Linear independence of the digamma function and a variant of a conjecture of Rohrlich
... Motivated by the above theorem, the authors, in the same paper, conjectured the following: Conjecture. Let K be any number field over which the qth cyclotomic polynomial is irreducible. Then the ϕ (q) numbers ψ(a/q) with 1 a q and (a, q) = 1 are linearly independent over K . In this context, they ...
... Motivated by the above theorem, the authors, in the same paper, conjectured the following: Conjecture. Let K be any number field over which the qth cyclotomic polynomial is irreducible. Then the ϕ (q) numbers ψ(a/q) with 1 a q and (a, q) = 1 are linearly independent over K . In this context, they ...
Proofs by induction - Australian Mathematical Sciences Institute
... one beneath it. The monks of the temple were assigned the task of transferring the stack of 64 discs to another pole, by moving discs one at a time between the poles, with one important proviso: a large disc could never be placed on top of a smaller one. When the monks completed their assigned task, ...
... one beneath it. The monks of the temple were assigned the task of transferring the stack of 64 discs to another pole, by moving discs one at a time between the poles, with one important proviso: a large disc could never be placed on top of a smaller one. When the monks completed their assigned task, ...
PEN A9 A37 O51
... square. We ommited this proof here, but not due to its complexity, but due to its length. Still, the most general claim wasn’t proven. Pillai’s work on relative primality of consecutive integers had some interesting results apart from the ’product-power problem’. Pillai has first shown that the leas ...
... square. We ommited this proof here, but not due to its complexity, but due to its length. Still, the most general claim wasn’t proven. Pillai’s work on relative primality of consecutive integers had some interesting results apart from the ’product-power problem’. Pillai has first shown that the leas ...
Full text
... for the pn . It is immediate that the two sequences coincide for n = 0, 1, so they must be the same for all n. Like Proposition 1, this is equivalent to a well-known statement about the Un. (b) Lemma 1 implies that 2 cos(tu/n) is a root of p for t = 1, 2, . .., n - 1 and, since the cosine is strictl ...
... for the pn . It is immediate that the two sequences coincide for n = 0, 1, so they must be the same for all n. Like Proposition 1, this is equivalent to a well-known statement about the Un. (b) Lemma 1 implies that 2 cos(tu/n) is a root of p for t = 1, 2, . .., n - 1 and, since the cosine is strictl ...
Sets, Functions and Euclidean Space
... such that b ≥ x for all x in X. This number b is called an upper bound for S. A set that is bounded above has many upper bounds. A least upper bound for the set X is a number b∗ that is an upper bound for S and is such that b∗ ≤ b for every upper bound b. The existence of a least upper bound is a ba ...
... such that b ≥ x for all x in X. This number b is called an upper bound for S. A set that is bounded above has many upper bounds. A least upper bound for the set X is a number b∗ that is an upper bound for S and is such that b∗ ≤ b for every upper bound b. The existence of a least upper bound is a ba ...
Answers.
... For any prime p the number of primitive roots is φ(p − 1). So the number of primitive roots modulo 19 is φ(18) = φ(2)φ(9) = 1 · 6 = 6. d. Name one primitive root modulo 19. There are 6 primitive roots; 2 is one of them, but so are 3, 6, 10, 14, and 15. Problem 5. [10; 2 points each part] On Euler’s ...
... For any prime p the number of primitive roots is φ(p − 1). So the number of primitive roots modulo 19 is φ(18) = φ(2)φ(9) = 1 · 6 = 6. d. Name one primitive root modulo 19. There are 6 primitive roots; 2 is one of them, but so are 3, 6, 10, 14, and 15. Problem 5. [10; 2 points each part] On Euler’s ...
20 primality
... m is not divisible by any prime pi, so m must be a multiple of a prime that is not in the “set of all primes”. Thanks to Euclid, we know p(n) as n . But how thickly distributed are the primes? That is, if we pick a random number in {1…n}, what’s the probability of getting a prime? We’ll see ...
... m is not divisible by any prime pi, so m must be a multiple of a prime that is not in the “set of all primes”. Thanks to Euclid, we know p(n) as n . But how thickly distributed are the primes? That is, if we pick a random number in {1…n}, what’s the probability of getting a prime? We’ll see ...
Full text
... (f* which will suggest a non-unitary analog. In particular, we may define 1 of n.
Similarly,
... (f* which will suggest a non-unitary analog. In particular, we may define
Chapter 4 Number theory - School of Mathematical and Computer
... divide b. If p does not divide a, then a and p are coprime. By Lemma 4.1.12, there exist integers x and y such that 1 = px + ay. Thus b = bpx + bay. Now p | bp and p | ba, by assumption, and so p | b, as required. Example 4.2.6. The above result is not true if p is not a prime. For example, 6 | 9 × ...
... divide b. If p does not divide a, then a and p are coprime. By Lemma 4.1.12, there exist integers x and y such that 1 = px + ay. Thus b = bpx + bay. Now p | bp and p | ba, by assumption, and so p | b, as required. Example 4.2.6. The above result is not true if p is not a prime. For example, 6 | 9 × ...
Elementary Number Theory
... These lecture notes grew out of a first course in number theory for second year students as is was given by the second author several times at the University of Siegen and by the first one in 2015/2016 at İstanbul Üniversitesi in Istanbul. There are many books on elementary number theory, most of ...
... These lecture notes grew out of a first course in number theory for second year students as is was given by the second author several times at the University of Siegen and by the first one in 2015/2016 at İstanbul Üniversitesi in Istanbul. There are many books on elementary number theory, most of ...
q - Personal.psu.edu - Penn State University
... Back to Naude’s Question One last comment on Naude’s question is in order. Naude asked: “How many ways can the number 50 be written as a sum of seven different positive integers?” The answer given by Euler is 522. ...
... Back to Naude’s Question One last comment on Naude’s question is in order. Naude asked: “How many ways can the number 50 be written as a sum of seven different positive integers?” The answer given by Euler is 522. ...
Divisor Goldbach Conjecture and its Partition Number
... where the product is over all primes p, and γc,p (n) is the number of solutions to the equation n = (q1 + ... + qc ) mod p in modular arithmetic, subject to the constraints q1 , ..., qc 6= 0 mod p [8]. It was known as Hardy-littlewood conjecture. Note that, the Goldbach conjecture is only for even n ...
... where the product is over all primes p, and γc,p (n) is the number of solutions to the equation n = (q1 + ... + qc ) mod p in modular arithmetic, subject to the constraints q1 , ..., qc 6= 0 mod p [8]. It was known as Hardy-littlewood conjecture. Note that, the Goldbach conjecture is only for even n ...
Lecture21.pdf
... In this lesson, we present a theorem without proof then use the theorem to find all the roots (real or non-real) of a polynomial equation. We start with the definition below. The complex number w a bi is an nth root of the complex n number z if a bi z . ...
... In this lesson, we present a theorem without proof then use the theorem to find all the roots (real or non-real) of a polynomial equation. We start with the definition below. The complex number w a bi is an nth root of the complex n number z if a bi z . ...
