Prime Numbers and Irreducible Polynomials
... The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry. There are certain conjectures indicating that the connection goes well beyond analogy. For example, there is a famous conjecture of Buniakowski formu ...
... The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry. There are certain conjectures indicating that the connection goes well beyond analogy. For example, there is a famous conjecture of Buniakowski formu ...
Unique representations of real numbers in non
... set. Note first that from Proposition 1 it follows that Aq is nowhere dense. Furthermore, if we endow Σ with its natural weak topology (the topology of coordinate-wise convergence), then the map πq : Σ → R is continuous. From Lemma 4 it is easy to deduce that Uq is closed, whence Aq is closed as wel ...
... set. Note first that from Proposition 1 it follows that Aq is nowhere dense. Furthermore, if we endow Σ with its natural weak topology (the topology of coordinate-wise convergence), then the map πq : Σ → R is continuous. From Lemma 4 it is easy to deduce that Uq is closed, whence Aq is closed as wel ...
here
... and that the only ones in the left half-plane {s : ℜ(s) < 0} are the negative even integers s = −2, −4, −6, . . .. (These real zeros had been found by Euler more than a century earlier—see [1].) Turning his attention to the zeros in the closed strip {s : 0 ≤ ℜ(s) ≤ 1}, Riemann proves that they are s ...
... and that the only ones in the left half-plane {s : ℜ(s) < 0} are the negative even integers s = −2, −4, −6, . . .. (These real zeros had been found by Euler more than a century earlier—see [1].) Turning his attention to the zeros in the closed strip {s : 0 ≤ ℜ(s) ≤ 1}, Riemann proves that they are s ...
Proofs and Proof Methods
... • Chomp is a two-player game played with a set of cookies laid out in an m by n rectangular array. The top left cookie is “poisoned”, and the player who takes it loses. A legal move is to take any cookie, along with all the cookies to the right of it and below it. • We know that one player must have ...
... • Chomp is a two-player game played with a set of cookies laid out in an m by n rectangular array. The top left cookie is “poisoned”, and the player who takes it loses. A legal move is to take any cookie, along with all the cookies to the right of it and below it. • We know that one player must have ...
Section 2
... Fact: Since 2 is the only even prime, every odd prime is congruent to either 1 (mod 4) or 3 (mod 4), that is, if p is prime, then p 1 (mod 4) or p 3 (mod 4) . Some p 1 (mod 4) primes: 5, 13, 17, 29, 37, 41, 53, 61, … Some p 3 (mod 4) primes: 3, 7, 11, 19, 23, 31, 43, 47, … Theorem 2: Primes ...
... Fact: Since 2 is the only even prime, every odd prime is congruent to either 1 (mod 4) or 3 (mod 4), that is, if p is prime, then p 1 (mod 4) or p 3 (mod 4) . Some p 1 (mod 4) primes: 5, 13, 17, 29, 37, 41, 53, 61, … Some p 3 (mod 4) primes: 3, 7, 11, 19, 23, 31, 43, 47, … Theorem 2: Primes ...
![arXiv:math/0408107v1 [math.NT] 9 Aug 2004](http://s1.studyres.com/store/data/015366745_1-b96e81e8ee635380ae70d20316f65919-300x300.png)
arXiv:math/0408107v1 [math.NT] 9 Aug 2004
... b being relatively prime, is a B-number. We start with the assumption that there is a factor that is not a B-number. We then prove that for any number with that property, we can find a smaller positive number with the same property. Now, by the principle of infinite descent, there is no such number. ...
... b being relatively prime, is a B-number. We start with the assumption that there is a factor that is not a B-number. We then prove that for any number with that property, we can find a smaller positive number with the same property. Now, by the principle of infinite descent, there is no such number. ...
Oliver Johnson and Christina Goldschmidt 1. Introduction
... πG (z) = i c(i)z i gives the number of ways to colour G using exactly z colours. Read [14] conjectured that (|c(i)|, i ≥ 0) is a unimodal sequence and Welsh [21] conjectured, more strongly, that it is log-concave. Hoggar proved Theorem 1.4 in a partial attempt to resolve this conjecture. It remains ...
... πG (z) = i c(i)z i gives the number of ways to colour G using exactly z colours. Read [14] conjectured that (|c(i)|, i ≥ 0) is a unimodal sequence and Welsh [21] conjectured, more strongly, that it is log-concave. Hoggar proved Theorem 1.4 in a partial attempt to resolve this conjecture. It remains ...
A new proof of Alexeyev`s Theorem
... measure realizes the maximal spectral type of U . The proof in [1] uses spectral theory and some arguments from the classical theory of analytic functions of one complex variable. Later, using the same idea, Fra̧czek [2] extended Alexeyev’s Theorem for realization of the maximal spectral type by fun ...
... measure realizes the maximal spectral type of U . The proof in [1] uses spectral theory and some arguments from the classical theory of analytic functions of one complex variable. Later, using the same idea, Fra̧czek [2] extended Alexeyev’s Theorem for realization of the maximal spectral type by fun ...
On the rational approximation to the binary Thue–Morse–Mahler
... Section 13.4 of [1]). Since the irrationality exponent of ξt,b is equal to 2 (see [3]), the transcendence of ξt,b cannot be proved by applying Roth’s theorem. In the present note, we focus on the so-called Thue–Morse constant ξt := ξt,2 = 0.412454 . . . Open Problem 9 on page 403 of [1] asks whether ...
... Section 13.4 of [1]). Since the irrationality exponent of ξt,b is equal to 2 (see [3]), the transcendence of ξt,b cannot be proved by applying Roth’s theorem. In the present note, we focus on the so-called Thue–Morse constant ξt := ξt,2 = 0.412454 . . . Open Problem 9 on page 403 of [1] asks whether ...
Even and Odd Permutations
... Theorem. Let τ1 , τ2 , τ3 , . . . , τm be m transpositions and let π be a permutation. Then N (τ1 τ2 · · · τm π) ≡ N (π) + m ...
... Theorem. Let τ1 , τ2 , τ3 , . . . , τm be m transpositions and let π be a permutation. Then N (τ1 τ2 · · · τm π) ≡ N (π) + m ...
Ch 8 - ClausenTech
... exactly n linear factors. Once we have these n linear factors, we can use the Zero Product Property to find the n roots or solutions of the polynomial. Conjugate Root Theorem for Complex Roots If a polynomial P(x) of degree greater than or equal to 1 (with real coefficients) has a complex number as ...
... exactly n linear factors. Once we have these n linear factors, we can use the Zero Product Property to find the n roots or solutions of the polynomial. Conjugate Root Theorem for Complex Roots If a polynomial P(x) of degree greater than or equal to 1 (with real coefficients) has a complex number as ...
http://www.cmi.ac.in/~vipul/studenttalks/liouvillenumbers.pdf
... PO. Padur,Siruseri-603 103 Tamilnadu,India. A Brief Introduction Louville’s theorem basically says that any algebraic number cannot be approximated 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 wel ...
... PO. Padur,Siruseri-603 103 Tamilnadu,India. A Brief Introduction Louville’s theorem basically says that any algebraic number cannot be approximated 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 wel ...
Ch8 - ClausenTech
... exactly n linear factors. Once we have these n linear factors, we can use the Zero Product Property to find the n roots or solutions of the polynomial. Conjugate Root Theorem for Complex Roots If a polynomial P(x) of degree greater than or equal to 1 (with real coefficients) has a complex number as ...
... exactly n linear factors. Once we have these n linear factors, we can use the Zero Product Property to find the n roots or solutions of the polynomial. Conjugate Root Theorem for Complex Roots If a polynomial P(x) of degree greater than or equal to 1 (with real coefficients) has a complex number as ...
Cubic Thue equations with many solutions
... where D is a polynomial in Z[t] of positive degree. For each F we shall show that there exists a polynomial D such that ED together with a Q(t) point determines an elliptic curve defined over Q(t), which is not isomorphic over Q to an elliptic curve defined over Q, and for which the rank of the grou ...
... where D is a polynomial in Z[t] of positive degree. For each F we shall show that there exists a polynomial D such that ED together with a Q(t) point determines an elliptic curve defined over Q(t), which is not isomorphic over Q to an elliptic curve defined over Q, and for which the rank of the grou ...
Answers to some typical exercises
... remainder 0 when divided by 6, then we have done. If none of them have remainder 0, then there are at most 5 cases (pigeonhole) of the remainder. Thus, at least two of them must have the same remainder. The positive difference of these two is a subsequence whose sum is divisible by 6. ...
... remainder 0 when divided by 6, then we have done. If none of them have remainder 0, then there are at most 5 cases (pigeonhole) of the remainder. Thus, at least two of them must have the same remainder. The positive difference of these two is a subsequence whose sum is divisible by 6. ...
Full text
... Therefore, a + & is even, so a and 2? are both even or both odd. In the former case, x and 5x + 4 are perfect squares. We claim this leads precisely to the class of solutions given by Theorem 1. In the latter case, it follows that a = b = 1. Thus, we seek integers s such that 5s 2 + 2 is a perfect s ...
... Therefore, a + & is even, so a and 2? are both even or both odd. In the former case, x and 5x + 4 are perfect squares. We claim this leads precisely to the class of solutions given by Theorem 1. In the latter case, it follows that a = b = 1. Thus, we seek integers s such that 5s 2 + 2 is a perfect s ...
