A rational approach to
... facts and stories has developed itself around this number. We can read about Archimedes' method to compute , Ludolf van Ceulen's record computation, Machin's formula, the arithmetic-geometric mean, Ramanujan's miraculous formulas, the impossibility of circle quadrature, computation of digits of w ...
... facts and stories has developed itself around this number. We can read about Archimedes' method to compute , Ludolf van Ceulen's record computation, Machin's formula, the arithmetic-geometric mean, Ramanujan's miraculous formulas, the impossibility of circle quadrature, computation of digits of w ...
41(2)
... The first assertion we shall disprove states that there are infinitely many pairs of positive coprime integers x, y such that 2\y, x2 + y2 E D, and ...
... The first assertion we shall disprove states that there are infinitely many pairs of positive coprime integers x, y such that 2\y, x2 + y2 E D, and ...
Mathematical writing - QMplus - Queen Mary University of London
... tries to convince someone else that the argument is correct. The act of exposition is inextricably linked to thinking, understanding, and self-evaluation. For this reason, undergraduate students should be encouraged to elucidate their thinking in writing, and to assume greater responsibility for the ...
... tries to convince someone else that the argument is correct. The act of exposition is inextricably linked to thinking, understanding, and self-evaluation. For this reason, undergraduate students should be encouraged to elucidate their thinking in writing, and to assume greater responsibility for the ...
Irrationality measures for some automatic real numbers
... irrationality exponent of almost all real numbers (1 ) is exactly equal to 2. A systematic study of rational approximations to automatic irrational numbers was started in [4]. These authors made the approach of [1, 3] quantitative in order to prove that automatic numbers all have a finite irrational ...
... irrationality exponent of almost all real numbers (1 ) is exactly equal to 2. A systematic study of rational approximations to automatic irrational numbers was started in [4]. These authors made the approach of [1, 3] quantitative in order to prove that automatic numbers all have a finite irrational ...
Introduction to mathematical reasoning Chris Woodward Rutgers
... world series, is a compound proposition. It might even be true! The statement The author of this book is an American and the Devil Rays are a baseball team is a compound proposition, which happens to be true, because of the two simple propositions that make up the compound the proposition both happe ...
... world series, is a compound proposition. It might even be true! The statement The author of this book is an American and the Devil Rays are a baseball team is a compound proposition, which happens to be true, because of the two simple propositions that make up the compound the proposition both happe ...
AN INVITATION TO ADDITIVE PRIME NUMBER THEORY A. V.
... generality (say, one may assume that A contains “many” integers), or it may be a particular sequence of some arithmetic interest (say, A may be the sequence of kth powers, the sequence of prime numbers, the values taken by a polynomial F (X) ∈ Z[X] at the positive integers or at the primes, etc.). I ...
... generality (say, one may assume that A contains “many” integers), or it may be a particular sequence of some arithmetic interest (say, A may be the sequence of kth powers, the sequence of prime numbers, the values taken by a polynomial F (X) ∈ Z[X] at the positive integers or at the primes, etc.). I ...
and let A,B be finitely generated graded S-modules. If T is a
... by n − q − 1 independent linear forms. In Section 7 we study powers of linearly presented ideals. The following conjecture sparked this entire paper: CONJECTURE 1.1 (Eisenbud and Ulrich). If I ⊂ S is a linearly presented mprimary ideal generated in degree d, then I n−1 = md(n−1) . We prove this conj ...
... by n − q − 1 independent linear forms. In Section 7 we study powers of linearly presented ideals. The following conjecture sparked this entire paper: CONJECTURE 1.1 (Eisenbud and Ulrich). If I ⊂ S is a linearly presented mprimary ideal generated in degree d, then I n−1 = md(n−1) . We prove this conj ...
Lecture 5 11 5 Conjectures and open problems
... Partial results are known (see § 2.3.5 and 3.3.3). Large transcendence degree results deal, more generally, with the values of the usual exponential function at products xi yj , when x1 , . . . , xd and y1 , . . . , y` are Q-linearly independent complex (or p-adic) numbers. The six exponentials Theo ...
... Partial results are known (see § 2.3.5 and 3.3.3). Large transcendence degree results deal, more generally, with the values of the usual exponential function at products xi yj , when x1 , . . . , xd and y1 , . . . , y` are Q-linearly independent complex (or p-adic) numbers. The six exponentials Theo ...
Distribution of Prime Numbers 6CCM320A / CM320X
... In this chapter we discuss divisibility of integers. The most important result is the fundamental theorem of arithmetic: every integer greater than 1 can be expressed as a product of prime numbers and apart from the order of the factors this product representation is unique. We also look at some com ...
... In this chapter we discuss divisibility of integers. The most important result is the fundamental theorem of arithmetic: every integer greater than 1 can be expressed as a product of prime numbers and apart from the order of the factors this product representation is unique. We also look at some com ...
More properties in Goldbach`s Conjecture
... The strong formulation of Goldbach conjecture, which is the subject of this paper, is much more difficult than the above weak one. Using the above method of Vinogradov [8], in separate works Chudakov [17], van der Corput [18] and Estermann [19] showed that almost all even number can be written as a ...
... The strong formulation of Goldbach conjecture, which is the subject of this paper, is much more difficult than the above weak one. Using the above method of Vinogradov [8], in separate works Chudakov [17], van der Corput [18] and Estermann [19] showed that almost all even number can be written as a ...
SOLUTIONS TO HOMEWORK 2
... if p is prime, ap−1 ≡ 1 mod p, and since 17 is a prime with 17 − 1 = 16, we must have, 6816 ≡ 1 mod 17. Taking a power of 2 on both sides of the congruence, we get, 6832 ≡ 1 mod 17. After you have answered what is the mistake above, write down the correct number between 0 and 16 that is 6832 mod 17. ...
... if p is prime, ap−1 ≡ 1 mod p, and since 17 is a prime with 17 − 1 = 16, we must have, 6816 ≡ 1 mod 17. Taking a power of 2 on both sides of the congruence, we get, 6832 ≡ 1 mod 17. After you have answered what is the mistake above, write down the correct number between 0 and 16 that is 6832 mod 17. ...
No Slide Title - Cloudfront.net
... you must prove it. To show that a conjecture is false, you have to find only one example in which the conjecture is not true. This case is called a counterexample. A counterexample can be a drawing, a statement, or a number. ...
... you must prove it. To show that a conjecture is false, you have to find only one example in which the conjecture is not true. This case is called a counterexample. A counterexample can be a drawing, a statement, or a number. ...
PDF
... Summary. The discriminant of a given polynomial is a number, calculated from the coefficients of that polynomial, that vanishes if and only if that polynomial has one or more multiple roots. Using the discriminant we can test for the presence of multiple roots, without having to actually calculate t ...
... Summary. The discriminant of a given polynomial is a number, calculated from the coefficients of that polynomial, that vanishes if and only if that polynomial has one or more multiple roots. Using the discriminant we can test for the presence of multiple roots, without having to actually calculate t ...
probtalk.pdf
... an object exists using probability. 1. Show that the probability that it exists is NONZERO. Hence there must be some set of random choices that makes it exist. We did this for the distinct-sums problem. 2. You want to show that an object of a size ≥ s exists. Show that if you do a probabilistic expe ...
... an object exists using probability. 1. Show that the probability that it exists is NONZERO. Hence there must be some set of random choices that makes it exist. We did this for the distinct-sums problem. 2. You want to show that an object of a size ≥ s exists. Show that if you do a probabilistic expe ...
34(4)
... In 1965, Brother Alfred Brousseau, under the auspices of The Fibonacci Association, compiled a twovolume set of Fibonacci entry points and related data for the primes 2 through 99,907. This set is currently available from The Fibonacci Association as advertised on the back cover of The Fibonacci Qua ...
... In 1965, Brother Alfred Brousseau, under the auspices of The Fibonacci Association, compiled a twovolume set of Fibonacci entry points and related data for the primes 2 through 99,907. This set is currently available from The Fibonacci Association as advertised on the back cover of The Fibonacci Qua ...
PPT
... Each person will have a unique number For each question, I will first give the class time to work out an answer. Then, I will call three different people at random They must explain the answer to the TAs (who are all the way in the back). If the TAs are satisfied, the class gets points. If the class ...
... Each person will have a unique number For each question, I will first give the class time to work out an answer. Then, I will call three different people at random They must explain the answer to the TAs (who are all the way in the back). If the TAs are satisfied, the class gets points. If the class ...
PPT
... Each person will have a unique number For each question, I will first give the class time to work out an answer. Then, I will call three different people at random They must explain the answer to the TAs and Gupta. If the TAs and Gupta are satisfied, your group gets points. The winning group will ge ...
... Each person will have a unique number For each question, I will first give the class time to work out an answer. Then, I will call three different people at random They must explain the answer to the TAs and Gupta. If the TAs and Gupta are satisfied, your group gets points. The winning group will ge ...
PPT - CMU School of Computer Science
... Each person will have a unique number For each question, I will first give the class time to work out an answer. Then, I will call three different people at random They must explain the answer to the TAs (who are all the way in the back). If the TAs are satisfied, the class gets points. If the class ...
... Each person will have a unique number For each question, I will first give the class time to work out an answer. Then, I will call three different people at random They must explain the answer to the TAs (who are all the way in the back). If the TAs are satisfied, the class gets points. If the class ...
Euclid(A,B)
... Rules of the Game Each person will have a unique number For each question, I will first give the class time to work out an answer. Then, I will call three different people at random They must explain the answer to me. If I’m satisfied, the class gets points. If the class gets 1,700 points, then you ...
... Rules of the Game Each person will have a unique number For each question, I will first give the class time to work out an answer. Then, I will call three different people at random They must explain the answer to me. If I’m satisfied, the class gets points. If the class gets 1,700 points, then you ...