Lecture 6 Induction

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• Lecture 6 Induction
• June 2nd, 2003

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Review of last class
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Induction

• Given that you are climbing an infinitely high
  ladder.
• Question How to you know whether you will be
  able to reach an arbitrarily high rung?

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Two assertions

• You can reach the first rung.
• Once you get to a rung, you can always climb to
  the next one up.
• Assertion 1 sets the starting point.
• Assertion2 guarantees that you can go beyond the
  first rung .
• This technique can be used to prove properties
  for all positive integers.

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• Two assertions to prove?

1 has property P

• P(1)
• For any positive integer k, P(k)?P(k1)

if we can prove this, P(n) holds for any positive
integer n
If any number has property p, so does the next
number.
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First Principle of Mathematical Induction
Inductive assumption or inductive hypothesis
Basis step

• P(1) is true
• (?k)P(k) true ? P(k1) true

P(n) true for all positive integers n.
to show that this is true
Inductive step
We show that 1 2 are true
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How do we prove 1 2?

• Prove 1
• We need only show that property P holds for the
• number 1.
• Prove 2 (Implication that must hold for all k)
• We need to assume that P(k) is true for an
• arbitrary integer k.
• Show that P(k1) is true based on the previous
• assumption.

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Mathematical induction

• Some issues to be cleared
• It is not an exploratory proof technique, it can
  only confirm a correct conjecture.
• It is actually a deductive technique, not a
  method of inductive.
• Even the conjecture is slightly incorrect, you
  might see the correct conclusion during the proof.

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Mathematical induction (cont.)

• Example Prove that the equation 135.(2n-1)
  n2
• Base Step
• P(1) 1 12
• Inductive hypothesis (Assumption).
• P(k) 135(2k-1) k2
• Prove the next one up.
• P(k1) 135(2(k1)-1) ? (k1)2

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Three steps for First Principle of induction

• Step 1---Prove the base case
• Step 2 Assume P(k)
• Step 3 Prove P(k1)

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One more Example

• 123n n(n1)/2.
• Base Step
• P(1) 1 12/2
• Inductive hypothesis (Assumption).
• P(k) 123kk(k1)/2
• Prove the next one up.
• P(k1) 123(k1) ? (k1)((k1)1)/2

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Second Principle of Induction

• P(1) is true
• (?k) P(r) true for all r,
• 1 ? r ? k ?P(k1) true

P(n) true for all positive integers n
In statement 2 we can assume that P(r) is true
for all integers r between 1 and an arbitrary
positive integer k in order to prove that P(k1)
is true. Both principles of induction are
equivalent.
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Assignment 2

• Exercise 2.2 -- 4, 13