1. Direct Proof and Counterexample III Part 2 Lecture 16 Section 3.3 Tue, Feb 13, 2007

2. Example: Direct Proof Theorem: Let a and b be integers. If a | b and b | a, then a =  b. Proof:

3. 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: A property ~ is called antisymmetry if a ~ b and b ~ a implies that a = b.

4. Example: Direct Proof Theorem: Let a, b, c be integers. If a | b and b | a + c, then a | c. Proof:

5. Example: Direct Proof Theorem: If n is odd, then 8 | (n 2 – 1). Proof:

6. Proving Biconditionals To prove a statement  x  D, P(x)  Q(x), we must prove both  x  D, P(x)  Q(x) and  x  D, Q(x)  P(x).

7. Proving Biconditionals Or we could prove both  x  D, P(x)  Q(x) and  x  D,  P(x)   Q(x).

8. Proving Biconditionals A half-integer is a number of the form n + ½, for some integer n. Theorem: Let a and b be real numbers. Then a + b and a – b are integers if and only if a and b are both integers or both half-integers.

9. Proving Biconditionals Proof (  ): Let a and b be real numbers and suppose that a + b and a – b are integers. …

10. Proving Biconditionals Case I: Suppose m and n are both even or both odd. Then …

11. Proving Biconditionals Case II: Suppose one of m and n is even and the other is odd. Then …

12. Proving Biconditionals Proof (  ): Let a and b be real numbers and suppose that a and b are both integers or both half- integers. Case I: Suppose that a and b are both integers. …

13. Proving Biconditionals Case II: Suppose a and b are both half- integers. …

14. The Fundamental Theorem of Arithmetic Theorem: Let a be a positive integer. Then a = p 1 a 1 p 2 a 2 …p k a k, where each p i is a prime and each a i is a nonnegative integer. Furthermore, this representation is unique except for the order of factors.

15. Application of the Fundamental Theorem of Arithmetic Let a and b be positive integers. Then a = p 1 a 1 p 2 a 2 …p k a k and b = p 1 b 1 p 2 b 2 …p k b k. Then the g.c.d. of a and b is gcd(a, b) = p 1 min(a 1, b 1 ) p 2 min(a 2, b 2 ) …p k min(a k, b k )

16. Application of the Fundamental Theorem of Arithmetic and the l.c.m. of a and b is lcm(a, b) = p 1 max(a 1, b 1 ) p 2 max(a 2, b 2 ) …p k max(a k, b k ).

17. Example: gcd’s and lcm’s Let a = 4200 and b = 1080. Then a = 2 3  3 1  5 2  7 1 and b = 2 3  3 2  5 1  7 0. Then gcd(a, b) = 2 3  3 1  5 1  7 0 = 120 and lcm(a, b) = 2 3  3 2  5 2  7 1 = 13600.

18. Example: gcd’s and lcm’s Corollary: Let a and b be positive integers. Then gcd(a, b)  lcm(a, b) = ab. Comment: The Euclidean algorithm is a lot faster.