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CSE 20 DISCRETE MATH Prof. Shachar Lovett Clicker frequency: CA

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Todays topics Proof by contraposition Proof by cases Section 3.5 in Jenkyns, Stephenson

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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)

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Truth table for implication 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

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Example Thm.: “Let x,y be numbers such that x 0. Then either x+y 0 or x-y 0.” Proof: Given (contrapositive form): Let WTS (contrapositive form): … Conclusion: … ???

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Example Thm.: “Let x,y be numbers 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

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Example Thm.: “Let x,y be numbers 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

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Example Thm.: “Let x,y be numbers 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: … Try it yourself first

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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))

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Breaking a proof into cases Sometimes it is useful to break a proof into a few cases, and then prove each one individually We will see an example demonstrating this principle 11

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Breaking a proof into cases 6 people at a party Any two people either know each other, or not (it is symmetric: if A knows B then B knows A) Theorem: among any 6 people, either there are 3 who all know each other (3 friends), or there are 3 who all don’t know each other (3 strangers) 12

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Breaking a proof into cases Theorem: Every 6 people includes 3 friends or 3 strangers Proof: The proof is by case analysis. Let x denote one of the 6 people. There are two cases: Case 1: x knows at least 3 of the other 5 people Case 2: x knows at most 2 of the other 5 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

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Breaking a proof into cases Theorem: Every 6 people includes 3 friends or 3 strangers Case 1: suppose x knows at least 3 other people. Case 1.1: No pair among these 3 people know each other. Case 1.2: Some pair among these people know each other. 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 possibilities within the case

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Breaking a proof into cases Theorem: Every 6 people includes 3 friends or 3 strangers Case 1: suppose x knows at least 3 other people Case 1.1: No pair among these people know each other. Case 1.2: Some pair among these people know each other. Proof for case 1.1: Let y,z,w be 3 friends of x. By assumption, none of them knows each other. So {y,z,w} is a set of 3 strangers.

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Breaking a proof into cases Theorem: Every 6 people includes 3 friends or 3 strangers Case 1: suppose x knows at least 3 other people Case 1.1: No pair among these people know each other. Case 1.2: Some pair among these people know each other. Proof for case 1.2: Let y,z be 2 friends of x who know each other. So {x,y,z} is a set of 3 friends.

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Breaking a proof into cases Theorem: Every 6 people includes 3 friends or 3 strangers Case 2: suppose x knows at most 2 of the other 5 people So, there are at least 3 people x doesn’t know Cases 2.1: All pairs among these people know each other. Case 2.2: Some pair among these people don’t know each other. Then this pair, together with x, form a group of 3 strangers. So the theorem holds in this subcase.

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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 18

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Next class Indirect proof techniques Read section 3.5 in Jenkyns, Stephenson

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