Direct Proof and Counterexample III

1. Direct Proof and Counterexample III Lecture 14 Section 3.3 Thu, Feb 3, 2005

2. Divisibility • Definition: An integer adivides an integer b if a 0 and there exists an integer c such that ac = b. • Write a | b to indicate that a divides b. • Divisibility is a positive property.

3. Prime Numbers • Definition: An integer p is prime if p 2 and the only positive divisors of p are 1 and p. • A prime number factors only in a trivial way: p = 1  p. • Prime numbers: 2, 3, 5, 7, 11, … • Is this a positive property? • Is there a positive characterization of primes?

4. Composite Numbers • Definition: An integer n is composite if there exist integers a and b such that 1 < a < n and 1 < b < n and n = ab. • A composite number factors in a non-trivial way : n = a  b, a > 1, b > 1. • Composite numbers: 4, 6, 8, 9, 10, 12, … • Is this a positive property?

5. Units and Zero • Definition: An integer u is a unit if u | 1. • The only units are 1 and –1. • Definition: 0 is zero.

6. Example: Direct Proof • Theorem: If a | b and b | c, then a | c. • Proof: • Let a, b, c be integers and assume a | b and b | c. • Since a | b, there exists an integer r such that ar = b. • Since b | c, there exists an integer s such that bs = c. • Therefore, a(rs) = (ar)s = bs = c. • So a | c.

7. Example: Direct Proof • Theorem: Let a and b be integers. If a | b and b | a, then a = b. • Proof: • Let a and b be integers. • Suppose a | b and b | a. • There exist integers c and d such that ac = b and bd = a. • Therefore, acd = bd = a.

8. Example: Direct Proof • Therefore, cd = 1. • Thus, c = d = 1 or c = d = -1. • Then a = b or a = -b.

9. Example: Direct Proof • Corollary: If a, bN and a | b and b | a, then a = b. • This is analogous to the set-theoretic statement that if A B and B  A, then A = B. • Preview: This property is called antisymmetry. • If a ~ b and b ~ a, then a = b.

10. Example: Direct Proof • Theorem: Let a, b, c be integers. If a | b and b | a + c, then a | c. • Proof: • Let a, b, and c be integers. • Suppose a | b and b | a + c. • There exist integers r and s such that ar = b and bs = a + c.

11. Example: Direct Proof • Substitute ar for b in the 2nd equation to get (ar)s = a + c. • Rearrange the terms and factor to get a(rs – 1) = c. • Therefore, a | c.

12. Example: Direct Proof • Theorem: If n is odd, then 8 | (n2 – 1). • Proof: • Let n be an odd integer. • Then n = 2k + 1 for some integer k. • So n2 – 1 = (2k + 1)2 – 1 = 4k2 + 4k = 4k(k + 1).

13. Example: Direct Proof • Either k or k + 1 is even. • Therefore, k(k + 1) is a multiple of 2. • Therefore, n2 – 1 is a multiple of 8. • Can you think of an alternate, simpler proof, based on the factorization of n2 – 1?

14. Example: Direct Proof • Theorem: If n is odd, then 24 | (n3 – n). • Proof: • ?

15. Example: If-and-Only-If Proof • Theorem: Let a, b be integers. Then a | b if and only if a2 | b2. • Proof: () • Suppose that a | b. • Then ac = b for some cZ. • So, a2c2 = b2 and therefore, a2 | b2.

16. Example: If-and-Only-If Proof • Proof: () • Suppose that a2 | b2. • Then a2d = b2 for some dZ. • Now what?

17. The Fundamental Theorem of Arithmetic • Theorem: Let n be a positive integer. Then n = p1a1p2a2…pkak, where each pi is prime and each ai is nonnegative. Furthermore, the primes and their exponents are unique.

18. The Fundamental Theorem of Arithmetic • Use the Fundamental Theorem of Arithmetic to complete the previous proof.