Sol 2 - D-MATH
... 1. (a) Is there an integral domain that contains exactly 15 elements ? Solution : The only additive abelian group of order 15 is the cyclic group Z15 (you can see this with Sylow, and conclude that all groups of order 15 are isomorphic to the product Z3 × Z5 of cyclic groups – cf. your notes of last ...
... 1. (a) Is there an integral domain that contains exactly 15 elements ? Solution : The only additive abelian group of order 15 is the cyclic group Z15 (you can see this with Sylow, and conclude that all groups of order 15 are isomorphic to the product Z3 × Z5 of cyclic groups – cf. your notes of last ...
Some facts about polynomials modulo m
... One can put Theorem 1 as follows: If r is a root of f , one can “factor out” the factor x − r from f . The next theorem says that if m is a prime number this can be done with several roots of a polynomial one after the other. The proof is a little more tricky. Theorem 2 Consider polynomials modulo m ...
... One can put Theorem 1 as follows: If r is a root of f , one can “factor out” the factor x − r from f . The next theorem says that if m is a prime number this can be done with several roots of a polynomial one after the other. The proof is a little more tricky. Theorem 2 Consider polynomials modulo m ...
THE CHINESE REMAINDER THEOREM CLOCK FIGURE 1. The
... Secondly, the Clock is a good excuse for practising divisions and for familiarizing with the periodic nature of remainders. Moreover, some small arithmetical problems may be visualized on the Clock dial (see Section 4). Last but not least the Clock can be entertaining (also for non-mathematicians): ...
... Secondly, the Clock is a good excuse for practising divisions and for familiarizing with the periodic nature of remainders. Moreover, some small arithmetical problems may be visualized on the Clock dial (see Section 4). Last but not least the Clock can be entertaining (also for non-mathematicians): ...
integers
... Integers a and b are said to be congruent modulo n, being n a positive integer if they have the same remainder when divided by n. This is denoted by writing a b (mod n). When working with congruence modulo n, the integer n is called the modulus. By the preceding proposition, a b (mod n) if and only ...
... Integers a and b are said to be congruent modulo n, being n a positive integer if they have the same remainder when divided by n. This is denoted by writing a b (mod n). When working with congruence modulo n, the integer n is called the modulus. By the preceding proposition, a b (mod n) if and only ...
Full text
... modulo m. Many moduli 777, characterized in [1], have the property that every residue modulo 777 occurs in each period. (Indeed, 8 and 11 are the smallest moduli which do not have this property.) However, moduli m with the property that all m residues modulo m appear in one period the some nwnhev of ...
... modulo m. Many moduli 777, characterized in [1], have the property that every residue modulo 777 occurs in each period. (Indeed, 8 and 11 are the smallest moduli which do not have this property.) However, moduli m with the property that all m residues modulo m appear in one period the some nwnhev of ...
Addition of polynomials Multiplication of polynomials
... Example 7. Find the greatest common divisor of a(x) = 2x3 +x2 −2x−1 and b(x) = x3 −x2 +2x−2. Solution. We use the Euclidean Algorithm: first divide a(x) by b(x), then divide b(x) by the remainder, then divide the first remainder by the new remainder, and so on. The last non-zero remainder is the gre ...
... Example 7. Find the greatest common divisor of a(x) = 2x3 +x2 −2x−1 and b(x) = x3 −x2 +2x−2. Solution. We use the Euclidean Algorithm: first divide a(x) by b(x), then divide b(x) by the remainder, then divide the first remainder by the new remainder, and so on. The last non-zero remainder is the gre ...
Week 10
... ∴ the smallest integer in the sum of 25 consecutive positive integers is 28. Note: The sum simplifies to 25n = 1000 because for each positive integer 1 to 12 in the sum, the corresponding integer of opposite sign −1 to −12 also appears. Then (1 + 2 + · · · + 11 + 12) + (−1 − 2 − · · · − 11 − 12) = 0 ...
... ∴ the smallest integer in the sum of 25 consecutive positive integers is 28. Note: The sum simplifies to 25n = 1000 because for each positive integer 1 to 12 in the sum, the corresponding integer of opposite sign −1 to −12 also appears. Then (1 + 2 + · · · + 11 + 12) + (−1 − 2 − · · · − 11 − 12) = 0 ...
Problem 1 Problem 2
... a with (a, n) = 1, are Carmichael numbers, and these are known to exist (e.g., n = 561). Remark: Note that the existence of pseudoprimes, i.e., composite numbers satisfying an−1 ≡ 1 mod n for a given base a, is not enough to disprove the statement, since pseudoprimes are defined with respect to a fi ...
... a with (a, n) = 1, are Carmichael numbers, and these are known to exist (e.g., n = 561). Remark: Note that the existence of pseudoprimes, i.e., composite numbers satisfying an−1 ≡ 1 mod n for a given base a, is not enough to disprove the statement, since pseudoprimes are defined with respect to a fi ...
MATHEMATICS CONTEST 2003 SOLUTIONS PART I: 30 Minutes
... b√ = 652 − a 2 . Now try different integers in place of a: 652 − 1 is not an integer, nor is 652 − 22 , etc. The first triples you get this way are 16, 63, 65 and 25, 60, 65. (There are two other answers as well: 33, 56, 65 and 39, 52, 65.) The trial and error process can be shortened by using the t ...
... b√ = 652 − a 2 . Now try different integers in place of a: 652 − 1 is not an integer, nor is 652 − 22 , etc. The first triples you get this way are 16, 63, 65 and 25, 60, 65. (There are two other answers as well: 33, 56, 65 and 39, 52, 65.) The trial and error process can be shortened by using the t ...
Fermat`s Last Theorem - UCLA Department of Mathematics
... known to many ancient cultures long before Pythagoras himself. There is evidence that the ancient Egyptians, Babylonians, Indians, and Chinese all knew the relationship among the sides of a right triangle. The ancients also knew a few Pythagorean triples, meaning threesomes of integers a, b, c with ...
... known to many ancient cultures long before Pythagoras himself. There is evidence that the ancient Egyptians, Babylonians, Indians, and Chinese all knew the relationship among the sides of a right triangle. The ancients also knew a few Pythagorean triples, meaning threesomes of integers a, b, c with ...
Full text
... —, an) denote a permutation of Zn. A rise of IT is a pair of consecutive elements^-, a[+i such that a\ < a{+i; in addition a conventional rise to the left of at is included. Then [6, Ch. 8] Antk is equal to the number of permutations of Zn with exactly k rises. To get a combinatorial interpretation ...
... —, an) denote a permutation of Zn. A rise of IT is a pair of consecutive elements^-, a[+i such that a\ < a{+i; in addition a conventional rise to the left of at is included. Then [6, Ch. 8] Antk is equal to the number of permutations of Zn with exactly k rises. To get a combinatorial interpretation ...
1 Factorization of Polynomials
... • Euclid’s Lemma for Polynomials: If p(x) is irreducible and p(x)|(f (x)g(x)), then p(x)|d(x) or p(x)|g(x). • Proof: Similar to proof for integers. If p(x) does not divide f (x), then gcd(p(x), g(x)) = 1 because the only factors of p(x) are 1 and p(x) (up to multiplication by units). So 1 = s(x)p(x) ...
... • Euclid’s Lemma for Polynomials: If p(x) is irreducible and p(x)|(f (x)g(x)), then p(x)|d(x) or p(x)|g(x). • Proof: Similar to proof for integers. If p(x) does not divide f (x), then gcd(p(x), g(x)) = 1 because the only factors of p(x) are 1 and p(x) (up to multiplication by units). So 1 = s(x)p(x) ...
01 - University of South Carolina
... 4. Explain what 43/2 means, and then calculate 43/2 in two different ways: ...
... 4. Explain what 43/2 means, and then calculate 43/2 in two different ways: ...
REVISED 3/30/14 Ms C. Draper lesson elements for Week of ___3
... expressions and equations. We will move forward with Pythagorean theorem and continue to review & practice basic skills. Students are now moving in and out of their “power groups” and returning to home teams to share and/or peer teach concepts at 4-teired readiness levels. This week tiered groups wi ...
... expressions and equations. We will move forward with Pythagorean theorem and continue to review & practice basic skills. Students are now moving in and out of their “power groups” and returning to home teams to share and/or peer teach concepts at 4-teired readiness levels. This week tiered groups wi ...
pdf-file
... 1.5. Factor Theorem Theorem 1.5. If x − a is a factor of a polynomial f (x), then f (a) = 0. Conversely, if f (a) = 0, then x − a is a factor of f (x). Example 1.6. Is x − 1 is a factor of f (x) = x5 − 1? Since f (1) = 0, the Factor Theorem tells us it must be. In fact x5 − 1 = (x − 1)(x4 + x3 + x2 ...
... 1.5. Factor Theorem Theorem 1.5. If x − a is a factor of a polynomial f (x), then f (a) = 0. Conversely, if f (a) = 0, then x − a is a factor of f (x). Example 1.6. Is x − 1 is a factor of f (x) = x5 − 1? Since f (1) = 0, the Factor Theorem tells us it must be. In fact x5 − 1 = (x − 1)(x4 + x3 + x2 ...
2 Congruences
... The integer m is called the modulus of the congruence. Proposition 2.2 (Elementary properties of congruences). Let a, b, c, d ∈ Z, m ∈ N. (i) If a ≡ b mod m and c ≡ d mod m, then a + c ≡ b + d mod m. (ii) If a ≡ b mod m and c ≡ d mod m, then ac ≡ bd mod m. (iii) If a ≡ b mod m, then an ≡ bn mod m fo ...
... The integer m is called the modulus of the congruence. Proposition 2.2 (Elementary properties of congruences). Let a, b, c, d ∈ Z, m ∈ N. (i) If a ≡ b mod m and c ≡ d mod m, then a + c ≡ b + d mod m. (ii) If a ≡ b mod m and c ≡ d mod m, then ac ≡ bd mod m. (iii) If a ≡ b mod m, then an ≡ bn mod m fo ...
Exam II Review Sheet Solutions
... The second exam will be on Thursday, March 29. The syllabus will consist of Chapter NT from the text, together with the two number theory supplements passed out in class (Divisibility and Congruences). For reference, I will refer to these as Supplement D and Supplement C. You should be able to do al ...
... The second exam will be on Thursday, March 29. The syllabus will consist of Chapter NT from the text, together with the two number theory supplements passed out in class (Divisibility and Congruences). For reference, I will refer to these as Supplement D and Supplement C. You should be able to do al ...
Math 154. Norm and trace An interesting application of Galois theory
... as follows: NL/k (a) = det(ma ), TrL/k (a) = trace(ma ) where ma : L → L is the k-linear map of multiplication by a. Since ma ◦ ma0 = maa0 , ma + ma0 = ma+a0 , and (for c ∈ k) mca = c · ma , the multiplicativity of determinants and the k-linearity of traces immediately imply that NL/k is multiplicat ...
... as follows: NL/k (a) = det(ma ), TrL/k (a) = trace(ma ) where ma : L → L is the k-linear map of multiplication by a. Since ma ◦ ma0 = maa0 , ma + ma0 = ma+a0 , and (for c ∈ k) mca = c · ma , the multiplicativity of determinants and the k-linearity of traces immediately imply that NL/k is multiplicat ...
Introduction to higher mathematics
... 17x ≤ 18x for any x |λ| · |µ| = |λµ| for any λ and µ π = 3.14 The first statement is false since both sides are zero. The second statement is true for positive x but false for x = −1. The last statement is false since π is only approximately equal to 3.14. But the third statement is true. The absolu ...
... 17x ≤ 18x for any x |λ| · |µ| = |λµ| for any λ and µ π = 3.14 The first statement is false since both sides are zero. The second statement is true for positive x but false for x = −1. The last statement is false since π is only approximately equal to 3.14. But the third statement is true. The absolu ...
2. EUCLIDEAN RINGS
... Then there’s the area of commutative rings, especially those where the cancellation law breaks down. This is a large and important subject but, so far, I haven’t written any notes on it. An important sub-department of this is the theory of integral domains and Euclidean rings where there is a divisi ...
... Then there’s the area of commutative rings, especially those where the cancellation law breaks down. This is a large and important subject but, so far, I haven’t written any notes on it. An important sub-department of this is the theory of integral domains and Euclidean rings where there is a divisi ...
Euler`s Formula and the Fundamental Theorem of Algebra
... to z. This angle is called an argument of z. (In particular, θ + 2kπ is another argument for z for any k ∈ Z) ...
... to z. This angle is called an argument of z. (In particular, θ + 2kπ is another argument for z for any k ∈ Z) ...
A coprimality condition on consecutive values of polynomials
... Erdős [5] was the first to prove the existence of Gs when s is the sequence of natural numbers. Later, the combined efforts of Pillai [14] and Brauer [3] gave a more explicit result, namely that gs = Gs = 17. We note that interest in such a problem is twofold. On one hand, Pillai aimed at the solut ...
... Erdős [5] was the first to prove the existence of Gs when s is the sequence of natural numbers. Later, the combined efforts of Pillai [14] and Brauer [3] gave a more explicit result, namely that gs = Gs = 17. We note that interest in such a problem is twofold. On one hand, Pillai aimed at the solut ...
Lecture 8 - Math TAMU
... For any integer n > 0 the prime factorisation of 10n is 2n ·5n . As follows from the Unique Factorisation Theorem, a positive integer A divides another positive integer B if and only if the prime factorisation of A is part of the prime factorisation of B. Hence 10n divides the given number if n ≤ 73 ...
... For any integer n > 0 the prime factorisation of 10n is 2n ·5n . As follows from the Unique Factorisation Theorem, a positive integer A divides another positive integer B if and only if the prime factorisation of A is part of the prime factorisation of B. Hence 10n divides the given number if n ≤ 73 ...
Chapter 4: Polynomials A polynomial is an expression of the form p
... according to what we have just learned, f (X) ≡ (X − a1 )(X − a2 ) · · · (X − am ) is a factor of p(X). This cannot happen because the degree of f (X) is m, which is greater than the degree of p(X), which is n. Indeed, the degree of a factor of p cannot be greater than the degree of p. Next we study ...
... according to what we have just learned, f (X) ≡ (X − a1 )(X − a2 ) · · · (X − am ) is a factor of p(X). This cannot happen because the degree of f (X) is m, which is greater than the degree of p(X), which is n. Indeed, the degree of a factor of p cannot be greater than the degree of p. Next we study ...
Note One
... We write equations in Zn as L = R and use the congruence notation when working in Z. 7. Z∗n = {a ∈ Zn : ∃b : ab = 1}, i.e. the set of elements with a multiplicative inverse. From Bezout’s lemma, this can be identified with the set of nonnegative integers less than or equal to n and relatively prime ...
... We write equations in Zn as L = R and use the congruence notation when working in Z. 7. Z∗n = {a ∈ Zn : ∃b : ab = 1}, i.e. the set of elements with a multiplicative inverse. From Bezout’s lemma, this can be identified with the set of nonnegative integers less than or equal to n and relatively prime ...
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?