Discrete Structures

# Lecture Notes, 10 Nov 2010

 Last Update: 10 November 2010 Note: or material is highlighted

## §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[nN  →  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 ≠ nN  →  (∃mN)[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 NN is 1–1 but not onto

• The successor function from NW 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, …, Pn, infer C"
∃ a corresponding tautology of the form "(P1 ∧ … ∧ Pn) → 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, …, Pn, infer C"
∃ a corresponding proof strategy: "to prove C, try to show P1 & … & try to show Pn"

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

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

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

3. Analogies:

• dominoes
• chain reaction

• 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) ∧ [(nN) → (n+1 ∈ N)]}

4. How to use math. induction to prove thms:

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

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

2. Gauss's Thm:

(∀nW)[Σi∈{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 Σi=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[ (Σi∈{1,…k}i = k(k+1)/2) → (Σi∈{1,…k+1}i = (k+1)(k+2)/2) ]:

Suppose Σi∈{1,…,k}i = k(k+1)/2   (inductive hypothesis for direct proof)

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

LHS = Σi∈{1,…,k+1}i = Σi∈{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