
By Garrett Birkhoff
ISBN-10: 0023100702
ISBN-13: 9780023100703
This vintage, written by way of younger teachers who turned giants of their box, has formed the knowledge of contemporary algebra for generations of mathematicians and continues to be a useful reference and textual content for self research and faculty classes.
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Babylonian mathematicians kept clay tablets on hand containing base 60 multiplication tables. ) Long division works in any base as well. In fact, in base 2 it is particularly easy, because in determining the correct digits of the quotient, no guesswork is involved. On the other hand, in base a where a is large it is harder to determine the digits of the quotient. A. L. Stein in 1960 that the closer the first digit of the divisor is to the base a, the easier guessing the digits of the quotient is.
Q j−1 , q j , q j+1 , . . , qm }. But then the two factorizations of a are the same, and the result is true for the number a. That would prove the theorem by complete induction. We prove that if p1 · . . · pn = q1 · . . · qm , then p1 = q j for some j by means of the following important lemma: 4 Unique Factorization 55 Lemma 3. If p is prime and p divides bc, then p divides b or p divides c. Proof. We use Corollary 8 of Chapter 3, an application of Bezout’s Identity, namely: if a divides bc and (a, b) = 1, then a divides c.
Then we see that 2 is a divisor of 1686: 1686 = 2 · 843, so 3372 = 2 · 2 · 843. Then we see that 3 divides 843: 843 = 3 · 281, so 3372 = 2 · 2 · 3 · 281. Finally, √ we check that 281 is not divisible by 2, 3, 5, 7, 11 or 13, and observe that 17 > 281, so 281 must be prime by Exercise 5, below, and we have the factorization of 3372 into a product of primes, 3372 = 2 · 2 · 3 · 281. There are obvious tricks for testing a number n for divisibility by 2, 3, or 5. Later we shall see tests for 7, 11, and 13.
A survey of modern algebra by Garrett Birkhoff
by Thomas
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