The Connectedness of Arithmetic Progressions in
... ξ ∈ {an + b} ∩ U1 = O1 . Now observe that from (24) we also have ξ = a2 + βaq1 r1 . By (18) and (19) we obtain ξ ∈ {rn + b2 } ⊆ U2 , whence ξ ∈ {an + b} ∩ U2 = O2 . Finally O1 ∩ O2 6= ∅, a contradiction. So, the progression {an + b} is D′ connected. The proof of Theorem 3.5 is complete. 5. Prime num ...
... ξ ∈ {an + b} ∩ U1 = O1 . Now observe that from (24) we also have ξ = a2 + βaq1 r1 . By (18) and (19) we obtain ξ ∈ {rn + b2 } ⊆ U2 , whence ξ ∈ {an + b} ∩ U2 = O2 . Finally O1 ∩ O2 6= ∅, a contradiction. So, the progression {an + b} is D′ connected. The proof of Theorem 3.5 is complete. 5. Prime num ...
Random geometric complexes in the thermodynamic regime
... CB (Φ, r) but with a homotopically equivalent abstract simplical complex with a natural combinatorial structure. This will be the Čech complex with radius r built over the point set Φ, denoted by C(Φ, r), and defined below in Section 2.1. The first, and perhaps most natural topological question to ...
... CB (Φ, r) but with a homotopically equivalent abstract simplical complex with a natural combinatorial structure. This will be the Čech complex with radius r built over the point set Φ, denoted by C(Φ, r), and defined below in Section 2.1. The first, and perhaps most natural topological question to ...
Use of Chinese Remainder Theorem to generate
... Normale in 1795. According to , many of the Chinese findings n mathematics ultimately made their way to Europe via India and Arabia. The Chinese Remainder Theorem became known in Europe through article, “Jottings on the science of Chinese arithmetic”, by Alex ...
... Normale in 1795. According to , many of the Chinese findings n mathematics ultimately made their way to Europe via India and Arabia. The Chinese Remainder Theorem became known in Europe through article, “Jottings on the science of Chinese arithmetic”, by Alex ...
CSNB143 – Discrete Structure
... being born in the same day (Monday to Sunday). Show that by using pigeonhole principle. Sol: Because there are 8 people and only 7 days per week, so Pigeonhole Principle says that, at least two or more people were being born in the same day. Note that Pigeonhole Principle provides an existence p ...
... being born in the same day (Monday to Sunday). Show that by using pigeonhole principle. Sol: Because there are 8 people and only 7 days per week, so Pigeonhole Principle says that, at least two or more people were being born in the same day. Note that Pigeonhole Principle provides an existence p ...
Numbers, proof and `all that jazz`.
... only our axioms. In fact, in these notes, we usually adopt a much looser standard. As the reader will see, proving everything directly from the axioms would take so long that we would never progress beyond this section! It is, however, important that the reader prove a number of basic number facts u ...
... only our axioms. In fact, in these notes, we usually adopt a much looser standard. As the reader will see, proving everything directly from the axioms would take so long that we would never progress beyond this section! It is, however, important that the reader prove a number of basic number facts u ...
on numbers equal to the sum of two squares in
... factorization property of Gaussian integers. This proof had the disappointing feature that it did not show how to find the numbers a, b, c, and d, but simply demonstrated their existence. Later we found a constructive proof and were surprised to see how elementary it was using only precalculus mathe ...
... factorization property of Gaussian integers. This proof had the disappointing feature that it did not show how to find the numbers a, b, c, and d, but simply demonstrated their existence. Later we found a constructive proof and were surprised to see how elementary it was using only precalculus mathe ...
ON FIBONACCI POWERS
... The Fibonacci numbers have engaged the attention of mathematicians for several centuries, and whilst many of their properties are easy to establish by very simple methods, there are several unsolved problems connected to them. Fibonacci numbers are defined with recurrence Fn = Fn−1 +Fn−2 , for n ≥ 2 ...
... The Fibonacci numbers have engaged the attention of mathematicians for several centuries, and whilst many of their properties are easy to establish by very simple methods, there are several unsolved problems connected to them. Fibonacci numbers are defined with recurrence Fn = Fn−1 +Fn−2 , for n ≥ 2 ...
Full text
... at random from {0, 1, 2, . . . , n} will end in 1 when written in binary is approximately onehalf. In fact, by taking n sufficiently large, the probability that a randomly-chosen number from {0, 1, 2, . . . , n} will have a 1 in some specified position can be made arbitrarily close to one-half. The ...
... at random from {0, 1, 2, . . . , n} will end in 1 when written in binary is approximately onehalf. In fact, by taking n sufficiently large, the probability that a randomly-chosen number from {0, 1, 2, . . . , n} will have a 1 in some specified position can be made arbitrarily close to one-half. The ...
Inequalities
... Case 4: kXkp > 0 and kY kq > 0 and at least one if infinite. Here kXkp kY kq is infinite, so (9) holds trivially. There is an addendum to Hölder’s inequality which can be established by pushing the arguments in the proof further. One says two random variables U and V are linearly dependent if there ...
... Case 4: kXkp > 0 and kY kq > 0 and at least one if infinite. Here kXkp kY kq is infinite, so (9) holds trivially. There is an addendum to Hölder’s inequality which can be established by pushing the arguments in the proof further. One says two random variables U and V are linearly dependent if there ...
Algebraic Symmetries I Just as we can factor z 3 − 1=(z − 1)(z + z + 1
... This is a statement proved by Gauss in the Disquisitiones. We will need it for p = 17. According to Bourbaki’s Éléments de l’histoire des mathématiques, this was the first general statement of this sort about polynomials ever proved. It seems to me that in one sense, it is also the last. In the n ...
... This is a statement proved by Gauss in the Disquisitiones. We will need it for p = 17. According to Bourbaki’s Éléments de l’histoire des mathématiques, this was the first general statement of this sort about polynomials ever proved. It seems to me that in one sense, it is also the last. In the n ...
Infinite Descent - but not into Hell!
... Everyone knows the principle of mathematical induction (PMI for short); it is now standard fare even a.t the high-school level. Curiously, very few seem to know its close relative - the principle of descent (PD for short); curious, because the two principles are complementary to one another. In this ...
... Everyone knows the principle of mathematical induction (PMI for short); it is now standard fare even a.t the high-school level. Curiously, very few seem to know its close relative - the principle of descent (PD for short); curious, because the two principles are complementary to one another. In this ...
Example Proofs
... when divided by 4. Since we know that k is odd, we have k=2*r+1 for some integer r. Now, we can compute k2. k2 = (2*r+1)2 = 4*r2+4*r+1 = 4(r2+r) + 1, which leaves a remainder of 1 when divided by 4 since 4 divides evenly into 4(r2+r). (Because r2+r must be an integer...) ...
... when divided by 4. Since we know that k is odd, we have k=2*r+1 for some integer r. Now, we can compute k2. k2 = (2*r+1)2 = 4*r2+4*r+1 = 4(r2+r) + 1, which leaves a remainder of 1 when divided by 4 since 4 divides evenly into 4(r2+r). (Because r2+r must be an integer...) ...
Full text
... It is not difficult to show that Hi is an infinite set9 i.e., there is an infinite set of PPT*s each one of which has a perimeter not shared by any other PPT. The surprising fact that E^ is also an infinite set is proved in [1]. It is the main purpose of this paper to prove that Hk is an infinite se ...
... It is not difficult to show that Hi is an infinite set9 i.e., there is an infinite set of PPT*s each one of which has a perimeter not shared by any other PPT. The surprising fact that E^ is also an infinite set is proved in [1]. It is the main purpose of this paper to prove that Hk is an infinite se ...
Overpseudoprimes, and Mersenne and Fermat Numbers as
... where p is prime. In this form numbers Mp , at the first time, were studied by Marin Mersenne (1588–1648) around 1644; see Guy [3, §A3] and a large bibliography there. In the next section, we introduce a new class of pseudoprimes and we prove that it just contains the odd numbers n such that |2|d is ...
... where p is prime. In this form numbers Mp , at the first time, were studied by Marin Mersenne (1588–1648) around 1644; see Guy [3, §A3] and a large bibliography there. In the next section, we introduce a new class of pseudoprimes and we prove that it just contains the odd numbers n such that |2|d is ...
T R I P U R A ... (A Central University) Syllabus for Three Year Degree Course
... theorem of LPP and their applications, theory and application of the simplex method of solution of LPP. Unit-5(15+2) (Linear Programming Problem-II) 5.1 Charne’s M-technique. The two phase method. 5.2 Duality theory. The dual of the dual is primal, relation between the objective function value of du ...
... theorem of LPP and their applications, theory and application of the simplex method of solution of LPP. Unit-5(15+2) (Linear Programming Problem-II) 5.1 Charne’s M-technique. The two phase method. 5.2 Duality theory. The dual of the dual is primal, relation between the objective function value of du ...
Full text
... n of degree m + l with coefficients involving Bernoulli numbers. See, for example, the papers by Christiano [6] and by de Bruyn and de Villiers [7]. Burrows and Talbot [2] treat this sum as a polynomial in (n + l/2), and Edwards [8] expresses the sums Sm{n) as polynomials in X& and Z £ 2 . Formulas ...
... n of degree m + l with coefficients involving Bernoulli numbers. See, for example, the papers by Christiano [6] and by de Bruyn and de Villiers [7]. Burrows and Talbot [2] treat this sum as a polynomial in (n + l/2), and Edwards [8] expresses the sums Sm{n) as polynomials in X& and Z £ 2 . Formulas ...
![[Michel Waldschmidt] Continued fractions](http://s1.studyres.com/store/data/018214733_1-ffc3ca0fc19190c3043e567af0ae13ba-300x300.png)
[Michel Waldschmidt] Continued fractions
... Remark 1. A variant of the algorithm of simple continued fractions is the following. Given two sequences (an )n≥0 and (bn )n≥0 of elements in a field K and an element x in K, one defines a sequence (possibly finite) (xn )n≥1 of elements in K as follows. If x = a0 , the sequence is empty. Otherwise x ...
... Remark 1. A variant of the algorithm of simple continued fractions is the following. Given two sequences (an )n≥0 and (bn )n≥0 of elements in a field K and an element x in K, one defines a sequence (possibly finite) (xn )n≥1 of elements in K as follows. If x = a0 , the sequence is empty. Otherwise x ...