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Unformatted text preview: Proof by Smallest Counterexample
Deﬁnitions: • log2 (n) is x such that 2x = n. log2 (n) is the unique i s.t. 2i ≤ n < 2i+1 e.g. log2 (2) = 1, log2 (3) = 1, log2 (4) = 2 log2 (31) = 4, log2 (32) = 5, log2 (33) = 5 • Prime factorization of n is the representation of n as multiplication of a list of primes. e.g. 12 = 2 × 2 × 3, 6! = 2 × 2 × 2 × 2 × 3 × 3 × 5 • Deﬁne SIZE (n) to be the number of prime factors in prime factorization of n. e.g. SIZE (12) = 3, SIZE (6!) = SIZE (720) = 7
1 Proof by Smallest Counterexample
Theorem: For any positive integer n, SIZE (n) ≤ log2 (n) . Proof: Let P (n) be the statement SIZE (n) ≤ log2 (n) . Assume the theorem is wrong. i.e. There is a smallest integer m s.t. P (m) is false. Let p be a prime factor of m. Then, SIZE (m) = SIZE (m/p × p) By deﬁnition = SIZE (m/p) + 1 m/p < m, so P (m/p) is true. ≤ log2 (m/p) + 1 By deﬁnition ≤ log2 (m/2) + 1 ≤ log2 (m) Contradiction! 2 Proof by Contradiction
Theorem: There are inﬁnitely many number of primes. Proof: Assume the number of primes is ﬁnite. Let m be the largest prime. Consider n = m! + 1, n mod 2 = 1 n mod 3 = 1 n mod 5 = 1 . . . n mod m = 1 ⇒ No prime is a factor of n . ⇒ n is a prime greater than m.
3 Contradiction! ...
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This note was uploaded on 08/25/2010 for the course COMP COMP170 taught by Professor M.j.golin during the Spring '10 term at HKUST.
 Spring '10
 M.J.Golin
 Computer Science

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