A Selection of Problems in the Theory of Numbers. Popular by Waclaw Sierpinski, I. N. Sneddon, M. Stark

By Waclaw Sierpinski, I. N. Sneddon, M. Stark

Show description

Read Online or Download A Selection of Problems in the Theory of Numbers. Popular Lectures in Mathematics PDF

Best mathematics_1 books

Mathematics, Affect and Learning: Middle School Students' Beliefs and Attitudes About Mathematics Education

This e-book examines the ideals, attitudes, values and feelings of scholars in Years five to eight (aged 10 to fourteen years) approximately arithmetic and arithmetic schooling. essentially, this booklet specializes in the improvement of affective perspectives and responses in the direction of arithmetic and arithmetic studying. additionally, it sounds as if scholars enhance their extra unfavourable perspectives of arithmetic throughout the heart college years (Years five to 8), and so the following we be aware of scholars during this severe interval.

Extra info for A Selection of Problems in the Theory of Numbers. Popular Lectures in Mathematics

Sample text

G n. Therefore Qn— 1 < n. But Qn— 1 is not divisible by any prime number < n (because these numbers are divisors of gM) and, as n ^ 4, the number gn—1 ^ g4— 1 = 5 > 1 has a prime divisor /? which must be ^ n. A fortiori, Qn—l}tn which gives a contradiction. Therefore Qn > n for n > 3. About the product Pn of all prime numbers ^ w we can prove that, for a natural number n, Pn < 4n (see Sierpinski [7], p. 396) and, for a natural number n Ξ> 29, P n > 2n. It has also been proved that, for a natural number n > 2, the sum of all prime numbers ^ n is > n.

E. the number p) has eleven digits. It is easy to prove that for integral « > 1 π(η—\) «—1 π(η) « where « is a prime number, and π(η— 1) π(ή) «—1 « where « is a composite number. We can prove in an elementary way that the ratio of π(«) to « tends to zero when « tends to infinity. η) = n for any natural n. It is easy to prove that there exist arbitrarily long sequences of natural numbers which contain no prime numbers. , and the last by m+l; thus they are all composite. For m = 100 the numbers would be gigantic, but between the prime numbers 370,261 and 370,373 there lie 111 successive composite numbers.

Thus in order that the natural number n > 1 be prime it is necessary and sufficient that the number (n—1)! + 1 be divisible by n. Theoretically, then, by only one division we can find out whether a number is prime or not. +1 has more than a hundred digits. +1 is divisible by/? 2 . For/? ^ 50,000, there are only three such numbers: 5, 13 and 563 (Fröberg [1]). We do not know if such numbers p are infinitely many. The theorems of Fermat and Wilson can be combined in the following theorem: If p is a prime number, then for every integer a the number a r(p—l)la is divisible by /?.

Download PDF sample

Rated 4.31 of 5 – based on 24 votes