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CSE 20: Discrete Mathematics for Computer Science Prof. Shachar Lovett
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Today’s Topics: 1. Proof by contraposition 2. Proof by cases 2
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1. Proof by contraposition 3
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Proof by contraposition To prove a statement of the form You can equivalently prove its contra-positive form Remember: (p→q) ( q→ p) 4
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Truth table for implication 5 pqp → q TTT TFF FTT FFT Rule this row out!
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Contrapositive proof of p→q Procedure: 1. Derive contrpositive form: ( q→ p) 2. Assume q is false (take it as “given”) 3. Show that p logically follows 6
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Example Thm.: “Let x,y be integers such that x 0. Then either x+y 0 or x-y 0.” Proof: Given (contrapositive form): Let WTS (contrapositive form): … Conclusion: … 7 ???
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Example Thm.: “Let x,y be integers such that x 0. Then either x+y 0 or x-y 0.” Proof: Given (contrapositive form): Let … A. x+y 0 or x-y 0 B. x+y=0 or x-y=0 C. x+y=0 and x-y=0 D. y 0 E. None/more/other 8 ???
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Example Thm.: “Let x,y be integers such that x 0. Then either x+y 0 or x-y 0.” Proof: Given (contrapositive form): Let x+y=0 and x-y=0 WTS (contrapositive form): A. x 0 B. x=0 C. x+y 0 or x-y 0 D. x+y=0 or x-y=0 E. None/more/other 9 ???
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Example Thm.: “Let x,y be integers such that x 0. Then either x+y 0 or x-y 0.” Proof: Given (contrapositive form): Let x+y=0 and x-y=0 WTS (contrapositive form): x=0 Conclusion: … 10 Try yourself first
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Example Thm.: “Let x,y be integers such that x 0. Then either x+y 0 or x-y 0.” Proof: Given (contrapositive form): Let x+y=0 and x-y=0 WTS (contrapositive form): x=0 Conclusion: x=0 11 By assumption x+y=0 and x-y=0. Summing the two equations together gives 0=0+0=(x+y)+(x-y)=2x So, 2x=0. Dividing by 2 gives that x=0.
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When should you use contra- positive proofs? You want to prove Which is equivalent to So, it shouldn’t matter which one to prove In practice, one form is usually easier to prove - depending which assumption gives more information (either P(x) or Q(x)) 12
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2. Proof by cases 13
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Breaking a proof into cases Theorem: for any integer n, n(n+1) is even Proof by cases: n is even or n is odd 14
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Breaking a proof into cases Theorem: for any integer n, n(n+1) is even Proof by cases: n is even or n is odd Case 1: n is even. By definition, n=2k for some integer k. Then n(n+1)=2k(2k+1). So n(n+1)=2a for a=k(2k+1). a is an integer, so n(n+1) is even. 15
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Breaking a proof into cases Theorem: for any integer n, n(n+1) is even Proof by cases: n is even or n is odd Case 2: n is odd. By definition, n=2k+1 for some integer k. Then n(n+1)=(2k+1)(2k+2). So n(n+1)=2a for a=(2k+1)(k+1). a is an integer, so n(n+1) is even. 16
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Breaking a proof into cases We will now see a more complicated example 6 friends, Alice, Bob, Charlie, Don, Eve and Frank meet at a party Each pair of friends either shakes hands, or not Theorem: one of these must be true: Either there are either 3 people who all shook hands with each other; Or there are 3 people who all did not shake hands with each other 17
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Breaking a proof into cases Theorem: one of these must be true: Either there exist 3 people who all shook hands with each other; or there exist 3 people who all did not shake hands with each other Proof: The proof is by case analysis. Case 1: Alice shook hands with at least 3 other people. Case 2: Alice shook hands with at most 2 other people. Notice it says “there are two cases” You’d better be right there are no more cases! Cases must completely cover possibilities Tip: you don’t need to worry about trying to make the cases “equal size” or scope Sometimes 99% of the possibilities are in one case, and 1% are in the other Whatever makes it easier to do each proof 18
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Breaking a proof into cases Theorem: one of these must be true: Either there exist 3 people who all shook hands with each other; or there exist 3 people who all did not shake hands with each other Case 1: Alice shook hands with at least 3 other people (lets call them X,Y,Z) Case 1.1: X,Y,Z did not shake hands with each other. In this case we proved the theorem (2 nd option) Case 1.2: Some pair among X,Y,Z shook hands (say X,Y); Then Alice,X,Y all shook hands with each other. In this case we also proved the theorem (1 st option) Notice it says: “This case splits into two subcases” Again, you’d better be right there are no more than these two! Subcases must completely cover the possibilites within the case 19
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Breaking a proof into cases Theorem: one of these must be true: Either there exist 3 people who all shook hands with each other; or there exist 3 people who all did not shake hands with each other Case 2: Alice shook hands with at most 2 other people. Let X,Y,Z be 3 people she did not shake hands with. Case 2.1: X,Y,Z all shook hands with each other. In this case we proved the theorem (1 st option) Case 2.2: Some pair among X,Y,Z did not shake hands (say X,Y); Then Alice,X,Y all did not shake hands with each other. In this case we also proved the theorem (2 nd option). 20
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Breaking a proof into cases Theorem: … Proof: There are two cases to consider Case 1: there are two cases to consider Case 1.1: Verify theorem directly Case 1.2: Verify theorem directly Case 2: there are two cases to consider Case 2.1: Verify theorem directly Case 2.2: Verify theorem directly 21
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Perspective: Theorem in language of graphs Graph: diagram which captures relations between pairs of objects Example: objects=people, relation=shook hands 22 A B C D A,B shook hands A,C shook hands A,D shook hands B,C shook hands B,D didn’t shake hands C,D didn’t shake hands
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Perspective: Theorem in language of graphs Graph terminology People = vertices Shook hands = edge Didn’t shake hands = no edge 23 A B C D
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Perspective: Theorem in language of graphs 24 Equivalent theorem Theorem: any graph with 6 vertices either contains a triangle (3 vertices with all edges between them) or an empty triangle (3 vertices with no edges between them) Same theorem as before, except that we ignore the meaning of edges and non-edges in the story This is math – isolating the core ingredients in a question, and then solving it generically.
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