Louville’s Theorem and Transcendence Proof and its application for constructing a class of Transcendental Numbers Swarnava Mukhopadhyay [email protected] Chennai Mathematical Institute Plot No. H1 SIPCOT IT Park, PO. Padur,Siruseri-603 103 Tamilnadu,India. A Brief Introduction Louville’s theorem basically says that any algebraic number cannot be ap- proximated by a sequence of numbers convering to it after a certain degree and thus this thoerem can be used to prove the existance of Transcendental Number as well as produce a class of Transcendental Numbers.We will look at a proof of Louville’s theorem and also constructed some class of transcen- dental numbers. The theorem states that If x is an algebraic number then there exists c ≥ 0 such that for all integer p and natural q we have, c p |x − | qd 6 q where d is the degree of the irreducible polynomial which is satisfied by x. I will give a proof of Louville’s Theorem and also an estimate of the value of the constant c given in the statementand use the theorem for constructing a certain class of Transcendental Numbers. There are certain class of number known as the Louville’s number which can be con- structed out of the Louville’s Theorem but all transcendental numbers are not Louville’s number. Louville proved this theorem in 1844 using countinued fractions and then in 1854 he reformulated it removing the dependence of the continued fractions which en- abled him to publish a class of transcendental numbers by which he proved the existance of transcendental numbers both real and complex. Claim 1. The inclusion of Q in R is strict Proof. The statement means that ∃ x such that x ∈ R but x∈ / Q. So we have to find a number which is in R but it is not in√Q. √ √So we start by taking the number 2. We know that 2 ∈ R but we will show that 2 ∈/ Q. We consider the map from R → R given as follows f : R → R x → x2 − 2 It is a continous function from R to R. Now f(2) ≥ 0 and f(0) ≤ 0. So by the Intermediate Value Theorem we can say that it will take the value 0 within 1 √ √ the interval [0,2]. If 2 is rational,then we can express 2 as √ p 2 = q where p and q are mutually coprime. Squaring we get the following, 2 = p2q2 ⇒ p2 = 2q2 ⇒ p2 is even. 2 2 It follows that q is also even. Since q is even we can say that q √is also even. This contradicts the fact that p and q are mutually co-prime. Therefore 2 ∈/ Q. Thus the inclusion of Q in R is strict. Let z be a complex algebraic number. It is the root of a polynomial with co-efficients in Q. So z is also a root of that polynomial. If a polynomial p(x) ∈ R[x] where R is a integral domain and let F be its fraction field. If the polynomial p(x) is reducible in its fraction field F [x] then by Gauss Lemma it is also reducible over the ring R[x]. Let the polynomial be P (x). P (x) = (x − z)(x − z)Q(x) Suppose P is a product of irreducible polynomials. So by assumption we can decompose P into a product of irreducible polynomials. say P = P1.P2.P3. ··· .Pn But P (z) = 0 ⇒∃ i such that Pi = 0. We can conclude that Pi is irreducible. After chang- ing the notations and multiplying P by an integer we assume that P ∈ Z is irreducible. Let d be its degree.We are going to prove the following. Claim 2. Let pn where n ∈ and p and q belongs to such that. qn N n n N p lim n = z n→∞ qn We will prove that 1 pn 0 d ≤ 2| − z||P (z)| qn qn Proof. If A is a polynomial and B is a polynomial in Q[x]. Then let D bethe gcd(A, B). So D divides both A and B. Now Q[x] is a Euclidean Domain. So by Bezout’s Theorem we have can write D as D = AM + BN where M and N are also polynomials. Now since P is a irreducible polynomial which leads us to conclude that P 0 and P are co-prime. So P (z) = 0 ⇒ P 0(z) 6= 0 ⇒ |P 0(z)| ≥ 0 Otherwise it contradicts the irreducibility of P .Therefore P is the polynomial of degree d such that P (z) = 0,where z is a complex algebric number. Now let pn where n ∈ be the sequence of numbers converging to z as defined earlier qn N and N is the set of natural numbers. By Taylor’s Expansion we have, (x − z)2 (x − z)d P (x) = (x − z)P 0(z) + P 00(z) + ··· + P d(z) 2! d! 2 Since the polynomial is of degree d and it is also a smooth function too. Its derivatives higher than d is zero. Putting x = pn in the above expression we have that. qn p p p p ( n − z)2 ( n − z)d P ( n ) = ( n − z)P 0(z) + qn P 00(z) + ··· + qn P d(z) qn qn 2! d! Taking modulus on both sides and applying triangle in-equality we get that the following expression. p p p p ( n − z)2 ( n − z)d |P ( n )| ≤ |( n − z)||P 0(z)| + | qn P 00(z) + ··· + qn P d(z)| qn qn 2! d! Since pn is close enough to z and since z is algebraic we have qn p P ( n ) 6= 0 qn d d−1 Let the irreducible polynomial satisfied by z is P (x) = a0x + a1x + ··· + ad. d Multiplying both sides of the above equation by qn we have that d pn d d−1 d qnP ( ) = a0pn + a1pn qn + ··· + adqn qn wherea0, a1, a2, a3, ··· , adare integers. So we have d pn qnP ( ) ∈ Z qn d d−1 d pn |a0pn + a1pn + ··· + anqn| |P ( )| = d qn qn pn 1 |P ( )| ≥ d qn qn d d . Since |a0pn + ··· + anqd| is a positive integer. Now p p n → z ⇒ P ( n ) → 0 qn qn d d X (x − z)i X |(x − z)i| | P i(z) | ≤ |P i(z)| i! i! n=2 n=2 Let pn = x and it converges to z. Suppose |x − z| ≤ δ qn d d X |(x − z)i| X δi |P i(z)| ≤ |P i(z)| i! i! n=2 n=2 |P i(z)| 2 ≤ (d − 1)max( i! )δ for δ ≤ 1 We choose δ such that |P i(z)| (d − 1)max( ).δ2 ≤ P 0(z).δ i! Since P 0(z) 6= 0 So we have a choice for δ and this is possible only because of the series coverges. |P 0(z)| δ ≤ |P i(z)| (d − 1)max( i! ) ∀ i ≥ 2 and we preferably choose δ 6= 0. |P 0(z)| δ ≤ min(1, ) |P i(z)| (d − 1)max( i! ) 3 Since pn converges to z. Given any δ ≥ 0 such that, qn p | n − z| ≤ δ qn Applying it the above equation and using x = pn we have, qn pn pn d d pn i ( q − z) ( q − z) X ( q − z) | n P 2(z) + ··· + n P d(z)| ≤ | n ||P i(z)| 2! d! i! i=1 p ≤ | n − z||P 0(z)| qn Replacing this result in the inequality derived by using Triangle Inequility we have the folowing, p p p |P ( n ) ≤ | n − z||P 0(z)| + | n − z||P 0(z)| qn qn qn p p |P ( n )| ≤ 2| n − z||P 0(z)| qn qn Using this we can write that, 1 pn pn 0 d ≤ |P ( )| ≤ 2| − z||P (z)| qn qn qn 1 1 pn d 0 ≤ | − z| qn 2|P (z)| qn Now from this Lemma the proof of the Louville’s theorem follows directly as some changing of terms is just requires.From the above Lemma we have also estimated the value of the constant c which we have proved that it can be found directly by computing the derivatives of the irreducible polynomial at z. We will now see some applications and use Louville’s theorem to construct some tran- scendental numbers. The essence and beauty of the Theorem lies in the fact that no algebraic number can’t be approximated after a certain degree by a sequence of numbers converging to it. P∞ 1 We consider the series n=1 10n! P∞ 1 Claim 3. The series n=1 10n! converges. Proof. We take the partial sums of the first n terms and check for the convergence. n X 1 s = n 10k! k=1 n X 1 ≤ 10k k=1 ∞ X 1 ≤ 10k k=1 1 10 = 1 1 − 10 4 1 = 9 The partial sums are therefore bounded and they are montonic thus they converge to a P∞ 1 finite value and the series n=1 10n! is thus convergent. Let ∞ X 1 z = 10n! n=0 Claim 4. z is transcendental. Proof. In the following expression, n X 1 s = n 10k! k=0 n n! P 10 k! s = k=0 n 10n! So we can replace s by pn where p and q are of the following form,we neednot need to n qn n n care what is pn. n! qn = 10 Assume that z is a algebraic number and then approach to prove it by contradiction that it is transcendental.