Lecture 6 Induction

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• Lecture 6 Induction

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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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Mathematical induction (cont.)

• Example Prove that the equation P(n)
  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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Details on our example

• In short, P(k1) has ONE MORE TERM on its LEFT
  SIDE than P(k) AND this term builds on P(k)
  .
• So, from P(n) 135.(2n-1) n2 with
  nk1
• P(k1) P(k) (2(k1)-1)
• k2 2k21 subst fo
  P(k) expand on rt
• k2 2k 1
  aft some arith.
• (k 1) 2
  QED

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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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One more Example

• Inductive hypothesis (Assumption).
• P(k1) P(k) (k1)
• SHORT WAY add LAST
  TERM
• from P(k1)s LT
  side expansion to P(k)
  
• (k(k1))/2 (k1)
  aft subst for P(k)
• (k1)k/2 (k1) aft
  switchg 1st term order
• (k1) (k/2 1) aft
  factorg out (k1)
• (k1) (k2)/2
  aft simple arith
• Youre done IF you recognize the RIGHT SIDE here!

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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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Second Principle BigTime Exs

• Bk ex 21 --- Fence problem
• Postage Stamp problems
• 1. Youre lookg at postage more than 8 cents
  (Why not 80 cents? --- Joke,
• daughter!) Show 5 and 8 cent stamps
  do all possible cases!
• Just prove it simple-mindedly for 8, 9,
  10 cents and then you can add 3
• cents to 8 to get 11, to 9 to get 12
  Obvious?
• 2. Prob 66 (the bk gives the answer) and 67
  4 and 5 cent stamps starting
• at 12 cents LIKE, demo 12, 13, 14, 15
  and play 4 games
• Others We tend to ignore TIME PRESSURE!
  Maybe too swift ?
• 1. Parens in programming lang p 103, Ex 22
• 2. For all n GE 2, n is either prime or
  product of primes

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Exercises 2.2 Prob 4, 13

• 4. Prove (simple inductn) 1 3 6
  (n(n1))/2 (n(n1)(n2))/6
• 1. Base case n1 1 (1 (11) (12))/6
  --- A-OK! (Dont think
• of n 0 aint THERE on the LEFT!)
• 2. Riot down DIRECTLY BELOW the problem
  statement with
• nk the case where nk1
• KEY POINT find the nk case embedded
  in the k1 case and
• do some arith (sometimes fancy
  footwork!!)
• Awesome thing to look at you have the answer in
  your book!
• 1. Again, the base case is n1 and is
  easy
• 2. You gotta solve k/(k1) 1/(k1)(k2)
  ? (k1)/(k2) via
• one of those common denominator
  thing (mult by (k2) on the first
• term on the left