Unformatted text preview: 62 1. ALGEBRAIC THEMES Proof. It suffices to show this for a a natural number (since the case a D is trivial and the case a < 0 follows from the case a > 0 ). The proof goes by induction on a . For a D 1 , the assertion is obvious. So assume that a > 1 and that .a 1/ p a 1 . mod p/ . Then a p D ..a 1/ C 1/ p .a 1/ p C 1 . mod p/ .a 1/ C 1 . mod p/ D a; where the first congruence follows from Lemma 1.9.8 , and the second from the induction assumption. The conclusion of part (a) is equivalent to OEaŁ p D OEaŁ in Z p . If a is not divisible by p , then by the previous proposition, OEaŁ has a multiplicative inverse OEaŁ 1 in Z p . Multiplying both sides of the equation by OEaŁ 1 gives OEaŁ p 1 D OE1Ł . But this is equivalent to a p 1 1 . mod p/ . n Inclusion–Exclusion Suppose you have three subsets A , B and C of a finite set U , and you want to count A [ B [ C . If you just add j A jCj B jCj C j , then you might have too large a result, because any element that is in more than one of...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
- Fall '08