CSE 20: Discrete Mathematics for Computer Science Prof. Shachar Lovett

Today’s Topics: 1. Proof by contraposition 2. Proof by cases 2

1. Proof by contraposition 3

Proof by contraposition  To prove a statement of the form  You can equivalently prove its contra-positive form  Remember: (p→q)  (  q→  p) 4

Truth table for implication 5 pqp → q TTT TFF FTT FFT 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 6

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 ???

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 ???

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 ???

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

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.

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

2. Proof by cases 13

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

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

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

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

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

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

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

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

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

Perspective: Theorem in language of graphs  Graph terminology  People = vertices  Shook hands = edge  Didn’t shake hands = no edge 23 A B C D

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.