EGYPTIAN FRACTIONS WITH EACH DENOMINATOR HAVING
... result that every rational number of the form m/n, where n is square-free, can be expressed as an Egyptian fraction whose denominators are the product of three primes. The problem is that the structure of Ln+3 (n) is not well understood at the ends of the interval. So the main challenge becomes in r ...
... result that every rational number of the form m/n, where n is square-free, can be expressed as an Egyptian fraction whose denominators are the product of three primes. The problem is that the structure of Ln+3 (n) is not well understood at the ends of the interval. So the main challenge becomes in r ...
Lecture 9: Integers, Rational Numbers and Algebraic Numbers
... The rational numbers Q and the real numbers R are examples of what is called an ordered field . More generally, a field is a set F upon which operations of “addition” and “multiplication” are defined and for which the following axioms are satisfied: ...
... The rational numbers Q and the real numbers R are examples of what is called an ordered field . More generally, a field is a set F upon which operations of “addition” and “multiplication” are defined and for which the following axioms are satisfied: ...
constant curiosity - users.monash.edu.au
... let’s spare a thought for a few of the lesser known mathematical constants — ones which might not permeate the various fields of mathematics but have nevertheless been immortalised in the mathematical literature in one way or another. In this seminar, we’ll consider a few of these numerical curios a ...
... let’s spare a thought for a few of the lesser known mathematical constants — ones which might not permeate the various fields of mathematics but have nevertheless been immortalised in the mathematical literature in one way or another. In this seminar, we’ll consider a few of these numerical curios a ...
Solutions
... number m > −1 is not a lower bound, so any lower bound is less that or equal to −1, which means that −1 is the infimum of E. 2 Show that if E is a nonempty bounded subset of Z, then both sup E and inf E exist and belong to E. For sup E, the proof of Theorem 1.21 can be applied here because nowhere i ...
... number m > −1 is not a lower bound, so any lower bound is less that or equal to −1, which means that −1 is the infimum of E. 2 Show that if E is a nonempty bounded subset of Z, then both sup E and inf E exist and belong to E. For sup E, the proof of Theorem 1.21 can be applied here because nowhere i ...
Methods of Proof
... Via indirect proof (Contrapositive) Need to show: If n is odd, then n3+5 is even Assume n is odd, and show that n3+5 is even n=2k+1 for some integer k (definition of odd numbers) n3+5 = (2k+1)3+5 = 8k3+12k2+6k+6 = 2(4k3+6k2+3k+3) As 2(4k3+6k2+3k+3) is 2 times an integer, it is even ...
... Via indirect proof (Contrapositive) Need to show: If n is odd, then n3+5 is even Assume n is odd, and show that n3+5 is even n=2k+1 for some integer k (definition of odd numbers) n3+5 = (2k+1)3+5 = 8k3+12k2+6k+6 = 2(4k3+6k2+3k+3) As 2(4k3+6k2+3k+3) is 2 times an integer, it is even ...
A Tail of Two Palindromes - Mathematical Association of America
... where (1810)(−305) − (−743)(743) = −1. Moreover, in these two cases, the indicated coefficients are the most petite ones that satisfy the conditions of (3); that is, all four of the coefficients in each case have the smallest possible absolute values. √ How would someone find those somewhat hefty in ...
... where (1810)(−305) − (−743)(743) = −1. Moreover, in these two cases, the indicated coefficients are the most petite ones that satisfy the conditions of (3); that is, all four of the coefficients in each case have the smallest possible absolute values. √ How would someone find those somewhat hefty in ...
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 ...
notes - Department of Computer Science and Engineering, CUHK
... The textbook states a more general version of the theorem accounting for the possibility that the subproblem sizes might not be exactly b1 n, . . . , bk n but perhaps deviate a bit from these values. Instead of showing the general version, let me just say that when this deviation is sufficiently sma ...
... The textbook states a more general version of the theorem accounting for the possibility that the subproblem sizes might not be exactly b1 n, . . . , bk n but perhaps deviate a bit from these values. Instead of showing the general version, let me just say that when this deviation is sufficiently sma ...
3 - Utrecht University Repository
... exaggerated way one might say that the values of σ are really eager to be perfect powers. The proof can be found in [5, Theorem 1.1], where an exponent slightly larger than 0.7 is given. We are thankful to the referee for pointing out Freiberg’s preprint to us. The paper [2] contains the ingredient ...
... exaggerated way one might say that the values of σ are really eager to be perfect powers. The proof can be found in [5, Theorem 1.1], where an exponent slightly larger than 0.7 is given. We are thankful to the referee for pointing out Freiberg’s preprint to us. The paper [2] contains the ingredient ...
ELEMENTARY NUMBER THEORY
... demanded in the way of specific prerequisites. A significant portion of the book can be profitably read by anyone who has taken the equivalent of a first-year college course in mathematics. Those who have had additional courses will generally be better prepared, if only because of their enhanced mat ...
... demanded in the way of specific prerequisites. A significant portion of the book can be profitably read by anyone who has taken the equivalent of a first-year college course in mathematics. Those who have had additional courses will generally be better prepared, if only because of their enhanced mat ...
Lecture 10 - 188 200 Discrete Mathematics and Linear Algebra
... Given a statement : ∃x P(x), how do we show it is true? We only have to show that P(x) is true for at least one x. There are two types of existential proofs: 1. Constructive proofs — find a specific value of x for which P(x) is true. 2. Non-constructive proofs — show that such x exists, but do not a ...
... Given a statement : ∃x P(x), how do we show it is true? We only have to show that P(x) is true for at least one x. There are two types of existential proofs: 1. Constructive proofs — find a specific value of x for which P(x) is true. 2. Non-constructive proofs — show that such x exists, but do not a ...
Some sufficient conditions of a given series with rational terms
... In series theory, there is well known Cauchy convergence test which is used to determine convergence of a given series,but Cauchy convergence test usually is not practical in most applications,so there come out various convergence test such as D’Alembert convergence test ,integral convergence test a ...
... In series theory, there is well known Cauchy convergence test which is used to determine convergence of a given series,but Cauchy convergence test usually is not practical in most applications,so there come out various convergence test such as D’Alembert convergence test ,integral convergence test a ...
without
... M328K Final Exam, May 10, 2003 1. “Bibonacci” numbers. The Bibonacci numbers b1 , b2 , . . . are defined by b1 = 1, b2 = 1, and, for n > 2, bn = bn−1 + 2bn−2 . a) Prove that, for all positive integers n, bn ≤ 2n−1 . b) Prove that, for all positive integers n, bn ≥ 2n−2 . 2. The following theorem-pro ...
... M328K Final Exam, May 10, 2003 1. “Bibonacci” numbers. The Bibonacci numbers b1 , b2 , . . . are defined by b1 = 1, b2 = 1, and, for n > 2, bn = bn−1 + 2bn−2 . a) Prove that, for all positive integers n, bn ≤ 2n−1 . b) Prove that, for all positive integers n, bn ≥ 2n−2 . 2. The following theorem-pro ...
On an Integer Sequence Related to a Product Combinatorial Relevance
... and C2 cannot have the same sum for every n ≥ 12. As in the previous cases all partitions produced in this way are distinct. This theorem has the following consequence: Corollary 2.4 We have S(n) > 6n/4 ≈ 1.56508n ...
... and C2 cannot have the same sum for every n ≥ 12. As in the previous cases all partitions produced in this way are distinct. This theorem has the following consequence: Corollary 2.4 We have S(n) > 6n/4 ≈ 1.56508n ...
A Property of 70
... p3 >p t g i (because p ; > q ;) . Thus all ak corresponding to n >289 are primes or powers of primes . The same conclusion holds for 70 < n < 289, and may be verified by direct computation . By more complicatedd methods, we can prove the following related result : THEOREM 2 . For all sufficiently la ...
... p3 >p t g i (because p ; > q ;) . Thus all ak corresponding to n >289 are primes or powers of primes . The same conclusion holds for 70 < n < 289, and may be verified by direct computation . By more complicatedd methods, we can prove the following related result : THEOREM 2 . For all sufficiently la ...
Keys GEO Openers 4-15
... CI s: 2 inside || lines on SAME side of transversal. CO s: 1 inside || lines & 1 outside || lines, on OPPOSITE sides of transversal. AI s: 2 inside || lines on OPPOSITE sides of transversal. AE s: 2 outside || lines on OPPOSITE sides of transversal. ...
... CI s: 2 inside || lines on SAME side of transversal. CO s: 1 inside || lines & 1 outside || lines, on OPPOSITE sides of transversal. AI s: 2 inside || lines on OPPOSITE sides of transversal. AE s: 2 outside || lines on OPPOSITE sides of transversal. ...
some remarks on number theory >t 6
... we obtain by a simple computation that for all n c l / loglog n < A(n, a) < c2 logloglog n . III Sivasankaranarayana, Pillai and Szekeres proved that for 1 < l _<_ 16 any sequence of l consecutive integers always contains one which is relatively prime to the others, but that this is in general not t ...
... we obtain by a simple computation that for all n c l / loglog n < A(n, a) < c2 logloglog n . III Sivasankaranarayana, Pillai and Szekeres proved that for 1 < l _<_ 16 any sequence of l consecutive integers always contains one which is relatively prime to the others, but that this is in general not t ...
Statistics of incomplete quotients of continued fractions of quadratic
... Otherwise, considering equality (11) modulo 4 or modulo p, we get the equation x2 ≡ −1; it is well known that the latter equation is neither solvable modulo 4 nor modulo p. Now Theorem 3.4 is reduced to the assertion that all primes in the form 4k + 1 belong to K. This fact was proved by Legendre (s ...
... Otherwise, considering equality (11) modulo 4 or modulo p, we get the equation x2 ≡ −1; it is well known that the latter equation is neither solvable modulo 4 nor modulo p. Now Theorem 3.4 is reduced to the assertion that all primes in the form 4k + 1 belong to K. This fact was proved by Legendre (s ...