Mathematical Induction

1
Mathematical Induction

• Section 4.1

2
Example

• Consider the predicate
• P(n) 1 3 (2n-1)
  is equal n2
• Lets verify the truth value of P(n) for some n
• P(1) 1 1 12
• P(2) 1 3 4 22
• P(3) 1 3 5 9 32
• ....
• P(9) 1 3 5 7 9 11 13 15 17 92
  

3
Example (no calculators!)

• Suppose, without verifying, that P(21)
  1 3 5 39 41 212
• Given above, can we prove that P(22) 1 3
  5 39 41 43 222 ?
• P(22) 1 3 5 39 41 43 222 ?
• P(22) P(21) 43
  222 ?
• P(22) 212 43 222 ?
• P(22) (22 1)2 (222 1) 222 ?
• P(22) 222 222 1 222 1 222 ?

4
Example (no calculators!)

• How general is our methods?
• Lets replace 21 with a place holder k.
• Suppose, without verifying, that P(k)
  1 3 5 (2k-1) k2
• Given this, can we prove that P(k1) 1 3
  5 (2(k1)-1) (k1)2 ?
• P(k1) 1 3 5 (2k-1) (2(k1)-1)
  (k1)2 ?
• P(k1) P(k)
  (2(k1)-1) (k1)2 ?
• P(k1) k2 (2(k1)-1) (k1)2 ?
• P(k1) ((k1) 1)2 (2(k1)-1) (k1)2 ?
• P(k1) (k1)2 2(k1) 1 2(k1) 1
  (k1)2 ?

5
Example

• What have we accomplished so far?
• We have shown that P(k) ? P(k1), for any k. For
  instance
• P(1) ? P(2),
• P(2) ? P(3),
• P(6) ? P(7),
• P(19) ? P(20),
• P(2000) ? P(2001), .
• Note that, we do NOT know if, for example, P(6)
  or P(19) is true. All we know is that IF, for
  example, P(6) is true, then P(7) is also true.
• What do we need to show that P(n) is true for all
  n?
• All we need is to PROVE that P(1) is indeed true.
  

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Mathematical Induction

• Let P(n) be some propositional function involving
  integer n.
• P(n) n (n3) is an even number
• P(n) 1 3 (2n-1) n2
• ..
• To prove that P(n) is true for all positive
  integers n, we can do the following
• Give a proof (usually a straight verification)
  that P(1) is true.
• Give a proof that for an arbitrary k, IF P(k) is
  true THEN P(k 1) is true. That is, we validate
  the logical implication P(k) ? P(k1)

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Mathematical Induction

• Mathematical induction amounts to the following
  rule of inference P(1) ? ?k ((k 1) ?
  P(k)) ? P(k 1)) ? ?n P(n)where our universe
  is the set of positive integers
• A proof using mathematical induction involves
• The Basis prove P(1) is true.
• The Hypothesis Assume P(k) is true for an
  arbitrary k 1.
• The Induction Step Establish that P(k 1)

A (direct) proof that ((k 1) ? P(k)) ? P(k 1)
8
Mathematical Induction.

• In proof by mathematical induction it is not a
  given that P(k) is true for all positive
  integers! It is only shown that IF P(k) is true,
  THEN P(k1) is also true. Otherwise the proof by
  mathematical induction was a case of begging the
  question or circular reasoning.

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Example

• Let P(n) 1 3 (2n-1) n2 . We want
  to prove by induction that P(n) is true for all
  n.
• Basis Prove that when n 1, the proposition is
  correct. P(1) is 1 121 thus, it is correct.
• Hypothesis We will assume that P(k) is true for
  some arbitrary k.
• Step Using the hypothesis, we want to prove that
  P(k 1) is true.

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Example
11
Example
12
Induction can start anywhere

• To show P(n) for any n b
• The Basis prove P(b) is true.
• The Hypothesis Assume P(k) is true and k b.
• The Induction Step Establish that P(k 1)

A (direct) proof that ((k b) ? P(k)) ? P(k 1)
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Example

• Suppose we have coins of two different
  denominations, namely, 3 cents and 5 cents. We
  want to show that it is possible to pay exactly
  for any purchase of 8 cents or higher. In other
  words, that we can make up 8, 9, 10, . cents
  using just these coins.
• Here is a proof by induction
• Basis 8 35 ?
• Hypothesis Suppose we can pay exactly with just
  3-cent coins and 5-cent coins for a purchase
  amount of k cents.

14
Example

• Want to show that it is we can pay exact for (k
  1) cents using just 3-cent coins and 5-cent
  coins.
• Proof If in making up k cents, we used at least
  one 5- cent coin, then replace that coin with two
  3-cent coins to make up (k 1) cents. Otherwise,
  we must have used at least three 3-cent coins to
  make up k since we are considering k gt 8. In the
  case, we will replace three 3-cent coins with two
  5-cent coins to make up (k 1) cents.

15
Strong induction

• To prove that P(k1) holds, we may use any of the
  following assumptions
• P(k) is true
• P(k 1) is true
• P(k 2) is true
• .
• P(1) is true
• We must make sure that all the assumptions are
  in fact true for the induction engine to start.

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Induction, recap

• Let P(n) be some propositional function about an
  integer n. To prove that P(n) is true for all
  positive integers n, we do the following
• Give a proof that P(1) is true
• Give a proof that if all of P(1), P(2), , P(k)
  are true and k 1 then P(k1) is also true.
• As before, the induction can start at any integer
  b
• Give a proof that P(b) is true
• Give a proof that if all of P(b), P(b1), , P(k)
  are true and k b then P(k1) is also true.

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Example

• Show that any positive integer n gt 1 is either a
  prime or a product of primes.
• Proof
• Basis n 2. Since 2 is prime.?
• Hypothesis Assume true for n k
• Step Need to show that, given the hypothesis,
  the proposition is true for k 1.
• Proof If n k 1 is a prime, then the
  proposition is true. If k 1 is not prime, then
  k 1pq, where p k and q k. By the
  hypothesis, p and q are either primes or product
  of primes, and thus so is pq.

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Example

• Show that for every nonnegative integer n
• 20 21 22 23 2n 2n1 1.
• Let P(n) be the corresponding predicate
• Basis P(0) is true since 20 1 21-1.
• Hypothesis Assume that P(k) is true and that k
  0.
• Step Need to show that P(k 1) is true given
  that the hypothesis is true.

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Example

• P(k1) 20212223 2k 2k1
• 2k1 1 2k1 by the induction
  hypothesis
• 2k2 1 by arithmetic
• 2(k1)1 1 by arithmetic
• Thus P(k) ? P(k 1)

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Example
21
Example
Hypothesis is used here
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Proof of Correctness of Mathematical Induction

• We want to show that P(1) ? ?k ((k 1) ? P(k))
  ? P(k 1)) ? ?n P(n) where the universe is
  the positive integers
• Proof is done by contradiction
• ?n P(n) Contrary assumption
• ?x P(x) 1, Demorgans
• P(r) 2, EI, select
  smallest r
• P(r -1) 3, choice of r
• P(1) Premise
• r gt 1 r is positive (by choice in 4) 5
• ?k ((k 1) ? P(k)) ? P(k 1)) Premise
• ((r-1 1) ? P(r-1)) ? P(r) UI, k
  r-1
• P(r) 4, 6 8, modus ponens
• Contradiction 3 9

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Example
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Example
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Proof
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Example

• Prove by induction that if S is a finite set with
  n elements then S has 2n subsets.
• Basis Assume n 0. There is one set with
  cardinality zero the empty set. Thus, S ?. The
  empty set has only one subset itself. Moreover,
  20 1. Thus the basis is verified.
• Hypothesis. Assume true for any set S with
  cardinality 1,2,, k
• Step. Want to show that if S has cardinality k1,
  the proposition still holds. Let S be such a set.
  Let r be an arbitrary element of S. Write S X ?
  r . Now, the subsets of S either contain r or
  they dont. For the ones that do not contain r ,
  those are also subsets of X, and there are 2k of
  them. To form the ones that do contain r, we take
  each subset of X and inset r in it. That would
  give us another 2k subsets. Thus, S has a total
  of 2k1 subsets.

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Induction and Mathematical Inductions

• Let P(n) denote the number of different ways to
  write n as a sum of positive integers when order
  is not important. For instance, integer 5 can be
  written in the following 7 ways111112111
  22131132415
• Verify that
• P(2) 2
• P(3) 3
• P(4) 5
• P(5) 7
• P(6) ?
• P(7) ..

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Whats wrong with the following proof?