The number of rational numbers determined by large sets of integers
... at least 1, α, β are real numbers in (0, 1]. A standard application of the Möbius inversion formula then shows that |A/B| αβXY . Our purpose is to investigate what might be deduced when in place of intervals we consider arbitrary subsets A and B of the integers in [1, X] and [1, Y ] respectively ...
... at least 1, α, β are real numbers in (0, 1]. A standard application of the Möbius inversion formula then shows that |A/B| αβXY . Our purpose is to investigate what might be deduced when in place of intervals we consider arbitrary subsets A and B of the integers in [1, X] and [1, Y ] respectively ...
Introduction to higher mathematics
... less than the cube root of two. This is a mathematical statement expressed first with symbols and then with English words. That it is a statement is evident from the symbol ⇒. The arrow states that if the left side is true then so is the right side. It is pronounced implies that. The left side start ...
... less than the cube root of two. This is a mathematical statement expressed first with symbols and then with English words. That it is a statement is evident from the symbol ⇒. The arrow states that if the left side is true then so is the right side. It is pronounced implies that. The left side start ...
Full text
... Since these results hold for all integers k J> ls we see that there are an infinite number of heptagonal numbers which are, at the same time9 the sums and differences of distinct heptagonal numbers. Q.E.D. For k = 1, 2, and 3 9 respectively9 Theorem 2 yields ...
... Since these results hold for all integers k J> ls we see that there are an infinite number of heptagonal numbers which are, at the same time9 the sums and differences of distinct heptagonal numbers. Q.E.D. For k = 1, 2, and 3 9 respectively9 Theorem 2 yields ...
Solution
... A Carmichael number is a composite number n satisfying xn−1 ≡ 1 (mod n) for all x ∈ Z× n . Carmichael number are exactly those numbers for which there exist no Fermat witnesses, although they are not prime. The Fermat test will always answer ‘probably prime’ for these numbers, in spite of the fact t ...
... A Carmichael number is a composite number n satisfying xn−1 ≡ 1 (mod n) for all x ∈ Z× n . Carmichael number are exactly those numbers for which there exist no Fermat witnesses, although they are not prime. The Fermat test will always answer ‘probably prime’ for these numbers, in spite of the fact t ...
Full text
... systems theory (which he called "iterated maps") concerning the number of period-/? points. As applications, Lin computed the number N(n) of period-w points of the maps B(ju, x) for some suitably chosen /i and obtained some interesting dividing formulas n\N(n) (i.e., formulas (4.23) in [8]) which ha ...
... systems theory (which he called "iterated maps") concerning the number of period-/? points. As applications, Lin computed the number N(n) of period-w points of the maps B(ju, x) for some suitably chosen /i and obtained some interesting dividing formulas n\N(n) (i.e., formulas (4.23) in [8]) which ha ...
1 Sets
... Sometimes it is difficult to prove directly a statement and is also difficult to find a counterexample. There are statements in mathematics (called open problems) that we know they are sure either true or false, but not both. However, we are just neither able to prove the statement nor to find a cou ...
... Sometimes it is difficult to prove directly a statement and is also difficult to find a counterexample. There are statements in mathematics (called open problems) that we know they are sure either true or false, but not both. However, we are just neither able to prove the statement nor to find a cou ...
June 2015 Question Paper 11 - Cambridge International Examinations
... Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included ...
... Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included ...
Congruence Properties of the Function that Counts Compositions
... Encyclopedia [8]; one can find numerous references there. Congruence properties of b(n) modulo powers of 2 were first observed by R. F. Churchhouse [5] (the main congruence was given without a proof as a conjecture). This conjecture was later proved by H. Gupta [6] and independently by Ø. Rødseth [7 ...
... Encyclopedia [8]; one can find numerous references there. Congruence properties of b(n) modulo powers of 2 were first observed by R. F. Churchhouse [5] (the main congruence was given without a proof as a conjecture). This conjecture was later proved by H. Gupta [6] and independently by Ø. Rødseth [7 ...
INFINITUDE OF ELLIPTIC CARMICHAEL NUMBERS
... In [AGP94.1], the authors prove there are infinitely many Carmichael numbers by constructing infinitely many squarefree composite numbers N such that p − 1|N − 1 for all primes p dividing N . They also mention that being able to construct infinitely many squarefree composite numbers N such that p + ...
... In [AGP94.1], the authors prove there are infinitely many Carmichael numbers by constructing infinitely many squarefree composite numbers N such that p − 1|N − 1 for all primes p dividing N . They also mention that being able to construct infinitely many squarefree composite numbers N such that p + ...
Inductive Reasoning
... 2.1 Inductive Reasoning Objectives: • I CAN use patterns to make conjectures. • I CAN disprove geometric conjectures using counterexamples. ...
... 2.1 Inductive Reasoning Objectives: • I CAN use patterns to make conjectures. • I CAN disprove geometric conjectures using counterexamples. ...
Chinese Reminder Theorem
... Why do this? This is answered in the text (T&W, p. 77). By breaking the problem into simultaneous congruences mod each prime factor of m, we can recombine the resulting information to obtain an answer for each prime factor power of m. The advantage is that it is often easier to analyze congruences m ...
... Why do this? This is answered in the text (T&W, p. 77). By breaking the problem into simultaneous congruences mod each prime factor of m, we can recombine the resulting information to obtain an answer for each prime factor power of m. The advantage is that it is often easier to analyze congruences m ...
Infinitely Many Carmichael Numbers for a Modified Miller
... At the time the best results for E and B allowed the exponent to be 2/7. The exponent has since been improved slightly; see [Har05, Har08]. To achieve this result, [AGP94] show there is an L (parameterized by X) where n((Z/LZ )∗ ) is relatively small compared to L. Ideally, they would have then show ...
... At the time the best results for E and B allowed the exponent to be 2/7. The exponent has since been improved slightly; see [Har05, Har08]. To achieve this result, [AGP94] show there is an L (parameterized by X) where n((Z/LZ )∗ ) is relatively small compared to L. Ideally, they would have then show ...
Solns
... • Suppose y ∈ Z ⇒ 3y ∈ Z. Then it must be the case that 2x ∈ / Z for the same argument. Therefore x ∈ /Z Therefore either x or y is not an integer Prove the following by contrapositive. 1. n(n + 1) is an even number. We want to show that: x is odd ⇒ x 6= n(n + 1) for any n ∈ N. Let x ∈ N, x odd, the ...
... • Suppose y ∈ Z ⇒ 3y ∈ Z. Then it must be the case that 2x ∈ / Z for the same argument. Therefore x ∈ /Z Therefore either x or y is not an integer Prove the following by contrapositive. 1. n(n + 1) is an even number. We want to show that: x is odd ⇒ x 6= n(n + 1) for any n ∈ N. Let x ∈ N, x odd, the ...
On the difference of consecutive primes.
... numbers such that 1 <= x < y < m, and N the number of primes p less than or equal to m such that p+ 1 is not divisible by any of the primes n, where x < P < y. Then ...
... numbers such that 1 <= x < y < m, and N the number of primes p less than or equal to m such that p+ 1 is not divisible by any of the primes n, where x < P < y. Then ...
Algebraic numbers of small Weil`s height in CM
... In the same paper it was shown that Cab cannot be replaced by any constant > (log 7)/12, since there exists an element α in a cyclotomic field such that h(α) = (log 7)/12; we note that (log 7)/12 < C. One may now ask whether the abelianity condition is necessary for this latter lower bound or if thi ...
... In the same paper it was shown that Cab cannot be replaced by any constant > (log 7)/12, since there exists an element α in a cyclotomic field such that h(α) = (log 7)/12; we note that (log 7)/12 < C. One may now ask whether the abelianity condition is necessary for this latter lower bound or if thi ...
![arXiv:1003.5939v1 [math.CO] 30 Mar 2010](http://s1.studyres.com/store/data/016290525_1-37241e0159a202e5bf035fd4e5a48270-300x300.png)
arXiv:1003.5939v1 [math.CO] 30 Mar 2010
... the representative element for the equivalence class under rotation of ω(x). If k does not occur in x, then λ(x) contributes exactly one integer k to the multiset Φ(n). Otherwise, λ(x) contributes m + 1 times to the count of c(n, k), where there are m ≥ 1 parts of x which equal k. In this case, ther ...
... the representative element for the equivalence class under rotation of ω(x). If k does not occur in x, then λ(x) contributes exactly one integer k to the multiset Φ(n). Otherwise, λ(x) contributes m + 1 times to the count of c(n, k), where there are m ≥ 1 parts of x which equal k. In this case, ther ...
Regular tetrahedra whose vertices
... A similar argument to the one in the first part of the proof shows that the fractions in the right-hand side of the equalities of (12) can be simplified only by a factor of 2, 3 or 6. Having four distinct possibilities in (12) for the denominators, exactly one of the fractions (simultaneously in the ...
... A similar argument to the one in the first part of the proof shows that the fractions in the right-hand side of the equalities of (12) can be simplified only by a factor of 2, 3 or 6. Having four distinct possibilities in (12) for the denominators, exactly one of the fractions (simultaneously in the ...
here
... (see [2]). The similar problem of finding the principal quadratic real fields is still an open problem. However, in the last years several progresses arose. A. Biró (see [3] and [4]) proved in 2003 two important results: Yokoi’s conjecture which asserts that h(m2 + 4) = 1 only for six values of m = ...
... (see [2]). The similar problem of finding the principal quadratic real fields is still an open problem. However, in the last years several progresses arose. A. Biró (see [3] and [4]) proved in 2003 two important results: Yokoi’s conjecture which asserts that h(m2 + 4) = 1 only for six values of m = ...