably yielding more owerful results. The results are roven using the same strategies, but more comlicated functions. Details and roofs of the following can be found in Baker [2]. Theorem 9 (Gelfond-Schneider). Let α be an algebraic number not equal to or, and β an algebraic number that is not rational. Then α β is transcendental. Theorem 2 (Baker). If β, β..., β n and α..., α n are algebraic numbers, then is either zero or transcendental. β + β log(α ) +... + β n log(α n ) Baker s work also allows lower bounds to be constructed which suffice to solve the class number one roblem. Some other results and aroaches are resented in Lang [5]. References. Martin Aigner and Günter M. Ziegler, Proofs from The Book, fifth ed., Sringer-Verlag, Berlin, 24, Including illustrations by Karl H. Hofmann. MR 32889 2. Alan Baker, Transcendental number theory, second ed., Cambridge Mathematical Library, Cambridge University Press, Cambridge, 99. MR 74572 (9f:49) 3. Henry Cohn, A short roof of the simle continued fraction exansion of e, Amer. Math. Monthly 3 (26), no., 57 62. MR 22292 (26j:8) 4. Michael Filaseta, Lecture notes for math 785, Available at htt://www.math.sc.edu/~filaseta. 5. Serge Lang, Introduction to transcendental numbers, Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., 966. MR 24547 (35 #5397)