Discrete Structures

Lecture Notes, 10 Nov 2010

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

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

§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
• 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) ∧ [(nN) → (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:

(∀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

Text copyright © 2010 by William J. Rapaport (rapaport@buffalo.edu)
Cartoon links and screen-captures appear here for your enjoyment. They are not meant to infringe on any copyrights held by the creators. For more information on any cartoon, click on it, or contact me.
http://www.cse.buffalo.edu/~rapaport/191/F10/lecturenotes-20101110.html-20101110