University of Florida Dept' of Computer

1
University of FloridaDept. of Computer
Information Science EngineeringCOT
3100Applications of Discrete StructuresDr.
Michael P. Frank

• Slides for a Course Based on the TextDiscrete
  Mathematics Its Applications (5th Edition)by
  Kenneth H. Rosen

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Module 11Proof Strategies

• Rosen 5th ed., 3.1
• 21 slides

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Module 11 Overview

• In module 2, we already saw
• Several types of proofs of implications p?q
• Vacuous, Trivial, Direct, Indirect
• Types of existence proofs
• Constructive vs. Nonconstructive.
• Some methods of proving general statements p
• Proof by cases, proof by contradiction.
• In this module, we will see examples of
• Forward vs. backward reasoning.
• Proof by cases.
• Adapting existing proofs.
• Turning conjectures into proofs.

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Forward Reasoning

• Have premises p, and want to prove q.
• Find a s1 such that p?s1
• Then, modus ponens gives you s1.
• Then, find an s2 ? (such that) s1?s2.
• Then, modus ponens gives you s2.
• And hope to eventually get to an sn ? sn?q.
• The problem with this method is
• It can be tough to see the path looking from p.

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Backward Reasoning

• It can often be easier to see the very same path
  if you just start looking from the conclusion q
  instead
• That is, first find an s-1 such that s-1?q.
• Then, find an s-2 ? s-2?s-1, and so on
• Working back to an s-n ? p?s-n.
• Note we still are using modus ponens to propagate
  truth forwards down the chain from p to s-n to
  to s1 to q!
• We are finding the chain backwards, but applying
  it forwards.
• This is not quite the same thing as an indirect
  proof
• In that, we would use modus tollens and q to
  prove s-1, etc.
• However, it it similar.

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Backward Reasoning Example
Example 1

• Theorem ?agt0,bgt0,a?b (ab)/2 gt (ab)1/2.
• Proof
• Notice it is not obvious how to go from the
  premises agt0, bgt0, a?b directly forward to the
  conclusion (ab)/2 gt (ab)1/2.
• So, lets work backwards from the conclusion,
  (ab)/2 gt (ab)1/2 !

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Steps of Example

• (ab)/2 gt (ab)1/2 ? (squaring both sides)
• This preserves the gt since both sides are
  positive.
• (ab)2/4 gt ab ? (multiplying through by 4)
• (ab)2 gt 4ab ? (squaring ab)
• a22abb2 gt 4ab ? (subtracting out 4ab)
• a2-2abb2 gt 0 ? (factoring left side)
• (a-b)2 gt 0
• Now, since a?b, (a-b)?0, thus (a-b)2gt0, and we
  can work our way back along the chain of steps

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Forwardized version of Example

• Theorem ?agt0,bgt0,a?b (ab)/2 gt (ab)1/2.
• Proof. Since a?b, (a-b)?0. Thus, (a-b)2gt0,
  i.e., a2-2abb2 gt 0. Adding 4ab to both sides,
  a22abb2 gt 4ab. Factoring the left side, we
  have (ab)2 gt 4ab, so (ab)2/4 gt ab. Since ab is
  positive, we can take the square root of both
  sides and get (ab)/2 gt (ab)1/2.
• This is just a simple proof proceeding directly
  from premises to conclusion, but if you dont
  realize how it was obtained, it looks magical.
• A common student reaction But how did you know
  to pick 4ab out of thin air, to add to both
  sides?
• Answer By working backwards from the conclusion!

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Stone Game Example
Example 2

• Game rules
• There are 15 stones in a pile. Two players take
  turns removing either 1, 2, or 3 stones. Whoever
  takes the last stone wins.
• Theorem There is a strategy for the first player
  that guarantees him a win.
• How do we prove this? Constructive proof
• Looks complicated How do we pick out the winning
  strategy from among all possible strategies?
• Work backwards from the endgame!

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Working Backwards in the Game

• Player 1 wins if it is player 2s turn and there
  are no stones
• P1 can arrange this ifit is his turn, and
  thereare 1, 2, or 3 stones
• This will be true aslong as player 2 had4
  stones on his turn
• And so on

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Forwardized version

• Theorem. Whoever moves first can always force a
  win.
• Proof. Player 1 can remove 3 stones, leaving 12.
  After player 2 moves, there will then be either
  11, 10, or 9 stones left. In any of these cases,
  player 1 can then reduce the number of stones to
  8. Then, player 2 will reduce the number to 7,
  6, or 5. Then, player 1 can reduce the number to
  4. Then, player 2 must reduce them to 3, 2, or
  1. Player 1 then removes the remaining stones
  and wins.

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Proof by Cases Example
Example 3

• Theorem ?n?Z (2n ? 3n) ? 24(n2-1)
• Proof Since 236, the value of n mod 6 is
  sufficient to tell us whether 2n or 3n. If (n
  mod 6)?0,3 then 3n if it is in 0,2,4 then
  2n. Thus (n mod 6)?1,5.
• Case 1 If n mod 6 1, then (?k) n6k1.
  n236k212k1, so n2-136k212k 12(3k1)k.
  Note 2(3k1)k since either k or 3k1 is even.
  Thus 24(n2-1).
• Case 2 If n mod 6 5, then n6k5. n2-1
  (n-1)(n1) (6k4)(6k6) 12(3k2)(k1).
  Either k1 or 3k2 is even. Thus, 24(n2-1).

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Proof by Examples?

• A universal statement can never be proven by
  using examples, unless the universe can be
  validly reduced to only finitely many examples,
  and your proof covers all of them!
• Theorem ?x,y?Z x23y2 8.
• Proof If x3 or y2 then x23y2 gt8. This
  leaves x2?0,1,4 and 3y2?0,3. The largest
  pair sum to 43 7 lt 8.

Example 4
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A Constructive Existence Proof
Example 7

• Theorem For any integer ngt0, there exists a
  sequence of n consecutive composite integers.
• Same statement in predicate logic?ngt0 ?x ?i
  (1?i?n)?(xi is composite)
• Proof follows on next slide

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The proof...

• Given ngt0, let x (n 1)! 1.
• Let i ? 1 and i ? n, and consider xi.
• Note xi (n 1)! (i 1).
• Note (i1)(n1)!, since 2 ? i1 ? n1.
• Also (i1)(i1). So, (i1)(xi).
• ? xi is composite.
• ? ?n ?x ?1?i?n xi is composite. Q.E.D.

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Nonconstructive Existence Proof

• Theorem There are infinitely many prime numbers.
• Any finite set of numbers must contain a maximal
  element, so we can prove the theorem if we can
  just show that there is no largest prime number.
• I.e., show that for any prime number, there is a
  larger number that is also prime.
• More generally For any number, ? a larger prime.
• Formally Show ?n ?pgtn p is prime.

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The proof, using proof by cases...

• Given ngt0, prove there is a prime pgtn.
• Consider x n!1. Since xgt1, we know that (x
  is prime)?(x is composite).
• Case 1 Suppose x is prime. Obviously xgtn, so
  let px and were done.
• Case 2 x has a prime factor p. But for q?n, q
  mod x 1. So pgtn, and were done.

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Adapting Existing Proofs
Example 5

• Theorem There are infinitely many primes of the
  form 4k3, where k?N. (Franks wrong proof)
• Recall we proved there are infinitely many primes
  because if p1,,pn were all the primes, then
  (?pi)1 must be prime or have a prime factor
  greater than pn, ? contradiction!
• Proof Similarly, suppose q1,,qn lists all
  primes of the form 4k3,
• and analogously consider Q 4(?qi)3.
• Unfortunately, since q1 3 is possible, 3Q and
  so Q does have a prime factor among the qi, so
  this doesnt work!
• So instead, consider Q 4(?qi)-1 4(?qi-1)3.
  This has the right form, and has no qi as a
  factor since ?i Q -1 (mod qi).

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and getting them right
Example 5

• Theorem There are infinitely many primes of the
  form 4k3, where k?N.
• The previous proof has a huge hole
• Just because no qi is a factor, doesnt mean
  there isnt a prime factor, or that it has the
  right form.
• In fact, we have to observe that all odd numbers
  (hence all primes except 2) have the form 4n1 or
  4n3
• But the product of numbers of the form 4n1
  itself has the form 4n1
• That is, (4k1)(4l1) 4(4kllk) 1
• And a product including 2 is of form 4n or 4n2
• So if Q is composite, it must have a prime factor
  of form 4k3
• And it cant be any of the qi
• C!
• (this means This is a contradiction)

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Conjecture and Proof
Example 6

• We know that some numbers of the form 2p-1 are
  prime when p is prime.
• These are called the Mersenne primes.
• Can we prove the inverse, that an-1 is composite
  whenever either agt2, or (a2 but n is composite)?
• All we need is to find a factor greater than 1.
• Note an-1 factors into (a-1)(an-1a1).
• When agt2, (a-1)gt1, and so we have a factor.
• When n is composite, ?r,sgt1 nrs. Thus, given
  a2, an 2n 2rs (2r)s, and since rgt1, 2r gt 2
  so 2n - 1 bs-1 with b 2r gt 2, which now fits
  the first case.

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Conjecture Counterexamples
Example 8

• Conjecture ? integers ngt0, n2-n41 is prime.
• Hm, lets see if we can find any
  counter-examples
• 12-141 41 (prime)
• 22-241 4-241 43 (prime)
• 32-341 9-341 47 (prime) Looking good so
  far!!
• Can we conclude after showing that it checks out
  in, say, 20 or 30 cases, that the conjecture must
  be true?
• NEVER NEVER NEVER NEVER NEVER!
• Of course, 412-4141 is divisible by 41!!

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Even Great Mathematicians Can Propose False
Conjectures!
Example 9

• Euler conjectured that for ngt2, the sum of n-1
  nth powers of positive integers is not an nth
  power.
• Remained true for all cases checked for 200
  years, but no proof was found.
• Finally, in 1966, someone noticed that 275 845
  1105 1335 1445.
• Larger counter-examples have also been found for
  n4, but none for ngt5 yet.

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Fermats Last Theorem

• Theorem xnynzn has no solutions in integers
  xyz ? 0 with integer ngt2.
• In the 1600s, Fermat famously claimed in a
  marginal note that he had a wondrous proof of
  the theorem.
• But unfortunately, if he had one, he never
  published it!
• The theorem remained a publicly unproven
  conjecture for the next 400 years!
• Finally, a proof that requires hundreds of pages
  of advanced mathematics was found by Wiles at
  Princeton in 1990.
• It took him 10 years of work to find it!
• Challenge Find a short, simple proof of Fermats
  last theorem, and you will become instantly
  famous!

Theorem 2
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Some Open Conjectures

• Conjecture There are infinitely many primes of
  the form n21, where n?Z.
• Conjecture (Twin Prime Conjecture) There are
  infinite pairs of primes of the form (p, p2).
• Conjecture (The Hailstone Problem) If h(x)
  x/2 when x is even, and 3x1 when x is odd, then
  ?x?N ?n?N hn(x) 1 (where the superscript
  denotes composition of h with itself n times).

Example 11
Example 12
Example 13
Prove any of these, and you can probably have a
lifetimecareer sitting around doing pure
mathematics