Binomial Coefficients, Congruences, Lecture 3 Notes
... Corollary 14. The power of prime p dividing nk is the number of carries when you add k to n − k in base p (and also the number of carries when you subtract k from n in ...
... Corollary 14. The power of prime p dividing nk is the number of carries when you add k to n − k in base p (and also the number of carries when you subtract k from n in ...
Characterizing integers among rational numbers
... 5. Defining rings of integers Theorem 5.1. There is a Π+ 2 -formula that in any number field k defines its ring of integers. Proof. Let ` be a prime greater than 11 such that (` − 1)/2 also is prime, e.g., ` = 23. Let ζ` be a primitive `th root of 1 in an algebraic closure of k. We will break into c ...
... 5. Defining rings of integers Theorem 5.1. There is a Π+ 2 -formula that in any number field k defines its ring of integers. Proof. Let ` be a prime greater than 11 such that (` − 1)/2 also is prime, e.g., ` = 23. Let ζ` be a primitive `th root of 1 in an algebraic closure of k. We will break into c ...
solutions 2 2. (i) I ran this 3 times, just for fun. Your answers will be
... the interval [n0 = 2249 + 1, n1 = 2250 −1]. So, using Heilbronn’s formula, we expect the average number of iterations to near the interval [.843 ln(n0 ) = 145.4 . . . , .843 ln(n1 ) = 146.0 . . .], and it is. (iv) The runtime will be in O(k 3 ). By assumption, there are Θ(k) iterations, and each ite ...
... the interval [n0 = 2249 + 1, n1 = 2250 −1]. So, using Heilbronn’s formula, we expect the average number of iterations to near the interval [.843 ln(n0 ) = 145.4 . . . , .843 ln(n1 ) = 146.0 . . .], and it is. (iv) The runtime will be in O(k 3 ). By assumption, there are Θ(k) iterations, and each ite ...
Bertrand`s Theorem - New Zealand Maths Olympiad Committee online
... New Zealand Mathematical Olympiad Committee Bertrand’s Theorem Arkadii Slinko ...
... New Zealand Mathematical Olympiad Committee Bertrand’s Theorem Arkadii Slinko ...
Math 3: Unit 1 – Reasoning and Proof Inductive, Deductive
... 12. Suppose it is true that “all members of the senior class are at least 5 feet 2 inches tall.” What, if anything, can you conclude with certainty about each of the following students? a. Darlene, who is a member of the senior class. ...
... 12. Suppose it is true that “all members of the senior class are at least 5 feet 2 inches tall.” What, if anything, can you conclude with certainty about each of the following students? a. Darlene, who is a member of the senior class. ...
(1) Find all prime numbers smaller than 100. (2) Give a proof by
... (1) Find all prime numbers smaller than 100. (2) Give a proof by induction (instead of a proof by contradiction given in class) that any natural number > 1 has a unique (up to order) factorization as a product of primes. (3) Give a proof by induction that if a ≡ b( mod m) then an ≡ bn ( mod m) for a ...
... (1) Find all prime numbers smaller than 100. (2) Give a proof by induction (instead of a proof by contradiction given in class) that any natural number > 1 has a unique (up to order) factorization as a product of primes. (3) Give a proof by induction that if a ≡ b( mod m) then an ≡ bn ( mod m) for a ...
Proof Example: The Irrationality of √ 2 During the lecture a student
... number which divides both of them. What is ment here by divides? A natural number d > 1 divides a natural number n iff there exists a natural k such that n = d ∗ k. Note that a divisor is automatically different from 0 and 1. Intuitively, this definition of rational numbers takes into account their ...
... number which divides both of them. What is ment here by divides? A natural number d > 1 divides a natural number n iff there exists a natural k such that n = d ∗ k. Note that a divisor is automatically different from 0 and 1. Intuitively, this definition of rational numbers takes into account their ...
Midterm #3: practice
... (a) Modulo 323, what do we learn from Euler's theorem? (b) Using the Chinese remainder theorem, show that x144 1 (mod 323) for all x coprime to 323. (c) Compare the two results! Bonus: Can you come up with a strengthening of Euler's theorem? ...
... (a) Modulo 323, what do we learn from Euler's theorem? (b) Using the Chinese remainder theorem, show that x144 1 (mod 323) for all x coprime to 323. (c) Compare the two results! Bonus: Can you come up with a strengthening of Euler's theorem? ...
Full text
... The object of this paper is to present a bracket function transform together with its inverse and some applications. The transform is the analogue of the binomial coefficient transform discussed in [2]. The inverse form will be used to give a short proof of an explicit formula in [1] for Rk (n), the ...
... The object of this paper is to present a bracket function transform together with its inverse and some applications. The transform is the analogue of the binomial coefficient transform discussed in [2]. The inverse form will be used to give a short proof of an explicit formula in [1] for Rk (n), the ...
38_sunny
... “Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich's Infinite Series and not fall into the pitfalls of divergent series. … I tol ...
... “Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich's Infinite Series and not fall into the pitfalls of divergent series. … I tol ...
Math 2710 (Roby) Practice Midterm #2 Spring 2013
... (d) If x ≡ a (mod m) and x ≡ a (mod n), then x ≡ a (mod mn). False. 5 ≡ 35 (mod 6) and 5 ≡ 35 (mod 10), but 5 6≡ 35 (mod 60). SALVAGE: True if (m, n) = 1. Proof. By hypothesis we have m | x − a and n | x − a. Since (m, n) = 1, this implies that mn | x − a =⇒ , which is what we want to show. ...
... (d) If x ≡ a (mod m) and x ≡ a (mod n), then x ≡ a (mod mn). False. 5 ≡ 35 (mod 6) and 5 ≡ 35 (mod 10), but 5 6≡ 35 (mod 60). SALVAGE: True if (m, n) = 1. Proof. By hypothesis we have m | x − a and n | x − a. Since (m, n) = 1, this implies that mn | x − a =⇒ , which is what we want to show. ...
What is. . . an L-function? - Mathematisch Instituut Leiden
... • Special values. There are many results and conjectures concerning values of L-functions at the integers. The prototypical example is Euler’s formula (1.6). There are essentially two ways of constructing L-functions: (1) from number theory and arithmetic geometry; (2) from automorphic forms and aut ...
... • Special values. There are many results and conjectures concerning values of L-functions at the integers. The prototypical example is Euler’s formula (1.6). There are essentially two ways of constructing L-functions: (1) from number theory and arithmetic geometry; (2) from automorphic forms and aut ...
Full text
... In the present note we shall give two proofs of a property of the poly-Bernoulli numbers, the closed formula for negative index poly-Bernoulli numbers given by Arakawa and Kaneko [1]. The first proof uses weighted Stirling numbers of the second kind (see [2], [3]). The second, much simpler, proof is ...
... In the present note we shall give two proofs of a property of the poly-Bernoulli numbers, the closed formula for negative index poly-Bernoulli numbers given by Arakawa and Kaneko [1]. The first proof uses weighted Stirling numbers of the second kind (see [2], [3]). The second, much simpler, proof is ...
Fermat Numbers in the Pascal Triangle
... C(a) depending only on a such that all solutions of equation (17) satisfies m < C(a). The fact that there are sometimes non-trivial solutions of (17) is illustrated by the example µ ¶ ...
... C(a) depending only on a such that all solutions of equation (17) satisfies m < C(a). The fact that there are sometimes non-trivial solutions of (17) is illustrated by the example µ ¶ ...
![[Part 2]](http://s1.studyres.com/store/data/008795795_1-c00648edd6f578e3e44ef8aca9f22ea2-300x300.png)
[Part 2]
... 1. H. W. Gould, "Equal Products of Generalized Binomial Coefficients," Fibonacci Quarterly, Vol. 9, No. 4 (1971), pp. 337-346. 2. H. W. Gould, D. C. Rine, and W. L. Scharff, "Algorithm and Computer P r o g r a m for the Determination of Equal Products of Generalized Binomial Coefficients, Tf to be p ...
... 1. H. W. Gould, "Equal Products of Generalized Binomial Coefficients," Fibonacci Quarterly, Vol. 9, No. 4 (1971), pp. 337-346. 2. H. W. Gould, D. C. Rine, and W. L. Scharff, "Algorithm and Computer P r o g r a m for the Determination of Equal Products of Generalized Binomial Coefficients, Tf to be p ...
![[Part 2]](http://s1.studyres.com/store/data/008795912_1-134f24134532661a161532d09dceadfe-300x300.png)
[Part 2]
... square over Z[a,b]. Even if that were surmounted, adapting Theorem 6 to quadratics would present some difficulties., For a and b integers, this paper does not add much to present knowledge except to place the problem in a larger setting. The Davenport-Baker result shows that in Theroem 9 when a = 1, ...
... square over Z[a,b]. Even if that were surmounted, adapting Theorem 6 to quadratics would present some difficulties., For a and b integers, this paper does not add much to present knowledge except to place the problem in a larger setting. The Davenport-Baker result shows that in Theroem 9 when a = 1, ...
Exponentiation: Theorems, Proofs, Problems
... Which, by definition, is just an−m . But what about these other two cases? First of all, what happens in the case that m = n? Well, if n m n m = n, then aam will just equal aan (or aam —either way is the same). But that clearly3 is just equal to 1. And I hope you will agree that if n and m are equal ...
... Which, by definition, is just an−m . But what about these other two cases? First of all, what happens in the case that m = n? Well, if n m n m = n, then aam will just equal aan (or aam —either way is the same). But that clearly3 is just equal to 1. And I hope you will agree that if n and m are equal ...
Exploration 14
... Proof. All positive integers are generated by repeatedly adding 1, so what we need to show is Claim. If x is a positive integer that is both even and odd then x + 1 is both even and odd. To prove this claim, suppose that x is a positive intger that is both even and odd. Since x is even and 1 is odd, ...
... Proof. All positive integers are generated by repeatedly adding 1, so what we need to show is Claim. If x is a positive integer that is both even and odd then x + 1 is both even and odd. To prove this claim, suppose that x is a positive intger that is both even and odd. Since x is even and 1 is odd, ...
Full text
... Thus no solution exists in this case. If n = 0 , 1 , 2, 3, or 4, we get pnq2 = 2, 8, 26, 76, 208, respectively, none of which are possible. The following theorems are stated without proof, for the proofs follow the same patterns as above. Theorem9. ...
... Thus no solution exists in this case. If n = 0 , 1 , 2, 3, or 4, we get pnq2 = 2, 8, 26, 76, 208, respectively, none of which are possible. The following theorems are stated without proof, for the proofs follow the same patterns as above. Theorem9. ...
![arXiv:1510.00735v3 [math.NT] 14 Oct 2015](http://s1.studyres.com/store/data/015647175_1-32c28f0e3a09596218fdb4a3c649b24f-300x300.png)
arXiv:1510.00735v3 [math.NT] 14 Oct 2015
... So we can view Ramanujan’s family of solutions as giving us two points on infinitely many cubic twists of E. However, we will first consider them all together as a single elliptic curve over Q(t). Theorem 1.1. The elliptic curve Em(t) /Q(t) : y 2 = x3 − 432m(t)2 has rank 2. This allows us to give a ...
... So we can view Ramanujan’s family of solutions as giving us two points on infinitely many cubic twists of E. However, we will first consider them all together as a single elliptic curve over Q(t). Theorem 1.1. The elliptic curve Em(t) /Q(t) : y 2 = x3 − 432m(t)2 has rank 2. This allows us to give a ...