# Factorization: Unique and Otherwise by Steven H. Weintraub By Steven H. Weintraub

The idea that of factorization, regular within the usual approach of complete numbers that may be written as a different made from top numbers, performs a primary function in smooth arithmetic and its functions. This exposition of the vintage conception leads the reader to an figuring out of the present wisdom of the topic and its connections to different mathematical thoughts, for instance in algebraic quantity concept. The ebook can be utilized as a textual content for a primary path in quantity thought or for self-study through stimulated highschool scholars or readers attracted to glossy arithmetic.

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Extra resources for Factorization: Unique and Otherwise

Example text

Since γ divides each αi , we see that δ divides each αi , so δ is in D({α } ∪ A). On the other hand, suppose δ is in D({α } ∪ A). Then δ divides α , and δ divides each αi . But by the deﬁnition of the gcd of A, δ divides γ. Hence δ is in D({α , γ}). 20. This proof only applies, however, to the case that A = {αi } is a ﬁnite set. 20: Suppose that A = {αi } = {α1 , . . , αn } is a ﬁnite set consisting of n elements. We prove the theorem by induction on n. First, suppose n = 1. Then the gcd of {α1 } is clearly γ = α1 .

Note that the points in the plane that represent elements of Q( D) are precisely the points (e, f ) where both e and f are rational numbers. We will call these points Q-points. , measure of distance) in the plane. We let (e, f ) D = |e2 − f 2 D| and this measures the distance√from the point (e, f ) to the origin. Note that (e, f ) corresponds to e + f D. More √ generally, if (s, t) is another point of f ) − (s, t) = (e − s, f √ − t) the plane, corresponding to s + t √D, then (e,√ corresponds to the diﬀerence (e + f D) − (s + t D) = (e − s) + (f − t) D, and then (e, f ) − (s, t) D = (e − s, f − t) D is the distance between these two points.

Consider a. We have just proved this claim if a = 0 or 0 < a < b , so we may restrict our attention to the case that b ≤ a . But in this case 0 < b ≤ a so 0 ≤ a − b < a. Set a = a − b. Then we can apply the inductive hypothesis to a to conclude that a = bq + r for some r with r = 0 or r < b . Substituting, we see that a − b = bq + r and hence that a = b(q + 1) + r = bq + r with q = q + 1 and r = r . But then also r = 0 or r < b , as required. Hence our claim is true for a, and so by complete induction we may conclude that our claim is true for every a ≥ 0.