Mathematical Induction

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Mathematical Induction
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Proving a divisibility property by mathematical
induction

• Proposition For any integer n1,
• 7n - 2n is divisible by 5. (P(n))
• Proof (by induction)
• 1) Basis step
• The statement is true for n1 (P(1))
• 71 21 7 - 2 5 is divisible by 5.
• 2) Inductive step
• Assume the statement is true for some k1
  (P(k))
• (inductive hypothesis)
• show that it is true for k1 . (P(k1))

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Proving a divisibility property by mathematical
induction

• Proof (cont.) We are given that
• P(k) 7k - 2k is divisible by 5.
  (1)
• Then 7k - 2k 5a for some a?Z . (by
  definition) (2)
• We need to show
• P(k1) 7k1 - 2k1 is divisible by 5.
  (3)
• 7k1 - 2k1 77k - 22k 57k 27k - 22k
• 57k 2(7k - 2k) 57k 25a (by (2))
• 5(7k 2a) which is divisible by 5. (by
  def.)
• Thus, P(n) is true by induction.

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Proving inequalities by mathematical induction

• Theorem For all integers n5,
• n2 lt 2n. (P(n))
• Proof (by induction)
• 1) Basis step
• The statement is true for n5 (P(5))
• 52 25 lt 32 25.
• 2) Inductive step
• Assume the statement is true for some k5
  (P(k))
• show that it is true for k1 .
  (P(k1))

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Proving inequalities by mathematical induction

• Proof (cont.) We are given that
• P(k) k2 lt 2k. (1)
• We need to show
• P(k1) (k1)2 lt 2k1.
  (2)
• (k1)2 k22k1 lt k2 2k (since k5)
• lt 2k 2k (based on (1))
• 22k 2k1.
• Thus, P(n) is true by induction.

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Proving inequalities by mathematical induction

• Theorem For all integers n4,
• 2n lt n! . (P(n))
• Proof (by induction)
• 1) Basis step
• The statement is true for n4 (P(4))
• 24 16 lt 24 4! .
• 2) Inductive step
• Assume the statement is true for some k4
  (P(k))
• show that it is true for k1 .
  (P(k1))

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Proving inequalities by mathematical induction

• Proof (cont.) We are given that
• P(k) 2k lt k! (1)
• We need to show
• P(k1) 2k1 lt (k1)! (2)
• 2k1 22k
• lt 2k! (based on (1))
• lt (k1)k! (since k4)
• (k1)!
• Thus, P(n) is true by induction.