Lecture Notes, 10 Nov 2010

Note: A username and password may be required to access certain documents. Please contact Bill Rapaport.

Index to all lecture notes
…Previous lecture

§4.1: Mathematical Induction (cont'd)

1. Math Induction as one of Peano's Axioms:
   1. Previous axioms:
      • Let S(n) = n+1 be the successor function.
        • Note that this is a 1–1 function.
      1. 0 ∈ N
         • i.e.) 0 is a natural number
         • This is a "base case"

      2. ∀n[n ∈ N  →  S(n) ∈ N]
         • i.e.) the successor of any natural number is also a natural number
         • This is an "inductive" (or "recursive") case

      3. ∀n[0 ≠ n ∈ N  →  (∃m ∈ N)[n = S(m)]]
         • i.e.) every natural number except 0 is the successor of a natural number.
         • This is a further constraint on S
         • It says that the successor function from N→N is 1–1 but not onto
           • The successor function from N→W is total, 1–1, and onto;
             ∴ it is a 1–1 correspondence.
             • That's why N is countable!

      ---

   2. The final axiom: the axiom of mathematical induction:
      1. Version #1: Set-theoretic version
         
         • This can be thought of as a "closure" clause:
           The first 2 axioms give a sufficient condition for being a natural number;
           this gives a necessary condition.

         • It says that the set S constructed by putting 0 in it
           and then, ∀n ∈ S, putting n+1 in it
           is N

         • (∀ set S ⊆ N)[ [ (0 ∈ S) ∧ (∀k ∈ N)[(k ∈ S) → (k+1 ∈ S) ] ] → (S = N)]

         • Notes:
           1. "0" could be "1", or any other starting point
           2. The structure of this is:
              • It is a universally quantified proposition
              • …whose scope is a conditional.
                • The conditional's antecedent is a conjunction
                  • …whose first conjunct is a set-membership proposition
                  • …& whose second conjunct is another universally quantified proposition
                    • …whose scope is another conditional
                      • …whose antecedent is another set-membership proposition
                      • …& whose consequent is yet another set-membership proposition

                • The conditional's consequent is an equality proposition

         ---

      2. Version #2:
         Based on facts that a property corresponds to the set of all objects that have the property, & vice versa:
         • i.e., based on the facts that:
           (∀ property P)(∃ set S)[S = {x | P(x)}
           
           &
           (∀ set S)(∃ property P)[(x ∈ S) ↔ P(x)]

         • (∀ property P)[ ( P(0) ∧ (∀k ∈ N)[P(k) → P(k+1)] ) → (∀n ∈ N)P(n) ]

           • Similar comments apply as for the "Notes" for Version #1

   ---

2. Math Induction as a rule of inference:
   • recall the relation between tautologies & rules!:
     • ∀ rule of the form "From P₁, …, P_n, infer C"
       ∃ a corresponding tautology of the form "(P₁ ∧ … ∧ P_n) → C",
       and vice versa
   1. Version #1 (as a rule):
      • 
        

        From P(0) and (∀k ∈ N)[P(k) → P(k+1)] infer (∀n ∈ N)P(n)

   2. Version #2 (as a proof strategy):
      (the most useful version)
      • Recall that, ∀ rule of the form "From P₁, …, P_n, infer C"
        ∃ a corresponding proof strategy: "to prove C, try to show P₁ & … & try to show P_n"
      • 
        

        To show: (∀n ∈ N)P(n): show: P(0) "base case" & show: (∀k ∈ N)[P(k) → P(k+1)] "inductive case"

        
        

        Note: The antecedent "P(k)" is called the

        "inductive hypothesis"

        • Note: To show (∀w ∈ W)P(w),
          let the base case be: P(1), etc.

   ---

3. Analogies:
   • dominoes
   • chain reaction
   • climbing ∞ ladder
     • a way to express ∞ instructions in a finite way
     • a way to express "…" in a finite way:
       • N = { 0,1,2,3,…} = {n | (0 ∈ N) ∧ [(n ∈ N) → (n+1 ∈ N)]}

   ---

4. How to use math. induction to prove thms:
   1. Read about Gauss

      •     1 +   2   + … + 100
        100 + 99   + … +     1
        101 + 101 + … + 101

        101 + … + 101 = (100*101)/2 = 10,100/2 = 5050
        ←…100 terms…→

   2. Gauss's Thm:
      (∀n ∈ W)[Σ_{i∈{1,…,n}}i = (n*(n+1))/2]
      • Because the only free variable in the summation equation is n,
        we can call the summation equation "P(n)".
        • (the only other variable is "i",
          but it is bound by the summation-sign)

      proof:

      Show ∀nP(n):
      Show P(1) & show ∀k[P(k) → P(k+1)]:
      Base Case:
      Show P(1):
      Show Σ_i=1i = 1(1+1)/2:
      Left-hand side of that equation (LHS) = 1
      Right-hand side of that equation (RHS) = 1*2/2 = 1
      ∴ LHS = RHS

      Inductive Case:

      Show ∀k[P(k) → P(k+1)]:
      Show ∀k[ (Σ_i∈{1,…k}i = k(k+1)/2) → (Σ_{i∈{1,…k+1}}i = (k+1)(k+2)/2) ]:
      Suppose Σ_{i∈{1,…,k}}i = k(k+1)/2   (inductive hypothesis for direct proof)

      and show Σ_{i∈{1,…,k+1}}i = (k+1)(k+2)/2:

      LHS = Σ_{i∈{1,…,k+1}}i = Σ_{i∈{1,…,k}}i + (k+1) = k(k+1)/2 + (k+1)   (by the inductive hypothesis)

      = [ k(k+1) + 2(k+1) ]/2

      = (k² + k + 2k + 2)/2

      = (k² + 3k + 2)/2

      = (k+1)(k+2)/2 = RHS

      
      QED

Next lecture…