Math 248A. Norm and trace An interesting application of Galois
... determinant of the multiplication map by a on L by first building a basis of K over k and then picking a basis of L over K, we get a block matrix with [L : K] blocks down the diagonal which are all just the matrix for multiplication by a on K. Thus, since the determinant of such a block form is the ...
... determinant of the multiplication map by a on L by first building a basis of K over k and then picking a basis of L over K, we get a block matrix with [L : K] blocks down the diagonal which are all just the matrix for multiplication by a on K. Thus, since the determinant of such a block form is the ...
MT 430 Intro to Number Theory MIDTERM 1 PRACTICE
... (2) Euclidean algorithm. Greatest common divisors. (3) Prime numbers. (4) Uniqueness of factorization. (5) Binomial coefficients. Binomial theorem. (6) Fermat’s little theorem. Wilson’s theorem. (7) Euler’s φ-function. (8) Chinese Remainder Theorem. (9) Algebraic congruences. Hensel’s lemma. (10) Eu ...
... (2) Euclidean algorithm. Greatest common divisors. (3) Prime numbers. (4) Uniqueness of factorization. (5) Binomial coefficients. Binomial theorem. (6) Fermat’s little theorem. Wilson’s theorem. (7) Euler’s φ-function. (8) Chinese Remainder Theorem. (9) Algebraic congruences. Hensel’s lemma. (10) Eu ...
Test 2 Working with Polynomials
... Donkey Kong is competing in a shot-put challenge at the Olympics. His throw can be modeled by the function h(t) = -5t2 + 8.5t + 1.8, where h is the height, in metres, of a shot-put t seconds after it is thrown. Determine the remainder when h(t) is divided by (t – 1.4). What does this value represent ...
... Donkey Kong is competing in a shot-put challenge at the Olympics. His throw can be modeled by the function h(t) = -5t2 + 8.5t + 1.8, where h is the height, in metres, of a shot-put t seconds after it is thrown. Determine the remainder when h(t) is divided by (t – 1.4). What does this value represent ...
UI Putnam Training Sessions Problem Set 4: Advanced Number
... x. (Hint: Convert the problem to one involving integer solutions.) ...
... x. (Hint: Convert the problem to one involving integer solutions.) ...
Full text
... Thus, the basic 2-transposable integers are 125, 251, 376. (Note that these numbers are expressed in base 9.) When k = 4, there is one 4-transposable integer, namely, 17. It is possible that, for a given g and k9 there will be more than one d which satisfies (i)-(iii) of Theorem 1. We illustrate thi ...
... Thus, the basic 2-transposable integers are 125, 251, 376. (Note that these numbers are expressed in base 9.) When k = 4, there is one 4-transposable integer, namely, 17. It is possible that, for a given g and k9 there will be more than one d which satisfies (i)-(iii) of Theorem 1. We illustrate thi ...
IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728, p-ISSN:2319-765X.
... Squares of integers can be expressed as sum of consecutive odd numbers. Is there a general case for all powers? After a thorough search with available materials I could not find one such theorem. Here is an attempt in that lines. Sum of consecutive odd numbers as powers of integers and sum of consec ...
... Squares of integers can be expressed as sum of consecutive odd numbers. Is there a general case for all powers? After a thorough search with available materials I could not find one such theorem. Here is an attempt in that lines. Sum of consecutive odd numbers as powers of integers and sum of consec ...
Exam II - U.I.U.C. Math
... By the Division Algorithm, 33 = 17 · 1 + 16 and so 233 = 217+16 = 217 · 216 ≡ 2 · 216 ≡ 217 ≡ 2(mod 17). Hence 2561 ≡ 2(mod 561). b) Consider the system of congruences: x ≡ 2(mod 3) x ≡ 2(mod 11) x ≡ 2(mod 17) Clearly x = 2 is a solution to the system. Moreover, we are given above that 2561 is also ...
... By the Division Algorithm, 33 = 17 · 1 + 16 and so 233 = 217+16 = 217 · 216 ≡ 2 · 216 ≡ 217 ≡ 2(mod 17). Hence 2561 ≡ 2(mod 561). b) Consider the system of congruences: x ≡ 2(mod 3) x ≡ 2(mod 11) x ≡ 2(mod 17) Clearly x = 2 is a solution to the system. Moreover, we are given above that 2561 is also ...
An Oscillation Theorem for a Sturm
... p. 2311). The main result of this paper is an oscillation formula ((OF) in Theorem 3.1 below) for the problem (E) under (A), which calculates the oscillation number Z,, i. e., the exact number of zeros of a given eigenfunction x, corresponding to an eigenvalue A, on the interval ( a ,b] for p E IN. ...
... p. 2311). The main result of this paper is an oscillation formula ((OF) in Theorem 3.1 below) for the problem (E) under (A), which calculates the oscillation number Z,, i. e., the exact number of zeros of a given eigenfunction x, corresponding to an eigenvalue A, on the interval ( a ,b] for p E IN. ...
Remainder Theorem
... 3) Concept of negative remainder Remainder can never be negative; its minimum value can only be 0. Consider an example of -30 / 7. Here, remainder is 5. It would not be (-28 – 2 / 7), but [(-35+5)/7] When you divide, you will get remainder of -2. Since remainder can never be negative, we subtract it ...
... 3) Concept of negative remainder Remainder can never be negative; its minimum value can only be 0. Consider an example of -30 / 7. Here, remainder is 5. It would not be (-28 – 2 / 7), but [(-35+5)/7] When you divide, you will get remainder of -2. Since remainder can never be negative, we subtract it ...
Cryptography Midterm Solutions
... a square root of a. On the other hand, if b2 = a modulo p then where e = logg (b), (so 1 ≤ e ≤ p − 1), we have that g 2e = b2 = a. Write 2e = c(p − 1) + r for some r with 1 ≤ r ≤ p − 1. Then g r = g 2e−c(p−1) = g 2e · g (p−1)·c ≡ g 2e ≡ a. So logg (a) = r, but r = 2e − c(p − 1), so because e and p − ...
... a square root of a. On the other hand, if b2 = a modulo p then where e = logg (b), (so 1 ≤ e ≤ p − 1), we have that g 2e = b2 = a. Write 2e = c(p − 1) + r for some r with 1 ≤ r ≤ p − 1. Then g r = g 2e−c(p−1) = g 2e · g (p−1)·c ≡ g 2e ≡ a. So logg (a) = r, but r = 2e − c(p − 1), so because e and p − ...
Full text
... Specifically, orthogonal Tchebycheff polynomials are used to calculate the numbers Hkj>n of histories of length n starting at level k and ending at level /, while the reciprocal Tchebycheff polynomials of degree h are used to derive generating functions for histories of height < h. In this context, ...
... Specifically, orthogonal Tchebycheff polynomials are used to calculate the numbers Hkj>n of histories of length n starting at level k and ending at level /, while the reciprocal Tchebycheff polynomials of degree h are used to derive generating functions for histories of height < h. In this context, ...
A polynomial of degree n may be written in a standard form:
... check if a particular number, x1, indeed is a zero of a polynomial we can divide the polynomial by the factor (x – x1). If the remainder is equal to zero than we can rewrite the polynomial in a factored form as ( x x1 ) f1 ( x) where f1 ( x) is a polynomial of degree ...
... check if a particular number, x1, indeed is a zero of a polynomial we can divide the polynomial by the factor (x – x1). If the remainder is equal to zero than we can rewrite the polynomial in a factored form as ( x x1 ) f1 ( x) where f1 ( x) is a polynomial of degree ...
The Remainder Theorem
... So we get x2 + 3x + 3 on top (this is q(x)), with a remainder of 16. As you recall, from long division of regular numbers from elementary school, that your remainder (if there is one) has to be less than whatever you divided by. In polynomial terms, since we're dividing by a linear factor (that is, ...
... So we get x2 + 3x + 3 on top (this is q(x)), with a remainder of 16. As you recall, from long division of regular numbers from elementary school, that your remainder (if there is one) has to be less than whatever you divided by. In polynomial terms, since we're dividing by a linear factor (that is, ...
Math 331: hw 7 Solutions 5.1.4 Show that, under congruence
... 5.2.6 Each element of the given congruence-class ring can be written in the form [ax + b] (Why?). Determine the rules for addition and multiplication of congruence classes. (In other words, if the product [ax + b][cx + d] is the class [rx + s], describe how to find r and s from a, b, c, d, and simil ...
... 5.2.6 Each element of the given congruence-class ring can be written in the form [ax + b] (Why?). Determine the rules for addition and multiplication of congruence classes. (In other words, if the product [ax + b][cx + d] is the class [rx + s], describe how to find r and s from a, b, c, d, and simil ...
The Euler characteristic of the moduli space of curves
... The complex Y we need is the " d u a l " to A; its existence is based on the fact that A - A ~ is a manifold. Explicitly, Y has a 6 g - 3 - n cell for each n - 1 cell
... The complex Y we need is the " d u a l " to A; its existence is based on the fact that A - A ~ is a manifold. Explicitly, Y has a 6 g - 3 - n cell for each n - 1 cell
a parallel code for solving linear system equations with multimodular
... However, a direct solution method should be used in the case of dense non structured matrices. If, in addition the coefficients and the right hand side are integers and an exact solution is desired, the only alternative is to use residual arithmetic. For this type of problems we observe a behavior v ...
... However, a direct solution method should be used in the case of dense non structured matrices. If, in addition the coefficients and the right hand side are integers and an exact solution is desired, the only alternative is to use residual arithmetic. For this type of problems we observe a behavior v ...
Solution
... (8) If an−1 ≡ 1 (mod n), then we say n passes Fermat’s primality test for a. Which one of the following statements is TRUE (A) If a number n passes Fermat’s test for some a, then it must be a prime number. (B) If a number n passes Fermat’s test for every a coprime with n, then n must be a prime numb ...
... (8) If an−1 ≡ 1 (mod n), then we say n passes Fermat’s primality test for a. Which one of the following statements is TRUE (A) If a number n passes Fermat’s test for some a, then it must be a prime number. (B) If a number n passes Fermat’s test for every a coprime with n, then n must be a prime numb ...
Factors, Zeros, and Roots: Oh My!
... There are only 6 possible roots: 1,2,4 . In the light of this fact, let’s take a look at the graph of this function. It should be apparent that none of these possible solutions are roots of the function. And without a little help at this point we are absolutely stuck. None of the strategies we h ...
... There are only 6 possible roots: 1,2,4 . In the light of this fact, let’s take a look at the graph of this function. It should be apparent that none of these possible solutions are roots of the function. And without a little help at this point we are absolutely stuck. None of the strategies we h ...
The Fundamental Theorem of Arithmetic: any integer greater than 1
... 1. The Fundamental Theorem of Arithmetic: Any integer greater than 1 can be written as a unique product (up to ordering of the factors) of prime numbers. eg. 13608 23 35 7 2. The remainder theorem: For any positive integers a b , we can find unique integers k and r such that a kb r , whe ...
... 1. The Fundamental Theorem of Arithmetic: Any integer greater than 1 can be written as a unique product (up to ordering of the factors) of prime numbers. eg. 13608 23 35 7 2. The remainder theorem: For any positive integers a b , we can find unique integers k and r such that a kb r , whe ...
Waldspurger formula over function fields
... correspond canonically to left ideal classes of R. Therefore we may identify Pic(X ) with the free abelian group generated by the double cosets in b n+ ,n− . D × \DA∞,× /R ...
... correspond canonically to left ideal classes of R. Therefore we may identify Pic(X ) with the free abelian group generated by the double cosets in b n+ ,n− . D × \DA∞,× /R ...
Final Exam conceptual review
... 14. Let R and S be rings, and let ϕ : R → S be a ring homomorphism. Show that if r is a unit in R, then ϕ(r) is a unit in S. 15. If R is a commutative ring with no zero divisors, show that the units of R[x] are exactly the units of R. 16. If F is an infinite field, show that f (α) = 0 for every α ∈ ...
... 14. Let R and S be rings, and let ϕ : R → S be a ring homomorphism. Show that if r is a unit in R, then ϕ(r) is a unit in S. 15. If R is a commutative ring with no zero divisors, show that the units of R[x] are exactly the units of R. 16. If F is an infinite field, show that f (α) = 0 for every α ∈ ...
How To Prove It
... 1. Analyze the logical forms of the following statements: (a) 3 is a common divisor of 6, 9, and 15. (Note: You did this in exercise 1 of Section 1.1, but you should be able to give a better answer now.) (Solution) Let D (x,3) stand for “ x is divisible by 3.” The entire statement would then be repr ...
... 1. Analyze the logical forms of the following statements: (a) 3 is a common divisor of 6, 9, and 15. (Note: You did this in exercise 1 of Section 1.1, but you should be able to give a better answer now.) (Solution) Let D (x,3) stand for “ x is divisible by 3.” The entire statement would then be repr ...
Pythagorean triples in elementary number theory
... 1. This does not look like our formula for where the leg and hypotenuse differ by 1. Prove that they give the same values. 2. A primitive Pythagorean triple is one where the gcd(a, b, c = 1). Find what constraints on y and z we need to have our general Pythagorean triple output primitive Pythagorean ...
... 1. This does not look like our formula for where the leg and hypotenuse differ by 1. Prove that they give the same values. 2. A primitive Pythagorean triple is one where the gcd(a, b, c = 1). Find what constraints on y and z we need to have our general Pythagorean triple output primitive Pythagorean ...
Chinese remainder theorem
The Chinese remainder theorem is a result about congruences in number theory and its generalizations in abstract algebra. It was first published in the 3rd to 5th centuries by the Chinese mathematician Sun Tzu.In its basic form, the Chinese remainder theorem will determine a number n that, when divided by some given divisors, leaves given remainders. For example, what is the lowest number n that when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 3, and when divided by 7 leaves a remainder of 2?