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CSE 20 DISCRETE MATH Prof. Shachar Lovett http://cseweb.ucsd.edu/classes/wi15/cse20-a/ Clicker frequency: CA Todays topics • Proof by contraposition • Proof by cases • Section 3.5 in Jenkyns, Stephenson Proof by contraposition • To prove a statement of the form (x)( P( x) Q( x)) • You can equivalently prove its contra-positive form (x)(Q( x) P( x)) • Remember: (p→q) (q→ p) Truth table for implication p T T F F q T F T F p→q T F T T Rule this row out! 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 Example • Thm.: “Let x,y be numbers such that x0. Then either x+y0 or x-y0.” • Proof: • Given (contrapositive form): Let • WTS (contrapositive form): … ??? • Conclusion: … Example • Thm.: “Let x,y be numbers such that x0. Then either x+y0 or x-y0.” • Proof: • Given (contrapositive form): Let … A. x+y0 or x-y0 B. x+y=0 or x-y=0 C. x+y=0 and x-y=0 D. y0 E. None/more/other Example • Thm.: “Let x,y be numbers such that x0. Then either x+y0 or x-y0.” • Proof: • Given (contrapositive form): Let x+y=0 and x-y=0 • WTS (contrapositive form): A. x0 B. x=0 C. x+y0 or x-y0 D. x+y=0 or x-y=0 E. None/more/other Example • Thm.: “Let x,y be numbers such that x0. Then either x+y0 or x-y0.” • Proof: • Given (contrapositive form): Let x+y=0 and x-y=0 • WTS (contrapositive form): x=0 Try it yourself first • Conclusion: … When should you use contra-positive proofs? • You want to prove (x)( P( x) Q( x)) • Which is equivalent to (x)(Q( x) P( x)) • 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)) 11 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 12 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) 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 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 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. 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. 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. 18 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 Next class • Indirect proof techniques • Read section 3.5 in Jenkyns, Stephenson