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CSE 20 – Discrete Mathematics CSE 20 – Discrete Mathematics

CSE 20 – Discrete Mathematics - PowerPoint Presentation

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CSE 20 – Discrete Mathematics - PPT Presentation

Dr Cynthia Bailey Lee Dr Shachar Lovett                             Peer Instruction in Discrete Mathematics by  Cynthia Lee is licensed under a  Creative Commons Attribution ID: 621460

theorem people cases proof people theorem proof cases form contrapositive club don

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Slide1

CSE 20 – Discrete Mathematics

Dr. Cynthia Bailey LeeDr. Shachar Lovett

 

                         

Peer Instruction in Discrete Mathematics by 

Cynthia

Lee

is

licensed under a 

Creative Commons Attribution-

NonCommercial

-

ShareAlike

4.0 International License

.

Based on a work at 

http://peerinstruction4cs.org

.

Permissions beyond the scope of this license may be available at 

http://peerinstruction4cs.org

.

Slide2

Today’s Topics:

Proof by contrapositionProof by cases

2Slide3

1. Proof by contraposition

3Slide4

Proof by contraposition

To prove a statement of the formYou can equivalently prove its contra-positive formRemember:

(p→q)  (q→  p)4Slide5

Contrapositive proof of

p→qProcedure:

Derive contrapositive form: (q→  p)Assume q is false (take it as “given”)Show that p logically follows5Slide6

Truth table for implication

6

pq~p~q~q → ~pTTFFTTFFT

FFTT

F

T

F

F

T

T

T

Rule this row out!Slide7

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: …7???Slide8

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 or x-y0x+y=0 or x-y=0x+y=0 and x-y=0y0None/more/other8

???Slide9

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=0WTS (contrapositive form):x0x=0x+y0 or x-y0x+y=0 or x-y=0None/more/other

9

???Slide10

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=0WTS (contrapositive form): x=0Conclusion: …10Try yourself firstSlide11

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=0WTS (contrapositive form): x=0Conclusion: x=0, which is what was to be shown, so the theorem is true.11By assumption x+y

=0 and x-y=0. Summing the two equations together gives0=0+0=(x+y)+(x-y)=2xSo, 2x=0. Dividing by 2 gives that x=0.Slide12

When should you use contra-positive proofs?

You want to proveWhich 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))12Slide13

2. Proof by cases

13Slide14

Breaking a proof into cases

Assume any two people either know each other or not (if A knows B then B knows A)We will prove the following theoremTheorem: among any 6 people

, there are 3 who all know each other (a club) OR 3 who don’t know each other (strangers)14Slide15

Breaking a proof into cases

Theorem: Every collection of 6 people includes a club of 3 people or a group of 3 strangersProof: 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 peopleCase 2: x knows at most 2 of the other 5 peopleNotice it says “there are two cases”You’d better be right there are no more cases!Cases must completely cover possibilitiesTip: you don’t need to worry about trying to make the cases “equal size” or scopeSometimes 99% of the possibilities are in one case, and 1% are in the otherWhatever makes it easier to do each proof15Slide16

Breaking a proof into cases

Theorem: Every collection of 6 people includes a club of 3 people or a group of 3 strangersCase 1:

suppose x knows at least 3 other peopleCases 1.1: No pair among these 3 people met each other. Then these are a group of 3 strangers. So the theorem holds in this subcase.Case 1.2: Some pair among these people know each other. Then this pair, together with x, form a club of 3 people. So the theorem holds in this subcase.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 case16Slide17

Breaking a proof into cases

Theorem: Every collection of 6 people includes a club of 3 people or a group of 3 strangersCase 2:

Suppose x knows at most 2 other people. So he doesn’t know at least 3 people.Cases 2.1: All pairs among these 3 people met each other. Then these are a club of 3. So the theorem holds in this subcase.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.17Slide18

Breaking a proof into cases

Theorem: …Proof: There are two cases to considerCase 1: there are two cases to considerCase 1.1: Verify theorem directlyCase 1.2: Verify theorem directlyCase 2: there are two cases to consider

Case 2.1: Verify theorem directlyCase 2.2: Verify theorem directly18Slide19

Perspective:Theorem in language of graphs

Graph: diagram which captures relations between pairs of objectsExample: objects=people, relation=know each other

19AB

CD

A,B know each other

A,C know each other

A,D know each other

B,C know each other

B,D don’t know each other

C,D don’t know each otherSlide20

Perspective:Theorem in language of graphs

Graph terminologyPeople = verticesKnow each other = edgeDon’t know each other = no edge

20AB

CD

A,B know each other

A,C know each other

A,D know each other

B,C know each other

B,D don’t know each other

C,D don’t know each otherSlide21

Perspective:Theorem in language of graphs

21

Theorem: Every collection of 6 people includes a club of 3 people or a group of 3 strangersEquivalently…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)

Instance of “Ramsey theory”