A Pisot number (or P.V. number) is an algebraic integer greater than
... properties of such polynomials Q. Note that Q is anti-reciprocal (that is, Q∗ (z) ≡ −Q(z)), hence 1 is a zero of Q. In [5], we showed that each of Q0 and Q00 has exactly one zero outside D and those two zeros must be real numbers greater than 1. So, if we could find a positive uniform lower bound fo ...
... properties of such polynomials Q. Note that Q is anti-reciprocal (that is, Q∗ (z) ≡ −Q(z)), hence 1 is a zero of Q. In [5], we showed that each of Q0 and Q00 has exactly one zero outside D and those two zeros must be real numbers greater than 1. So, if we could find a positive uniform lower bound fo ...
REPRESENTATIONS OF INTEGERS BY QUADRATIC FORMS As
... is an inhomogeneous polynomial equation (i.e., d ̸= 0), then the determination of whether it has an integer solution is considerably more subtle than whether it has a rational solution. Perhaps the best single example of this is the proven nonexistence of an algorithm to determine whether a polynomi ...
... is an inhomogeneous polynomial equation (i.e., d ̸= 0), then the determination of whether it has an integer solution is considerably more subtle than whether it has a rational solution. Perhaps the best single example of this is the proven nonexistence of an algorithm to determine whether a polynomi ...
DIOPHANTINE APPROXIMATION OF COMPLEX NUMBERS
... many approximations. Of course, this remark also applies to quadratic number fields with slightly different rings of integers. 3. Application to a General Lattice Considering a lattice λ generated by two arbitrary R-linearly independent vectors ω1 = a + ib, ω2 = c + id, where a, b, c, d ∈ R, we obta ...
... many approximations. Of course, this remark also applies to quadratic number fields with slightly different rings of integers. 3. Application to a General Lattice Considering a lattice λ generated by two arbitrary R-linearly independent vectors ω1 = a + ib, ω2 = c + id, where a, b, c, d ∈ R, we obta ...
Lecture 06
... Theorem: Prove that is irrational. Proof: p: is irrational. ¬p : is rational. Assume that ¬p is true. Then = a/b where b ≠ 0, and gcd(a, b) = 1. Therefore, b2 is even. Again then b must be even as well. Why? Exercise! But then 2 must divide both a and b. This contradicts our assumption that a and b ...
... Theorem: Prove that is irrational. Proof: p: is irrational. ¬p : is rational. Assume that ¬p is true. Then = a/b where b ≠ 0, and gcd(a, b) = 1. Therefore, b2 is even. Again then b must be even as well. Why? Exercise! But then 2 must divide both a and b. This contradicts our assumption that a and b ...
Additive properties of even perfect numbers
... 1) is an even perfect number. Two millennia later, Euler proved the converse. Theorem 1. An even positive integer n is perfect if and only if it is of the form n = 2p−1 (2p − 1), where 2p − 1 is prime. For a proof, see [1, Theorem 1.3.3, p. 22]. So we have a characterization of even perfect numbers. ...
... 1) is an even perfect number. Two millennia later, Euler proved the converse. Theorem 1. An even positive integer n is perfect if and only if it is of the form n = 2p−1 (2p − 1), where 2p − 1 is prime. For a proof, see [1, Theorem 1.3.3, p. 22]. So we have a characterization of even perfect numbers. ...
An identity involving counting sums of square and triangular numbers
... The result (1) has an interpretation in terms of sums of square and triangular numbers, which we see by considering the coefficient of q n on each side: Theorem The number of ways of expressing an integer n as the sum of a square and twice a triangular number is the same as the number of ways of exp ...
... The result (1) has an interpretation in terms of sums of square and triangular numbers, which we see by considering the coefficient of q n on each side: Theorem The number of ways of expressing an integer n as the sum of a square and twice a triangular number is the same as the number of ways of exp ...
Primes and Modular Arithmetic
... If p1 is equal to q1, then we can divide both factorizations by that number, and we have a smaller value with two distinct factorizations (contrary to our assumption). So choose the smaller of p1, q1: let’s say it’s p1. Then by our earlier lemma 2, p1 must go into one of the q factors, contrary to o ...
... If p1 is equal to q1, then we can divide both factorizations by that number, and we have a smaller value with two distinct factorizations (contrary to our assumption). So choose the smaller of p1, q1: let’s say it’s p1. Then by our earlier lemma 2, p1 must go into one of the q factors, contrary to o ...
(n!)+
... “There are infinitely many prime numbers.” Any finite set of numbers must contain a maximal element, so we can prove the theorem if we can just show that there is no largest prime number. i.e., show that for any prime number, there is a larger number that is also prime. More generally: For any numbe ...
... “There are infinitely many prime numbers.” Any finite set of numbers must contain a maximal element, so we can prove the theorem if we can just show that there is no largest prime number. i.e., show that for any prime number, there is a larger number that is also prime. More generally: For any numbe ...
CONDITIONAL INDEPENDENCE 1. Introduction
... probability theory is that of independence. This is the condition which serves as part of the hypothesis of most of the major theorems of probability, most notably the laws of large numbers. In many cases, this condition is weakened to stationarity or uncorrelatedness, and usually there is a corresp ...
... probability theory is that of independence. This is the condition which serves as part of the hypothesis of most of the major theorems of probability, most notably the laws of large numbers. In many cases, this condition is weakened to stationarity or uncorrelatedness, and usually there is a corresp ...
4-6 - Mr. Idea Hamster
... 1. Fill in the blanks in the following proof by contradiction that there is no least positive real number. Proof: Suppose not. That is, suppose that there is a least positive real number x. [We must deduee~] Consider the number x/2. Since x is a positive real number, x/2 is also ~. In addition, we ...
... 1. Fill in the blanks in the following proof by contradiction that there is no least positive real number. Proof: Suppose not. That is, suppose that there is a least positive real number x. [We must deduee~] Consider the number x/2. Since x is a positive real number, x/2 is also ~. In addition, we ...
Mathematical Induction Proof by Weak Induction
... collection satisfy some property, we have to choose as base cases the sizes that are meaningful for these objects. In these cases, we actually apply an induction principle to prove some paraphrase of the original instead of the original. However, the paraphrase must imply the original. • A statement ...
... collection satisfy some property, we have to choose as base cases the sizes that are meaningful for these objects. In these cases, we actually apply an induction principle to prove some paraphrase of the original instead of the original. However, the paraphrase must imply the original. • A statement ...
MAT 300/371 Mathematical Structures/Advanced Calculus (Why) is
... removed, and show that the number of these endpoints is countable. However, the Cantor set is uncountable. Challenge: Explicitly determine some elements of the Cantor sets that are not endpoints of any removed intervals. MAT 371: Show that the Cantor set is a perfect set: It is closed and every poin ...
... removed, and show that the number of these endpoints is countable. However, the Cantor set is uncountable. Challenge: Explicitly determine some elements of the Cantor sets that are not endpoints of any removed intervals. MAT 371: Show that the Cantor set is a perfect set: It is closed and every poin ...
CHAP08 Multiplicative Functions
... whenever m, n are coprime. Functions with this property are called multiplicative functions. Of course there are trivial examples such as the constant functions F(n) = 1 or G(n) = n. Less trivial examples are the number of divisors and the sum of the divisors. It is clear that if F(n) is multiplicat ...
... whenever m, n are coprime. Functions with this property are called multiplicative functions. Of course there are trivial examples such as the constant functions F(n) = 1 or G(n) = n. Less trivial examples are the number of divisors and the sum of the divisors. It is clear that if F(n) is multiplicat ...
How to Guess What to Prove Example
... Proof: By strong induction. Let P (n) be the statement that n has a unique factorization. We prove P (n) for n > 1. Basis: P (2) is clearly true. Induction step: Assume P (2), . . . , P (n). We prove P (n + 1). If n + 1 is prime, we are done. If not, it factors somehow. Suppose n + 1 = rs r, s > 1. ...
... Proof: By strong induction. Let P (n) be the statement that n has a unique factorization. We prove P (n) for n > 1. Basis: P (2) is clearly true. Induction step: Assume P (2), . . . , P (n). We prove P (n + 1). If n + 1 is prime, we are done. If not, it factors somehow. Suppose n + 1 = rs r, s > 1. ...