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Direct Proof and Counterexample III Part 2 Lecture 16 Section 3.3 Tue, Feb 13, 2007
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Example: Direct Proof Theorem: Let a and b be integers. If a | b and b | a, then a = b. Proof:
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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.
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Example: Direct Proof Theorem: Let a, b, c be integers. If a | b and b | a + c, then a | c. Proof:
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Example: Direct Proof Theorem: If n is odd, then 8 | (n 2 – 1). Proof:
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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).
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Proving Biconditionals Or we could prove both x D, P(x) Q(x) and x D, P(x) Q(x).
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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.
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Proving Biconditionals Proof ( ): Let a and b be real numbers and suppose that a + b and a – b are integers. …
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Proving Biconditionals Case I: Suppose m and n are both even or both odd. Then …
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Proving Biconditionals Case II: Suppose one of m and n is even and the other is odd. Then …
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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. …
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Proving Biconditionals Case II: Suppose a and b are both half- integers. …
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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.
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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 )
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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 ).
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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.
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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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