College Algebra Exam Review 51

College Algebra Exam Review 51 - 1.9. COUNTING 61 Lemma...

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Unformatted text preview: 1.9. COUNTING 61 Lemma 1.9.8. Let p be a prime number. Âà p (a) If 0 < k < p , then is divisible by p . k (b) For all integers a and b , .a C b/p Á ap C b p .mod p/. Proof. For part (a), we have pŠ D Âà p kŠ .p k k /Š. Now p divides pŠ, Âà p but p does not divide kŠ or .n k /Š. Consequently, p divides . Âà k Pp p knk The binomial theorem gives .a C b/p D ab . By k D0 k part (a), all the terms for 0 < k < p are divisible by p , so .a C b/p is congruent modulo p to the sum of the terms for k D 0 and k D p , namely to ap C b p . I I hope you already discovered the first part of the next lemma while doing the exercises for Section 1.7. Proposition 1.9.9. (a) Let n 2 be a natural number. An element Œa 2 Zn has a multiplicative inverse if, and only if, a is relatively prime to n. (b) If p is a prime, then every nonzero element of Zp is invertible. Proof. If a is relatively prime to n, there exist integers s; t such that as C nt D 1. But then as Á 1 .mod n/, or ŒaŒs D Œ1 in Zn . On the other hand, if a is not relatively prime to n, then a and n have a common divisor k > 1. Say kt D a and ks D n. Then as D k t s D nt . Reducing modulo n gives ŒaŒs D Œ0, so Œa is a zero divisor in Zn , and therefore not invertible. If p is a prime, and Œa ¤ 0 in Zp , then a is relatively prime to p , so Œa is invertible in Zp by part (a). I Proposition 1.9.10. (Fermat’s little theorem). Let p be a prime number. (a) For all integers a, we have ap Á a .mod p/. (b) If a is not divisible by p , then ap 1 Á 1 .mod p/. ...
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