1. COT 3100, Spring 2001 Applications of Discrete Structures
Section #1089X - MWF 4th period
Dr. Michael P. Frank Lecture #26 Fri., Mar. 16, 2001
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Lecture #26

2. Administrivia Exam regrade issues: Today:
In question #1, “one-to-one correspondence” in place of “bijection” was often mistakenly graded wrong.
All regrade requests for Exam #1 are due in writing (on a separate page) by class-time Monday.
Today:
Quiz #7 on §2.4 & 2.6.
Finish §3.1: Proof Methods.
Start §3.2: Induction.
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Lecture #26

3. Review: Proof Methods So Far
Direct, indirect, vacuous, and trivial proofs of statements of the form pq.
Proof by contradiction of any statements.
Constructive and nonconstructive existence proofs.
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4. Proving Existentials
A proof of a statement of the form x P(x) is called an existence proof.
If the proof demonstrates how to actually find or construct a specific element a such that P(a) is true, then it is a constructive proof.
Otherwise, it is nonconstructive.
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5. A Constructive Existence Proof
(Example 23, p.179)
Show that for any n>0 there exists a sequence of n consecutive composite integers.
Same statement in predicate logic: n>0 x i (1in)(x+i is composite)
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6. The proof... Given n>0, let x = (n + 1)! + 1.
Let i  1 and i  n, and consider x+i.
Note x+i = (n + 1)! + (i + 1).
Note (i+1)|(n+1)!, since 2  i+1  n+1.
Also (i+1)|(i+1). So, (i+1)|(x+i).
 x+i is composite.
 n x 1in : x+i is composite. Q.E.D.
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7. Nonconstructive Existence Proof
(Example 24, p. 180)
Show that there are infinitely many primes.
Show there is no largest prime.
Show that for any prime number, there is a larger number that is also prime.
Show that for any number,  a larger prime.
Show that n p>n : p is prime.
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8. Da proof... Given n>0, prove there is a prime p>n.
Consider x=n!+1. Since x>1, we have (x is prime)(x is composite).
Case 1: x is prime. Obviously x>n, so let p=x and we’re done.
Case 2: x has a prime factor p. But if pn, then p mod x = 1. So p>n, and we’re done.
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9. The Halting Problem (Turing‘36)
Involves a non-existence proof.
The first mathematical function proven to have no algorithm that computes it!
The desired function is Halts(P,I) = the truth value of the statement ‘Program P, given input I, eventually halts’.
Implies general impossibility of predictive analysis of arbitrary computer programs.
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10. The Proof Given any arbitrary program H(P,I),
Consider algorithm Breaker, defined as: procedure Breaker(P: a program) halts := H(P,P) if halts then while T begin end
Note that Breaker(Breaker) halts iff H(Breaker,Breaker) = F.
So H does not compute the function Halts!
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11. Limits on Proofs
Some very simple statements of number theory haven’t been proved or disproved!
E.g. Goldbach’s conjecture: Every integer n>2 is exactly the average of some two primes.
n>2  primes p,q : n=(p+q)/2.
There are true statements of number theory (or any sufficiently powerful system) that can never be proved (or disproved) (Gödel).
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12. §3.2: Mathematical Induction
A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large.
Essentially a “domino effect” principle.
Based on the predicate-logic inference rule: (P(0)n (P(n)P(n+1))  n P(n)
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13. The Well-Ordering Property
The validity of the inductive inference rule can be proved using the well-ordering property, which says:
Every non-empty set of non-negative integers has a minimum element.
 SN : mS : nS : mn
Implies {n|P(n)} has a min. element m, but then P(m-1)P((m-1)+1) contradicted.
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14. Outline of an Inductive Proof
Want to prove n P(n).
Base case: Prove P(0).
Inductive step: Prove n P(n)P(n+1).
E.g. use a direct proof:
Let nN, assume P(n). (inductive hypothesis)
Under this assumption, prove P(n+1).
Inductive inference rule then gives n P(n).
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