### slides

```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 x0. Then either
x+y0 or x-y0.”
• Proof:
• Given (contrapositive form): Let
• WTS (contrapositive form): …
???
• Conclusion: …
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
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
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
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 -
(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.
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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
Next class
• Indirect proof techniques
• Read section 3.5 in Jenkyns, Stephenson
```