Full text
... generates partitions into at most 2777+1 parts where each part has a value less than or equal to n - 777. Multiplication of iz^+l J by ^ m Z = ^ 1+ 3 + ''" +2-m_ 1 means that we are adding 2777 - 1 to the largest part, 2777 - 3 to the next largest part, 2m - 5 to the next largest part, etc. Since th ...
... generates partitions into at most 2777+1 parts where each part has a value less than or equal to n - 777. Multiplication of iz^+l J by ^ m Z = ^ 1+ 3 + ''" +2-m_ 1 means that we are adding 2777 - 1 to the largest part, 2777 - 3 to the next largest part, 2m - 5 to the next largest part, etc. Since th ...
Formal Power Series and Algebraic Combinatorics S´ eries Formelles et Combinatoire Alg´ ebrique
... we have proven that the initial conditions agree. Note that the sets of (3.1) yielding the terms of these sums are respectively {1}, {2}, { } and ...
... we have proven that the initial conditions agree. Note that the sets of (3.1) yielding the terms of these sums are respectively {1}, {2}, { } and ...
Full text
... black and white. It is known that D ∗ (n) = 1 + D(n) [4]. Definition 3. Let D ∗∗ (n) denote the total number of necklaces of n beads in two colors, with the proviso that at least one bead must be black. Clearly, D ∗∗ (n) = D(n), but we will prove more, namely, we will demonstrate a bijection between ...
... black and white. It is known that D ∗ (n) = 1 + D(n) [4]. Definition 3. Let D ∗∗ (n) denote the total number of necklaces of n beads in two colors, with the proviso that at least one bead must be black. Clearly, D ∗∗ (n) = D(n), but we will prove more, namely, we will demonstrate a bijection between ...
SUMS OF DISTINCT UNIT FRACTIONS PAUL ERDŐS AND
... a, 's in (2 kg, 2k ,+1 ) and by definition to more than half of them there does not correspond any d i ; thus to those a ; 's to which no d corresponds we can make correspond the f X0 .
The proof is then completed as for Theorem 1 . Note tha ...
... a, 's in (2 kg, 2k ,+1 ) and by definition to more than half of them there does not correspond any d i ; thus to those a ; 's to which no d corresponds we can make correspond the f
Square roots
... There are many ways to see this fact, already accessible to us after the first lecture! (In case you are wondering why this might be interesting, it is rumoured that Hippassus of Metapontum was killed over his discovery of this fact, so this was a very astonishing and abrasive discovery at one time. ...
... There are many ways to see this fact, already accessible to us after the first lecture! (In case you are wondering why this might be interesting, it is rumoured that Hippassus of Metapontum was killed over his discovery of this fact, so this was a very astonishing and abrasive discovery at one time. ...
PDF
... Proof. As described above, if S is a set of prime factors of an abundant number, then we may bound each term in the inequality of the previous theorem to obtain the inequality in the current theorem. Assume, then that S is a finite set of prime numbers which satisfies said inequality. Then, by conti ...
... Proof. As described above, if S is a set of prime factors of an abundant number, then we may bound each term in the inequality of the previous theorem to obtain the inequality in the current theorem. Assume, then that S is a finite set of prime numbers which satisfies said inequality. Then, by conti ...
a theorem in the theory of numbers.
... L A G R A N G E has shown that if the indeterminate equation x2 — Ry2 = ± D is resolvable in integers, D being less than VR, and x and y being relative primes, then D is a denominator of a complete quotient in the expansion of VR in a continued fraction. (For a proof of this theorem, see ChrystaPs A ...
... L A G R A N G E has shown that if the indeterminate equation x2 — Ry2 = ± D is resolvable in integers, D being less than VR, and x and y being relative primes, then D is a denominator of a complete quotient in the expansion of VR in a continued fraction. (For a proof of this theorem, see ChrystaPs A ...
NJDOE MODEL CURRICULUM PROJECT CONTENT AREA
... Compare rational and irrational numbers to demonstrate that the decimal expansion of irrational numbers do not repeat; show that every rational number has a decimal expansion which eventually repeats and covert such decimals into rational numbers. Use rational numbers to approximate and locate irrat ...
... Compare rational and irrational numbers to demonstrate that the decimal expansion of irrational numbers do not repeat; show that every rational number has a decimal expansion which eventually repeats and covert such decimals into rational numbers. Use rational numbers to approximate and locate irrat ...
Full text
... Thus, given any number from the collection j 3, 9, 27, 13, 39, 39 \ there exists an/?GT such that/?M is the given number. The following theorem verifies that every positive integer can be obtained in this manner. Before stating the theorem, the following conventions are adopted. The set qf non-negat ...
... Thus, given any number from the collection j 3, 9, 27, 13, 39, 39 \ there exists an/?GT such that/?M is the given number. The following theorem verifies that every positive integer can be obtained in this manner. Before stating the theorem, the following conventions are adopted. The set qf non-negat ...
Read each question carefully. Use complete sentences. Above all
... 2. Why do students need to understand the cancellation property? How would you show them its importance? 3. What role did elliptic curves play in Wiles’ proof of Fermat’s Last Theorem? 4. Show by example that sometimes it is hard to tell whether a number is real. 5. What is Diophantos’ “chord and ta ...
... 2. Why do students need to understand the cancellation property? How would you show them its importance? 3. What role did elliptic curves play in Wiles’ proof of Fermat’s Last Theorem? 4. Show by example that sometimes it is hard to tell whether a number is real. 5. What is Diophantos’ “chord and ta ...
1.16. The Vector Space Cn of n-Tuples of Complex Numbers
... 1.16. The Vector Space Cn of n-Tuples of Complex Numbers The set of all n-tuples of complex numbers is denoted by Cn. By replacing Rn with Cn in the definition of section 1.2, we can turn Cn into a vector space. Note that either R or C can be used as scalars, thus giving rise to a complex vector spa ...
... 1.16. The Vector Space Cn of n-Tuples of Complex Numbers The set of all n-tuples of complex numbers is denoted by Cn. By replacing Rn with Cn in the definition of section 1.2, we can turn Cn into a vector space. Note that either R or C can be used as scalars, thus giving rise to a complex vector spa ...
Full text
... Since composing the article, I have corresponded with Professor Shanks and others whose interest in this topic came to light in that correspondence. It seems that everyone has his own favorite proof of this identity, usually reflecting the individual's background in classical algebra. It also appear ...
... Since composing the article, I have corresponded with Professor Shanks and others whose interest in this topic came to light in that correspondence. It seems that everyone has his own favorite proof of this identity, usually reflecting the individual's background in classical algebra. It also appear ...
THE DIVISOR PROBLEM ON SQUARE
... the sharp asymptotic formula. That is, we will prove the following theorem. This work is supported by Basic Research Fund of Northwestern Polytechnical University of P.R.China(JC201123). ...
... the sharp asymptotic formula. That is, we will prove the following theorem. This work is supported by Basic Research Fund of Northwestern Polytechnical University of P.R.China(JC201123). ...
MODULE 5 Fermat`s Theorem INTRODUCTION
... and consider the linear congruence ax 1(mod p). Then gcd(a, p) = 1. By Theorem 4-4, this congruence admits a unique solution modulo p; hence, there is a unique integer a’, with 1 ≤ a’ ≤ p – 1, satisfying aa’ 1(mod p). Since p is a prime, a = a’ if and only if a = 1 or a = p – 1. Indeed, the cong ...
... and consider the linear congruence ax 1(mod p). Then gcd(a, p) = 1. By Theorem 4-4, this congruence admits a unique solution modulo p; hence, there is a unique integer a’, with 1 ≤ a’ ≤ p – 1, satisfying aa’ 1(mod p). Since p is a prime, a = a’ if and only if a = 1 or a = p – 1. Indeed, the cong ...
Direct Proofs (continued)
... In this lecture Number Theory ● Rational numbers ● Divisibility Proofs ● Direct proofs (cont.) ● Common mistakes in proofs ● Disproof by counterexample ...
... In this lecture Number Theory ● Rational numbers ● Divisibility Proofs ● Direct proofs (cont.) ● Common mistakes in proofs ● Disproof by counterexample ...
Congruent numbers with many prime factors
... of the Theory of Numbers (Chelsea, New York), Vol 2, Chap 16]. In modern language, this object is to find a rational point of infinite order on the elliptic curve my 2 = x 3 − x. Heegner constructed such rational points in the case that m are primes congruent to 5,7 modulo 8 or twice primes congruent ...
... of the Theory of Numbers (Chelsea, New York), Vol 2, Chap 16]. In modern language, this object is to find a rational point of infinite order on the elliptic curve my 2 = x 3 − x. Heegner constructed such rational points in the case that m are primes congruent to 5,7 modulo 8 or twice primes congruent ...
Sperner`s Lemma and its application
... This turns out be link with a very deep result in topology: Hopf’s index theorem. In dimension one, the question is the same as how many times a function will come cross the x-coordinated if we know f (a) < 0 and f (b) > 0. Then if counted by the order of the zero points, we know it will always be o ...
... This turns out be link with a very deep result in topology: Hopf’s index theorem. In dimension one, the question is the same as how many times a function will come cross the x-coordinated if we know f (a) < 0 and f (b) > 0. Then if counted by the order of the zero points, we know it will always be o ...
