13(3)
... The California Mathematics Council All subscription correspondence should be addressed to Professor Leonard Klosinski, Mathematics Department University of Santa Clara, Santa Clara, California 95053* All checks ($12.00 per year) should be made out to the Fibonacci Association or the Fibonacci Quarte ...
... The California Mathematics Council All subscription correspondence should be addressed to Professor Leonard Klosinski, Mathematics Department University of Santa Clara, Santa Clara, California 95053* All checks ($12.00 per year) should be made out to the Fibonacci Association or the Fibonacci Quarte ...
Introduction to the Theory of Computation
... Jack sees Jill, who has just come in from outdoors Proof by Induction dry. Jack knows that it is not raining. For any n, there exist n consecutive composite integers. His proof: if it were raining (assuming the statement Proof: here Jill are the n be consecutive composite integers is false), woud we ...
... Jack sees Jill, who has just come in from outdoors Proof by Induction dry. Jack knows that it is not raining. For any n, there exist n consecutive composite integers. His proof: if it were raining (assuming the statement Proof: here Jill are the n be consecutive composite integers is false), woud we ...
Chapter 5 - Set Theory
... • In looking at the truth table, we consider the cases where the hypothesis is false to yield vacuous results. The interesting cases are when the hypothesis is true. ...
... • In looking at the truth table, we consider the cases where the hypothesis is false to yield vacuous results. The interesting cases are when the hypothesis is true. ...
Public Key Encryption
... – given n = n1n2…nk then the structure of Zn is identical to Zn1 x Zn2 x … x Znk – this can give efficient algorithms since Zn can be decomposed into smaller systems ...
... – given n = n1n2…nk then the structure of Zn is identical to Zn1 x Zn2 x … x Znk – this can give efficient algorithms since Zn can be decomposed into smaller systems ...
Why Is the 3X + 1 Problem Hard? - Department of Mathematics, CCNY
... what he meant and the conversation meandered off to other subjects. Over the next two weeks I finally saw the ergodic theory perspective which Dennis had been pointing along. This viewpoint, worked out by him and David Ruelle over lunch one day, does not solve the problem. Instead, it suggests why i ...
... what he meant and the conversation meandered off to other subjects. Over the next two weeks I finally saw the ergodic theory perspective which Dennis had been pointing along. This viewpoint, worked out by him and David Ruelle over lunch one day, does not solve the problem. Instead, it suggests why i ...
MORE ON THE TOTAL NUMBER OF PRIME FACTORS OF AN ODD
... point approximation for σ−1 (N ). It should be noted that when we start with a prime p other than 3, and we have already proven a contradiction for all primes between 3 and p, then we may assume that P N for all 3 ≤ P < p. This is taken into account in contradiction (1). This procedure is done on ...
... point approximation for σ−1 (N ). It should be noted that when we start with a prime p other than 3, and we have already proven a contradiction for all primes between 3 and p, then we may assume that P N for all 3 ≤ P < p. This is taken into account in contradiction (1). This procedure is done on ...
THE p–ADIC ORDER OF POWER SUMS, THE ERD
... Proof. (i). Assume that m ≡ −1 (mod p). Then by Remark 7 we have Vp (m) = vp (m + 1). If p − 1 | n, then using Theorem 4 and applying vp to both sides of equation (2) gives Vp (m) − 1 = vp (Sn (m)) = vp (a) + nvp (m + 1) = vp (a) + nVp (m), contradicting vp ≥ 0 and Vp ≥ 0. Therefore p − 1 - n. (ii). ...
... Proof. (i). Assume that m ≡ −1 (mod p). Then by Remark 7 we have Vp (m) = vp (m + 1). If p − 1 | n, then using Theorem 4 and applying vp to both sides of equation (2) gives Vp (m) − 1 = vp (Sn (m)) = vp (a) + nvp (m + 1) = vp (a) + nVp (m), contradicting vp ≥ 0 and Vp ≥ 0. Therefore p − 1 - n. (ii). ...
Exceptional real Lucas sequences
... it suffices to take I > 0. For if (U) and (Uf) are generated by #2 — Iz + m and z2 + Iz + m, respectively, then Z7W = (—ly^Ul. In all that follows we therefore suppose I > 0. If i2 > 4m, (C7) will be called real. Birkhoff and Vandiver [1] have shown that when a and /3 are coprime rational integers t ...
... it suffices to take I > 0. For if (U) and (Uf) are generated by #2 — Iz + m and z2 + Iz + m, respectively, then Z7W = (—ly^Ul. In all that follows we therefore suppose I > 0. If i2 > 4m, (C7) will be called real. Birkhoff and Vandiver [1] have shown that when a and /3 are coprime rational integers t ...
Polynomials with integer values.
... decomposition in algebraic number fields. Start with any (complex) root α of f and look at the field K = Q(α) of all those complex numbers which can be written as polynomials in α with coefficients from Q. The basic fact that we will be using (without proof) is that any nonzero ideal in ‘the ring of ...
... decomposition in algebraic number fields. Start with any (complex) root α of f and look at the field K = Q(α) of all those complex numbers which can be written as polynomials in α with coefficients from Q. The basic fact that we will be using (without proof) is that any nonzero ideal in ‘the ring of ...
On the Distribution of Counter-Dependent Nonlinear Congruential
... It is obvious that the sequence (1) eventually becomes periodic with some period t ≤ M 2 . Throughout this paper we assume that this sequence is purely periodic, that is, un = un+t beginning with n = 0, otherwise we consider a shift of the original sequence. In the case that f (X, Y ) = h(X) ∈ ZZM [ ...
... It is obvious that the sequence (1) eventually becomes periodic with some period t ≤ M 2 . Throughout this paper we assume that this sequence is purely periodic, that is, un = un+t beginning with n = 0, otherwise we consider a shift of the original sequence. In the case that f (X, Y ) = h(X) ∈ ZZM [ ...
On nonexistence of an integer regular polygon∗
... a regular polygon with n sides, whereby n is not a prime, it suffices to study all regular polygons having the number of sides equal to any divisor of the number n, different from 2. If such polygons with integer coordinates do not exist, then an integer polygon with n sides does not exist either. W ...
... a regular polygon with n sides, whereby n is not a prime, it suffices to study all regular polygons having the number of sides equal to any divisor of the number n, different from 2. If such polygons with integer coordinates do not exist, then an integer polygon with n sides does not exist either. W ...
Full text
... The smallest F.Psp. is Qi = 705. It was discovered by M. Pettet in 1966 [9] who discovered also Q2 - 2465 and Q3 = 2 7 3 7 , but we cannot forget the unbelievable misfortune of D. Lind [10] who in 1967 limited his computer experiment for disproving the converse of (1.6) to n = 700, thus missing the ...
... The smallest F.Psp. is Qi = 705. It was discovered by M. Pettet in 1966 [9] who discovered also Q2 - 2465 and Q3 = 2 7 3 7 , but we cannot forget the unbelievable misfortune of D. Lind [10] who in 1967 limited his computer experiment for disproving the converse of (1.6) to n = 700, thus missing the ...
Powerpoint of Notes
... How can you tell You can tell whether a conjecture is reasonable by making a prediction. whether a conjecture Example: Is the conjecture that average hourly earnings in 2013 will be abou is reasonable? $18.25 reasonable? ...
... How can you tell You can tell whether a conjecture is reasonable by making a prediction. whether a conjecture Example: Is the conjecture that average hourly earnings in 2013 will be abou is reasonable? $18.25 reasonable? ...
Number Theory Learning Module 2 — Prime Numbers and the
... The last number was factored by Morrison and Brillhart in 1970, using one of the first algorithms for factoring especially designed for modern electronic computers. Notice that our definition explicitly excludes negative numbers. This convention is not universal, but facilitates the statement of som ...
... The last number was factored by Morrison and Brillhart in 1970, using one of the first algorithms for factoring especially designed for modern electronic computers. Notice that our definition explicitly excludes negative numbers. This convention is not universal, but facilitates the statement of som ...
3.4 Complex Zeros and the Fundamental Theorem of Algebra
... takes time to digest. Don’t be overly concerned if it doesn’t seem to sink in all at once, and pace yourself in the Exercises or you’re liable to get mental cramps. But before we get to the Exercises, we’d like to offer a bit of an epilogue. Our main goal in presenting the material on the complex ze ...
... takes time to digest. Don’t be overly concerned if it doesn’t seem to sink in all at once, and pace yourself in the Exercises or you’re liable to get mental cramps. But before we get to the Exercises, we’d like to offer a bit of an epilogue. Our main goal in presenting the material on the complex ze ...
Diophantine approximation with primes and powers of two
... coefficient-free version discussed above. On the other hand, certain aspects of our analysis are actually simpler because we have less need for information about the distribution of primes in arithmetic progressions. We prove Theorem 1 using the Davenport-Heilbronn version of the HardyLittlewood metho ...
... coefficient-free version discussed above. On the other hand, certain aspects of our analysis are actually simpler because we have less need for information about the distribution of primes in arithmetic progressions. We prove Theorem 1 using the Davenport-Heilbronn version of the HardyLittlewood metho ...
New York Journal of Mathematics Diophantine approximation with primes and
... coefficient-free version discussed above. On the other hand, certain aspects of our analysis are actually simpler because we have less need for information about the distribution of primes in arithmetic progressions. We prove Theorem 1 using the Davenport-Heilbronn version of the HardyLittlewood metho ...
... coefficient-free version discussed above. On the other hand, certain aspects of our analysis are actually simpler because we have less need for information about the distribution of primes in arithmetic progressions. We prove Theorem 1 using the Davenport-Heilbronn version of the HardyLittlewood metho ...
Induction
... color”, referring to something that is quite different from normal or common expectation. The famous mathematician George Polya (who was also a great expositor of mathematics for the lay public) gave the following proof to show that there is no horse of a different color! Theorem: All horses are the ...
... color”, referring to something that is quite different from normal or common expectation. The famous mathematician George Polya (who was also a great expositor of mathematics for the lay public) gave the following proof to show that there is no horse of a different color! Theorem: All horses are the ...