Inductive Proofs

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Inductive Proofs

• Rosen 6th ed., 4.1-4.3

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Mathematical Induction

• A powerful, rigorous technique for proving that a
  predicate P(n) is true for every natural number
  n, no matter how large.
• Based on a predicate-logic inference rule
• P(0)?n?0 (P(n)?P(n1))??n?0 P(n)

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Outline of an Inductive Proof

• Want to prove ?n P(n)
• Base case (or basis step) Prove P(0).
• Inductive step Prove ?n P(n)?P(n1).
• e.g. use a direct proof
• Let n?N, assume P(n). (inductive hypothesis)
• Under this assumption, prove P(n1).
• Inductive inference rule then gives ?n P(n).

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Induction Example

• Prove that

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Another Induction Example

• Prove that ? , nlt2n. Let P(n)(nlt2n)
• Base case P(0)(0lt20)(0lt1)T.
• Inductive step For prove P(n)?P(n1).
• Assuming nlt2n, prove n1 lt 2n1.
• Note n 1 lt 2n 1 (by inductive hypothesis)
  lt 2n 2n (because 1lt22?20?2?2n-1 2n)
  2n1
• So n 1 lt 2n1, and were done.

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Validity of Induction

• Prove if ?n?0 (P(n)?P(n1)), then ?k?0
  P(k)
• (a) Given any k?0, ?n?0 (P(n)?P(n1)) implies
• (P(0)?P(1)) ? (P(1)?P(2)) ? ?
  (P(k?1)?P(k))
• (b) Using hypothetical syllogism k-1 times we
  have
• P(0)?P(k)
• (c) P(0) and modus ponens gives P(k).
• Thus ?k?0 P(k).

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The Well-Ordering Property

• The validity of the inductive inference rule can
  also be proved using the well-ordering property,
  which says
• Every non-empty set of non-negative integers has
  a minimum (smallest) element.
• ? S?N ?m?S ?n?S m?n

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Use Well-Ordering Property and Proof by
Contradiction

• Suppose P(0) is true and that P(k)?P(k1) is true
  for all positive k.
• Assume P(n) is false for some positive n.
• Implies Sn?P(n) is non-empty and has a
  minimum element m (i.e., P(m)false)
• But then, P(m-1)?P((m-1)1)P(m) which is a
  contradiction!

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Generalizing Induction

• Can also be used to prove ?n?c P(n) for a given
  constant c?Z, where maybe c?0, then
• Base case prove P(c) rather than P(0)
• The inductive step is to prove
• ?n?c (P(n)?P(n1)).

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Induction Example

• Prove that the sum of the first n odd positive
  integers is n2. That is, prove
• Proof by induction.
• Base case Let n1. The sum of the first 1 odd
  positive integer is 1 which equals 12.(Cont)

P(n)
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Example cont.

• Inductive step Prove ?n?1 P(n)?P(n1).
• Let n?1, assume P(n), and prove P(n1).

By inductivehypothesis P(n)
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Strong Induction

• Characterized by another inference
  ruleP(0)?n?0 (?0?k?n P(k)) ? P(n1)??n?0
  P(n)
• Difference with previous version is that the
  inductive step uses the fact that P(k) is true
  for all smaller , not just for kn.

P is true in all previous cases
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Example 1

• Show that every ngt1 can be written as a product
  p1p2ps of some series of s prime numbers. Let
  P(n)n has that property
• Base case n2, let s1, p12.
• Inductive step Let n?2. Assume ?2?k?n P(k).
  Consider n1. If prime, let s1, p1n1.Else
  n1ab, where 1lta?n and 1ltb?n.Then ap1p2pt and
  bq1q2qu. Then n1 p1p2pt q1q2qu, a product
  of stu primes.

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Example 2

• Prove that every amount of postage of 12 cents or
  more can be formed using just 4-cent and 5-cent
  stamps.
• Base case 123(4), 132(4)1(5), 141(4)2(5),
  153(5), so ?12?n?15, P(n).
• Inductive step Let n?15, assume ?12?k?n P(k).
  Note 12?n?3?n, so P(n?3), so add a 4-cent stamp
  to get postage for n1.

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Example 3

• Suppose a0, a1, a2, is defined as follows
• a01, a12, a23,
• ak ak-1ak-2ak-3 for all integers k3.
• Then an 2n for all integers n0. P(n)
• Proof (by strong induction)
• 1) Basis step
• The statement is true for n0 a01 120 P(0)
• for n1 a12 221 P(1)
• for n2 a23 422 P(2)

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Example 3 (contd)

• 2) Inductive hypothesis For any kgt2,
• Assume P(i) is true for all i with 0iltk
• ai 2i for all 0iltk .
• 3) Inductive step
• Show that P(k) is true ak 2k
  ak ak-1ak-2ak-3
• 2k-12k-22k-3 (using inductive
  hypothesis)
• 20212k-32k-22k-1
• 2k-1 (as a sum of geometric sequence)
• 2k
• Thus, P(n) is true by strong induction.