By Goodman F.M.
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Proof. For part (a), note that neither u nor v can be zero. Suppose first that both are positive. We have uv maxfu; vg 1. If equality holds, then u D v D 1. In general, if uv D 1, then also jujjvj D 1, so juj D jvj D 1. Thus both u and v are ˙1. Since their product is positive, both have the same sign. For (b), let u; v be integers such that b D ua and a D vb. Then b D uvb. uv 1/b. Now the product of nonzero integers is nonzero, so either b D 0 or uv D 1. In the former case, a D b D 0, and in the latter case, v D ˙1 by part (a), so a D ˙b.
Imagine taking a number 38 1. ALGEBRAIC THEMES line, with the integer points marked, and wrapping it around the circumference of the clock, with 0 on the number line coinciding with 0 on the clock face. Then the numbers : : : ; 3n; 2n; n; 0; n; 2n; 3n; : : : on the number line all overlay 0 on the clock face. The numbers : : : ; 3n C 1; 2n C 1; n C 1; 1; n C 1; 2n C 1; 3n C 1; : : : on the number line all overlay 1 on the clock face. In general, for 0 Ä k Ä n 1, the numbers : : : ; 3n C k; 2n C k; n C k; k; n C k; 2n C k; 3n C k; : : : on the number line all overlay k on the clock face.
21. The greatest common divisor of nonzero integers a1 ; a2 ; : : : ; an exists, and is an integer linear combination of a1 ; a2 ; : : : ; an . Proof. d; 0; : : : ; 0/Q 1 : It follows that d is the greatest common divisor of a1 ; : : : ; an . a1 ; : : : ; an /. 1. 7 using the well–ordering principle. (a) (b) If a ¤ 0, consider the set S of nonnegative integers that can be written in the form a sd , where s is an integer. Show that S is nonempty. By the well–ordering principle, S has a least element, which we write as r D a qd .
Algebra. Abstract and concrete by Goodman F.M.