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lecture09(complete)

# lecture09(complete) - MA1100 Lecture 9 Mathematical Proofs...

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1 MA1100 Lecture 9 Mathematical Proofs Proof by Contradiction Existence Proof Chartrand: section 5.2, 5.4

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Lecture 9 2 Announcement ± Homework set 2 due next Tuesday ± Homework 1 solutions & scores in IVLE ± Lecture Quiz scores (last week) in IVLE
Lecture 9 3 Knights or Knaves knights always tell the truth and knaves always lie . (1) A says : B is a knight. (2) B says : A and I are of opposite type. Are A and B knights or knaves? Assumption : A is a knight (a) (1) implies B is a knight (b) (2) implies A and B are of opposite type (c) Conclusion :A i s a k n a v e (1) implies B is a knave (c) contradicts (a) and (b) this assumption is false since A is telling the truth since B is telling the truth since A is lying

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Lecture 9 4 Knights or Knaves knights always tell the truth and knaves always lie . (1) A says : B and I are knaves. Are A and B knights or knaves? Assumption : A is a knight (a) (1) implies A and B are knaves (b) Conclusion :A i s a k n a v e (1) implies A or B is a knight implies B is a knight this assumption is false since A is telling the truth (b) contradicts (a) since A is lying since A is not a knight
Lecture 9 5 Proof by Contradiction To prove statement R is true Assume ~R is true e.g. R is true or other absurdity To prove universal statement ( " x) R(x) is true Assume ( \$ x) ~R(x) is true . Try to get a contradiction Conclude that R must be true Try to get a contradiction Conclude that ( " x) R(x) must be true To prove the implication ( " x) P(x) Q(x) is true Assume ( \$ x) P(x) ~Q(x) is true . Try to get a contradiction Conclude that ( " x) P(x) Q(x) must be true e.g. P(x) is false, Q(x) is true,…

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Lecture 9 6 Proof by Contradiction Backward-Forward-table Step Statement Reason P P1 : Q1 Q R1 R2 R3 : C We only ask forward questions assumption ( \$ x) [P(x) ~Q(x)] contradiction
Lecture 9 7 Proof by Contradiction Proposition For all positive real numbers x and y, 2 x y y x > + if x y, then Proof by contradiction: Assume ( \$ x \$ y) [P(x,y) ~Q(x,y)] is true: x y and 2 x y y x + P(x,y) Q(x,y) for some positive real numbers x, y ( " x " y) [P(x,y) Q(x,y)]

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Lecture 9 8 Proof by Contradiction Let x, y be positive real numbers Proof such that x y + xy 2 yx By multiplying xy on both sides, we get +≤ 22 xy2 x y Hence beware of inequality −+ x2 x y y 0 −≤ 2 (x y) 0 and = 2 0 Since square can’t be negative we must have = 0 This happens only if This contradicts the fact the x y Start with P ~Q , get ~P Suppose Hence we conclude that + > 2 . (assumption)
Lecture 9 9 Proving “There is no” Proposition Proof by contradiction: ~( \$ n) P(n) Symbolic form There is an integer n such that 2n ª 1 mod 4 Then 2n = 1 + 4k for some integer k Now 2n is even but 1 + 4k is odd.

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lecture09(complete) - MA1100 Lecture 9 Mathematical Proofs...

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