On Sets Which Are Measured bar Multiples of Irrational Numbers
... for almost all ~ satisfying the condition a(~) E V,-V, . Obviously, to prove the inequality C,:-, = Cr-, it is sufficient to shoe- that the outer Lebesgue measure of the set S= {~ : a($) EV,,-V,} is positive .Supose the contrary, i . e . ...
... for almost all ~ satisfying the condition a(~) E V,-V, . Obviously, to prove the inequality C,:-, = Cr-, it is sufficient to shoe- that the outer Lebesgue measure of the set S= {~ : a($) EV,,-V,} is positive .Supose the contrary, i . e . ...
Infinite Descent - but not into Hell!
... 1 E S, and (b) if k E S, then k + 1 E S too. Then it must be the case that S = N. It can be stated in 'contrapositive form' too. Let S be a subset of the set of natural numbers N with the property that (a) 1 E S, (b) if k rt S for some kEN, k > 1, then k ·- 1 rt S. Then it must be the case that S = ...
... 1 E S, and (b) if k E S, then k + 1 E S too. Then it must be the case that S = N. It can be stated in 'contrapositive form' too. Let S be a subset of the set of natural numbers N with the property that (a) 1 E S, (b) if k rt S for some kEN, k > 1, then k ·- 1 rt S. Then it must be the case that S = ...
MATH10040: Numbers and Functions Homework 5: Solutions
... (a) Let Z≥0 denote the set of nonnegative integers an let Sn,m = {(a1 , . . . , an ) ∈ Zn≥0 | a1 + · · · + an = m}. (For example, S3,2 = {(0, 0, 2), (0, 1, 1), (0, 2, 0), (1, 0, 1), (1, 1, 0), (2, 0, 0)}.) Recall that B`,s is the set of binary strings of length ` containing exactly s 1s. Describe a ...
... (a) Let Z≥0 denote the set of nonnegative integers an let Sn,m = {(a1 , . . . , an ) ∈ Zn≥0 | a1 + · · · + an = m}. (For example, S3,2 = {(0, 0, 2), (0, 1, 1), (0, 2, 0), (1, 0, 1), (1, 1, 0), (2, 0, 0)}.) Recall that B`,s is the set of binary strings of length ` containing exactly s 1s. Describe a ...
![arXiv:1406.2183v1 [math.NT] 9 Jun 2014](http://s1.studyres.com/store/data/017014777_1-03be99f26e7ff20d7b1e4737d0f677b5-300x300.png)
arXiv:1406.2183v1 [math.NT] 9 Jun 2014
... Proof. This is the case b = 6 of Theorem 3. If the Euler prime, q, of n divides both A and B it must divide 2b − 1 = 11, so q = 11 which is impossible since q ≡ 1 (mod 4). If q | B, then A = 2p + 5 = dx2 , with d = 1 or 11. Since x is odd we get 5 ≡ d (mod 8) which is absurd. We conclude that q | A ...
... Proof. This is the case b = 6 of Theorem 3. If the Euler prime, q, of n divides both A and B it must divide 2b − 1 = 11, so q = 11 which is impossible since q ≡ 1 (mod 4). If q | B, then A = 2p + 5 = dx2 , with d = 1 or 11. Since x is odd we get 5 ≡ d (mod 8) which is absurd. We conclude that q | A ...
CSE 20 * Discrete Mathematics
... Assume that for some n2, all integers 2kn are divisible by a prime. WTS that n+1 is divisible by a prime. Proof by cases: Case 1: n+1 is prime. n+1 divides itself so we are done. Case 2: n+1 is composite. Then n+1=ab with 1
... Assume that for some n2, all integers 2kn are divisible by a prime. WTS that n+1 is divisible by a prime. Proof by cases: Case 1: n+1 is prime. n+1 divides itself so we are done. Case 2: n+1 is composite. Then n+1=ab with 1
1 Unique Factorization of Integers
... This tells us, from last time, that there exist integers x, y such that px + ay = 1. This in turn implies that pxb + ayb = b. Now, p|pxb and p|ayb (as p|ab), and thus p must divide b. Note that this is not true for composite numbers: An example to illustrate this is that 6|150 = (10 × 15) but 6 does ...
... This tells us, from last time, that there exist integers x, y such that px + ay = 1. This in turn implies that pxb + ayb = b. Now, p|pxb and p|ayb (as p|ab), and thus p must divide b. Note that this is not true for composite numbers: An example to illustrate this is that 6|150 = (10 × 15) but 6 does ...
pdf-file - Institut for Matematiske Fag
... In 1998, the second author raised the problem of classifying the irreducible characters of Sn of prime power degree. Zalesskii proposed the analogous problem for quasi-simple groups, and he has, in joint work with Malle, made substantial progress on this latter problem. With the exception of the alt ...
... In 1998, the second author raised the problem of classifying the irreducible characters of Sn of prime power degree. Zalesskii proposed the analogous problem for quasi-simple groups, and he has, in joint work with Malle, made substantial progress on this latter problem. With the exception of the alt ...
MATH 363 Discrete Mathematics SOLUTIONS : Assignment 3 1
... to the assumption that p and q have no common factors. Therefore we conclude that 3 2 is irrational. 3. (4pt) Prove that either 2 · 3100 + 5 or 2 · 3100 + 6 is not a perfect square. Let n and n + 1 be a positive integers, we compute the difference between their squares: (n + 1)2 − n2 = n2 + 2n + 1 − ...
... to the assumption that p and q have no common factors. Therefore we conclude that 3 2 is irrational. 3. (4pt) Prove that either 2 · 3100 + 5 or 2 · 3100 + 6 is not a perfect square. Let n and n + 1 be a positive integers, we compute the difference between their squares: (n + 1)2 − n2 = n2 + 2n + 1 − ...
On the prime factors of the number 2 p-1 - 1
... prime factor of the number 2 ~ -1 is the second prime factor of the form (/? - l)x +1. The primep which divided 2(p~1)lq— 1 is the third prime factor of the form(/?—l)x+ 1. Now suppose that/? = 5 (mod 8). In view of a theorem of A. Schinzel [8], there exist two primitive prime factors q and r of the ...
... prime factor of the number 2 ~ -1 is the second prime factor of the form (/? - l)x +1. The primep which divided 2(p~1)lq— 1 is the third prime factor of the form(/?—l)x+ 1. Now suppose that/? = 5 (mod 8). In view of a theorem of A. Schinzel [8], there exist two primitive prime factors q and r of the ...
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"".