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Lecture 6 Induction

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It is actually a deductive technique, not a method of inductive. ... an arbitrary positive integer k in order to prove that P(k 1) is true. ... – PowerPoint PPT presentation

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Title: Lecture 6 Induction


1
  • Lecture 6 Induction

2
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?

3
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.

4
  • 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.
5
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
6
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.

7
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.

8
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

9
Three steps for First Principle of induction
  • Step 1---Prove the base case
  • Step 2 Assume P(k)
  • Step 3 Prove P(k1)

10
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

11
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

12
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

13
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!

14
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.
15
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

16
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
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