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Direct Proof and Counterexample III Part 2

Direct Proof and Counterexample III Part 2. Lecture 16 Section 3.3 Tue, Feb 13, 2007. Example: Direct Proof. Theorem: Let a and b be integers. If a | b and b | a , then a =  b . Proof:. Example: Direct Proof.

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Direct Proof and Counterexample III Part 2

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  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 | (n2 – 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 = p1a1p2a2…pkak, where each pi is a prime and each ai 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 = p1a1p2a2…pkak and b = p1b1p2b2…pkbk. • Then the g.c.d. of a and b is gcd(a, b) = p1min(a1,b1)p2min(a2,b2)…pkmin(ak,bk)

  16. Application of the Fundamental Theorem of Arithmetic and the l.c.m. of a and b is lcm(a, b) = p1max(a1,b1)p2max(a2,b2)…pkmax(ak,bk).

  17. Example: gcd’s and lcm’s • Let a = 4200 and b = 1080. • Then a = 23315271 and b = 23325170. • Then gcd(a, b) = 23315170 = 120 and lcm(a, b) = 23325271 = 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.

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