Lecture Notes, 12 Nov 2010

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§4.1: Mathematical Induction (cont'd)

General Notes:
• To make my typing easier, I will no longer italicize variables, so I'll write "∀x" instead of "∀x", etc.

• Also, please remember that when I write something like:
  Σ_{i ∈ {1,…,n}}

  that's just an easier way of typing:

  i = n
  Σ
  i = 1

1. Mathematical Induction as a Proof Strategy (reminder):

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

   • Notes:
     1. To show P(0):
        Identify P(n), & let n:=0; then use algebra

     2. Sometimes you'll need to show (∀w ∈ W)P(w) instead,
        so then just let the base case be: P(1), etc.

     3. To show the inductive case,
        just realize that it's a quantified conditional
        and we already know how to prove those:
        Choose an arbitrary k, and show P(k) → P(k+1)
        And how do we prove a conditional? Using Direct Proof:

        Suppose P(k), & try to show P(k+1).
        

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

        "inductive hypothesis"

   ---

2. Another example of a proof by mathematical induction:

   Thm (General Distributive Law)

   Let a,b₁,…,b_n ∈ R = {x | x is a real number}.
   Then: (∀n ∈ N)[a(b₁+…+b_n) = ab₁+…+ab_n]
   • Just to help you get used to the notation, here's the theorem re-stated using Σ notation:
     (∀n ∈ N)[aΣ_{i ∈ {1,…,n}}b_i = Σ_{i ∈ {1,…,n}}ab_i]

     proof:

     The "property" P that we're interested in proving that it holds for all natural numbers is:

     P(n): "a(b₁+…+b_n) = ab₁+…+ab_n"

     Base Case (first (failed) attempt):

     Show P(0):
     Hmmm, what on earth could P(0) be?
     I have no idea, but math inductions can begin with 1,
     so let's re-state the theorem as:
     (∀ n ∈ W)P(n)

     Base Case (second (slightly more successful) attempt):

     Show P(1):
     i.e., show ab₁ = ab₁

     That's trivially true! That was easy!

     There's only one problem: It won't be of any help to us :-|

     So, let's re-state the theorem one more time, as:

     (∀ n ≥ 2)P(n)

     (This is the only interesting case anyway.)

     Base Case (third (finally successful) attempt):
     Show P(2):
     i.e., show a(b₁+b₂) = ab₁+ab₂:
     Almost trivial(!), because…
     …this is the distributive law of multiplication over addition, which is an axiom.
     • Or it can be proved from the "recursive" definitions of + and *, which we'll get to later in the semester.

     • If you want a geometric "proof",
       it says that the area of a large rectangle consisting of two smaller rectangles piled one on top of the other,
       one with dimensions a×b₁ and one with dimensions a×b₂,
       equals the sum of the areas of the two smaller rectangles.

     So now we're ready for the…

     …Inductive Case:

     Show ∀k[P(k) → P(k+1)]:

     i.e., show ∀k[ (a(b₁+…+b_k) = ab₁+…+ab_k) → (a(b₁+…+b_k+1) = ab₁+…+ab_k+1) ]

     To do this, suppose the antecedent: a(b₁+…+b_k) = ab₁+…+ab_k
     • (that's the "inductive hypothesis";
       but really all it is is the assumption for a direct proof)

     & show the consequent: a(b₁+…+b_k+1) = ab₁+…+ab_k+1:

     Up to here, it's all been fairly mechanical; now some "insight" is needed.
     • (But it's only high-school-level, algebraic insight)

     Let's compute a(b₁+…+b_k+1):

     a(b₁+…+b_k+1) = a(b₁+…+b_k+b_k+1)
     • (just spelling out the "…")

     = a( (b₁+…+b_k) + b_k+1)

     • (just putting some parentheses in, to clarify the structure)

     = a(    B   + b_k+1)

     • (just letting B = (b₁+…+b_k) )

     = aB + ab_k+1, by the base case!

     = a(b₁+…+b_k) + ab_k+1

     • (by the definition of B)

     = (ab₁+…+ab_k) + ab_k+1

     • (by the inductive hypothesis)

     QED

   ---

3. Another Example (which we did not have time for in class, so I'll record the entire proof here):

   Thm (the cardinality of the power set of S = 2 raised to the cardinality of S)

   • i.e., if S has n members, then it has 2ⁿ subsets.
   (∀ n ∈ N)(∀ set S)[ |S|=n → |℘(S)| = 2ⁿ ]

   proof:

   Here, P(n) is: "(∀ set S)[ |S|=n → |℘(S)| = 2ⁿ ]"

   Base Case:

   Show P(0):
   Show (∀ set S)[ |S|=0 → |℘(S)| = 2⁰ ]:
   Choose arb set S.
   Suppose |S|=0, i.e., S = ∅
   Show |℘(S)| = 2⁰:
   ℘(S) = ℘(∅) = {∅}
   ∴ |℘(S)| = |{∅}| = 1 = 2⁰.

   [This is as far as we got in lecture; now for the rest of the proof:]

   Inductive Case:

   Show ∀k[P(k) → P(k+1)]:
   i.e., show ∀k[ ∀S[ |S]=k → |℘(S)|=2^k] → ∀S′[ |S′|=k+1 → |℘(S′)|=2^k+1] ]:
   • (I'm only using "S′" so that you can keep the S-in-the-antecedent separate from the S-in-the-consequent.)
   Choose arb k
   Suppose ∀S[ |S]=k → |℘(S)|=2^k] (ind. hyp.)
   & show: ∀S′[ |S′|=k+1 → |℘(S′)|=2^k+1]:
   Choose arb S′
   Suppose |S′|=k+1
   & show |℘(S′)|=2^k+1:
   We are assuming |S′|=k+1;
   ∴ we have the following picture:
   S′, which has k+1 elements, consists of
   a subset S′′ with only k elements,
   unioned with one more element; call it "e".
   • i.e., S′ = S′′ ∪ {e},
     where |S′| = |S′′| + |{e}| = k+1

   

   Describing this in FOL, we have:

   ∃S′′[S′=S′′∪{e} ∧ |S′′|=k]

   ∴ |℘(S′′)|=2^k, by the inductive hypothesis

   But to each subset of S′′, there correspond two subsets of S′:

   

   i.e., in (sort of) FOL:

   (∀ subset S^* ⊆ S′′)(∃ 2 subsets of S′=S′′∪{e}), namely: S^* itself & S^*∪{e}.

   ∴ |℘(S′)| = |℘(S′&prime)|*2 (i.e., doubled)
   = 2^k*2 = 2^k+1

   QED

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