College Algebra Exam Review 29

# College Algebra Exam Review 29 - c 2 ŒaŁ ŒbŁ Then a ³...

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1.7. MODULAR ARITHMETIC 39 Also denote by rem n .a/ the unique number r such that 0 ± r < n and a ² r is divisible by n . (Proposition 1.6.7 ). Thus rem n .a/ is the unique element of ŒaŁ that lies in the interval f 0;1;:::;n ² 1 g . Equivalently, rem n .a/ is the label on the circumference of the clock face at the point where a falls when the number line is wrapped around the clock face. Lemma 1.7.3. For a;b 2 Z , the following are equivalent: (a) a ³ b . mod n/ . (b) ŒaŁ D ŒbŁ . (c) rem n .a/ D rem n .b/ . (d) ŒaŁ \ ŒbŁ ¤ ; . Proof. Suppose that a ³ b . mod n/ . For any c 2 Z , if c ³ a . mod n/ , then c ³ b . mod n/ , by the previous lemma, part (c). This shows that ŒaŁ ´ ŒbŁ . Likewise, if c ³ b . mod n/ , then c ³ a . mod n/ , so ŒbŁ ´ ŒaŁ . Thus ŒaŁ D ŒbŁ . This shows that (a) implies (b). Since for all integers x , rem n .x/ is characterized as the unique element of ŒxŁ \f 0;1;:::;n ² 1 g , we have (b) implies (c), and also (c) implies (d). Finally, if ŒaŁ \ ŒbŁ ¤ ; , let
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Unformatted text preview: c 2 ŒaŁ \ ŒbŁ . Then a ³ c . mod n/ and c ³ b . mod n/ , so a ³ b . mod n/ . This shows that (d) implies (a). n Corollary 1.7.4. There exist exactly n distinct residue classes modulo n , namely Œ0Ł;Œ1Ł;:::;Œn ² 1Ł . These classes are mutually disjoint. Congruence respects addition and multiplication, in the following sense: Lemma 1.7.5. Let a;a ;b;b be integers with a ³ a . mod n/ and b ³ b . mod n/ . Then a C b ³ a C b . mod n/ and ab ³ a b . mod n/ . Proof. By hypothesis, a ² a and b ² b are divisible by n . Hence .a C b/ ² .a C b / D .a ² a / C .b ² b / is divisible by n , and ab ² a b D .ab ² a b/ C .a b ² a b / D .a ² a /b C a .b ² b / is divisible by n . n...
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