Large gaps between consecutive prime numbers
... log x p6z p log x log z of the remaining numbers. There are alternative approaches using explicit choices for ap ; we will choose our ap at random. (The set V of numbers surviving this second sieving has about the same size in each case.) If |V | 6 π(x) − π(δx), the number of “very large” primes, th ...
... log x p6z p log x log z of the remaining numbers. There are alternative approaches using explicit choices for ap ; we will choose our ap at random. (The set V of numbers surviving this second sieving has about the same size in each case.) If |V | 6 π(x) − π(δx), the number of “very large” primes, th ...
On the non-existence of constants of derivations: the proof of a
... Let us now make precise some notions that date back to Darboux’s memoir [2]. We are of course responsible for the names given to these notions. A non-trivial solution F of equation (2.4) will be called a Darboux polynomial of derivation dV and the algebraic hypersurface {F = 0} in C I n a Darboux m ...
... Let us now make precise some notions that date back to Darboux’s memoir [2]. We are of course responsible for the names given to these notions. A non-trivial solution F of equation (2.4) will be called a Darboux polynomial of derivation dV and the algebraic hypersurface {F = 0} in C I n a Darboux m ...
Chapter 12: Ruler and compass constructions
... In 1796, nineteen-year-old Carl Friedrich Gauß, who was undecided about whether to study mathematics or languages, discovered how to construct a regular 17-gon. Gauß was so pleased with his discovery that he dedicated his life to mathematics. He also came up with the following theorem about which n- ...
... In 1796, nineteen-year-old Carl Friedrich Gauß, who was undecided about whether to study mathematics or languages, discovered how to construct a regular 17-gon. Gauß was so pleased with his discovery that he dedicated his life to mathematics. He also came up with the following theorem about which n- ...
1.2 Inductive Reasoning
... 3. If the product of two numbers is even, the numbers must be even. 4. If a shape has two sides the same length, it must be a rectangle. ...
... 3. If the product of two numbers is even, the numbers must be even. 4. If a shape has two sides the same length, it must be a rectangle. ...
1.2 Inductive Reasoning
... 3. If the product of two numbers is even, the numbers must be even. 4. If a shape has two sides the same length, it must be a rectangle. ...
... 3. If the product of two numbers is even, the numbers must be even. 4. If a shape has two sides the same length, it must be a rectangle. ...
Introduction to Number Theory
... facts that you will be able to easily verify: a. If a|b, then a|kb, ka|kb. b. If ka|kb and k 6= 0, then a|b. c. If a|b and b|c, then a|c. d. If a|b and a|c, then a|(mb + nc). e. If a|b and b|a, then a = ±b. f. If a|b, a, b > 0, then a ≥ b. g. For any choice of a, b, there exists a unique q, r, 0 ≤ r ...
... facts that you will be able to easily verify: a. If a|b, then a|kb, ka|kb. b. If ka|kb and k 6= 0, then a|b. c. If a|b and b|c, then a|c. d. If a|b and a|c, then a|(mb + nc). e. If a|b and b|a, then a = ±b. f. If a|b, a, b > 0, then a ≥ b. g. For any choice of a, b, there exists a unique q, r, 0 ≤ r ...
A STUDY OF EULERIAN NUMBERS FOR PERMUTATIONS IN THE
... canonical permutations of the form a1 a2 · · · an−1 n by (i) and (ii) of Section 1. It suffices to deal only with canonical ones in counting orbits. If A = a1 a2 · · · an−1 n ∈ Ee− (n, k), we see that a1 a2 · · · an−1 ∈ Ee (n − 1, k − 1), since inv(a1 a2 · · · an−1 ) = inv(A) and n is deleted. There ...
... canonical permutations of the form a1 a2 · · · an−1 n by (i) and (ii) of Section 1. It suffices to deal only with canonical ones in counting orbits. If A = a1 a2 · · · an−1 n ∈ Ee− (n, k), we see that a1 a2 · · · an−1 ∈ Ee (n − 1, k − 1), since inv(a1 a2 · · · an−1 ) = inv(A) and n is deleted. There ...
Fermat's Last Theorem
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. The cases n = 1 and n = 2 were known to have infinitely many solutions. This theorem was first conjectured by Pierre de Fermat in 1637 in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. The first successful proof was released in 1994 by Andrew Wiles, and formally published in 1995, after 358 years of effort by mathematicians. The theretofore unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century. It is among the most notable theorems in the history of mathematics and prior to its proof it was in the Guinness Book of World Records for ""most difficult mathematical problems"".