Bureaucrats write memoranda both because they appear to be busy when they are writing and because th

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• Bureaucrats write memoranda both because they
  appear to be busy when they are writing and
  because the memos, once written, immediately
  become proof that they were busy.
• Charles Peters

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CSE 502NFundamentals of Computer Science

• Fall 2004
• Lecture 6
• Mathematical Induction (Rosen 3.3)
• Inductive Proof Techniques

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Mathematical Induction

• Mathematical induction is a technique for proving
  propositions of the form "nP(n) where the
  universe of discourse is the set of integers
  b
• Essentially, we prove that the propositional
  function is true for the first element and the
  implication that if the propositional function is
  true for an element k then it must be true for
  k1
• Therefore, the universal quantifier must be true
• Basic Step show the proposition P(b) to be true
• Inductive Step show the implication P(k)
  P(k1) to be true for every integer k b, b1,
  b2,
• Inductive hypothesis assume P(k) is true for a
  fixed integer k
• Rule of inference P(1) Ù P(k) P(k1))
  "nP(n)

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Strong Induction

• Strong induction strengthens the inductive
  hypothesis by assuming that P(j) is true not just
  for a fixed integer k, but for j 1, 2, , k
• Basic Step show the proposition P(b) to be true
• Inductive Step show the implicationP(1) Ù P(2)
  Ù Ù P(k) P(k1) to be true for every integer
  k b, b1, b2,

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Well-Ordering Property

• Every nonempty set of nonnegative integers has a
  least element
• The well-ordering property is useful in proving
  many theorems, as well as the validity of
  mathematical induction
• Proof by contradiction
• Suppose we know P(1) is true and P(k) P(k1) is
  true for all positive integers k
• Assume that there is one positive integer for
  which P(n) is false
• Therefore, the set S of positive integers for
  which P(n) is false is nonempty
• By the well-ordering property S contains a least
  element m
• m cannot be 1, because P(1) is true
• Since m gt 1, then m 1 gt 0
• Since m 1 lt m, m 1 is not in S and P(m 1)
  must be true
• Since P(m 1) P(m) is true, then P(m) must be
  true
• This is a contradiction which means P(n) must be
  true for all positive integers