Induction Practice

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Induction Practice

• CS1050

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Prove that
whenever n is a positive integer.

• Proof
• Basic Case Let n 1, then

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Prove that
whenever n is a positive integer.

• Inductive Case
• Assume that the expression is true for n, i.e.,
  that
• Then we must show that

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(No Transcript)
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If S is a finite set with n elements, then S has
2n subsets (1of 2)

• We will do a proof by induction.
• Basis Step Assume that the set is empty, I.e.,
  that it has 0 elements, then 20 1 subset which
  is ?
• Inductive Step Assume that a set with n elements
  has 2n subsets. Then we must show that this
  implies that a set with n1 elements has 2n1
  subsets.

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(2 of 2)

• Let S be the set with n elements and T be the set
  with n1 elements. Let a be the element that is
  added to S to get T. So T S?a or S T-a.
  For every subset, x, of S there must be exactly
  two subsets of T x and x?a.
• 22n 2n1

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Let p(n) be the proposition that there exists t,e
? N such that 3t 8e n. Prove ?n ? 14?N, p(n)
is true.

• We will do a proof by induction.
• Basis Step Prove p(14)
• Let t 2 and e 1, then 3(2) 8(1) 14.
• Inductive Step Assume p(k) is true. Thus there
  exist t,e ?N such that 3t8e k. We need to
  show that there exist u,f ?N such that 3u 8f
  k1. There are two cases e 0 (there are no
  8s) or e ?1 (there is at least one 8).

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Case 1 e ? 1

• Let u t3 and f e-1. Since t,e ? N and e ?
  1, u,f ? N. Now 3u 8f 3(t3) 8(e-1) 3t
  8e 1 k1.
• (Informally what we have done is remove one 8 and
  replace it with 3 threes to get a net increase of
  1)

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Case 2 e 0

• Since k ? 14 and e 0, then t ? 5. Let u t-5
  and f 2. Since t ? 5, u ? N and clearly f ? N.
• Now 3u 8f 3(t-5) 8(2) 3t1 k 1.
• (Informally, take out three fives and replace
  with 2 eights.)