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Unformatted text preview: roduct of primes. To prove this, let p(n) denote the statement (n = 1) or (n is a product of primes). First of all, clearly p(1) is true. Also 2 is a prime, and so is a product of primes, so that p(2) is true. Suppose now that n > 2 and that p(m) is true for every 1 ≤ m < n. Then in particular, every m ∈ N satisfying 2 ≤ m < n is representable as a product of primes. If n is a prime, then it is obviously representable as a product of primes. If n is not a prime, then there exist n1 , n2 ∈ N satisfying 2 ≤ n1 < n and 2 ≤ n2 < n such that n = n1 n2 . By our induction hypothesis, both n1 and n2 are representable as products of primes, so that n must be representable as a product of primes, whence p(n) is true. It now follows from the Principle of induction (Strong form) that every natural number n > 1 is representable as a product of primes. 1.3. Completeness of the Real Numbers The set Z of all integers is an extension of the set N of all natural numbers to include 0 and all numbers of the form −n, where n ∈ N....
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This note was uploaded on 02/01/2009 for the course MATH 2343124 taught by Professor Staff during the Fall '08 term at UCSD.
 Fall '08
 staff
 Math, Calculus

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