Mathematical Induction and Other Proof Methods

(数学的帰納法などの証明方法)

Discrete Mathematics I

14th lecture, January 17, 2020

https://www.sw.it.aoyama.ac.jp/2019/Math1/lecture14.html

Martin J. Dürst

© 2006-20 Martin J. Dürst Aoyama Gakuin University

Today's Schedule

• Remaining Schedule
• Summary of last lecture
• Digit sums and digital roots
• Proof Methods
• Mathematical Induction

Remaining Schedule

• January 24: 15th lecture
• January 31, 11:10-12:35: Term final exam

About makeup classes: The material in the makeup class is part of the final exam.

Questions about Final Exam

Summary of Last Lecture

• Congruence (≡) is at the core of Modular Arithmetic
• Two integers are congruent (modulo n) if they have the same remainder when divided by n
• The Congruence Relation is an equivalence relation, with the remainder as the class representative
• Congruence relations and modulo operations have many useful properties
• The result of the modulo operation for negative operands depends on the definition (programming language). In this lecture, we use the Mathematical definition.
• Modular division is defined as multiplication by the modular multiplicative inverse (where defined)

Digit Sum and Digital Root

Digit sum: Sum of all of the digits of a number

Digital root: Single-digit result of repeatedly calculating the digit sum

Example in base 10:
The digit sum of 1839275 is 1+8+3+9+2+7+5 = 35
The digit sum of 35 is 3+5 = 8
The digital root of 1839275 is 8

Example in base 16:
The digit sum of A8FB is A+8+F+B (10+8+15+11) = (44) 2C
The digit sum of 2C is 2+C (2+12) = (14) E
The digital root of A8FB is E

Application of Congruence: Casting out Nines

• The digital root of a number n in base b is equal to n mod (b-1)
  (dr(n_b) = n mod (b-1))
• Reason: For n = d_k...d₁d₀ = d_k·b^k+ d _(k-1)·b ^(k-1)+...+d₁·b¹+d₀·b⁰,
  because b^m ≡_{(mod (b-1))} 1^m = 1, n ≡_{(mod (b-1))} d_k+ d_(k-1)+...+d₁+d₀
  
  
• If b=10, then b-1=9
• Example in base 10: 1839275 mod 9 = 8
• Example in base 16: A8FB mod F = E
• This can be used for cross-checking the result of an arithmetic operation:
  x · y = z ⇒ dr(dr(x) · dr(y)) ≡_{mod (b-1)} dr(z)
  x · y ≠ z ⇐ dr(dr(x) · dr(y)) ≢_{mod (b-1)} dr(z)
  
• Example (homework): Only one of the two equations below is correct. Which one?
  1346021 · 6292041 = 8469219318861
  3615137 · 8749019 = 31628902349603

About the Handout

• From a (famous) book about (formal) language theory
  Please ignore pieces about language theory and automata theory
• Selected because it is general and easily readable
• Sometimes, is uses a different mathematical "dialect"
• The contents is part of the final exam

Importance of Proofs

• Very important tool for Mathematics
  (the goal of Mathematics is to prove as many useful and interesting theorems and properties from very few axioms and definitions)
• For computer science and information technology:
  • Proofs of properties of data structures
  • Proofs of correctness or other properties of algorithms
  • Proofs of correctness or other properties (e.g. speed) of a program
  • Proofs of correctness of program transformations

How Detailled Should a Proof Be?

• Intuition is not a proof
• (Program) test: Individual test cases are not enough for a proof
• Level of detail of a proof:
  • Rough proof
    Example: x + (y + 1) = (x + 1) + y
  • Detailled proof
    Example: x + (y + 1)
    = ^{(commutativity of addition)}
    x + (1 + y)
    =^{(associativity of addition)}
    (x + 1) + y

Ways to Express a Proof

• Mostly textual proof
• Proof using formulæ
• Automatically verified proof
• Mixtures of the above

Proof Methods

• Deductive proof (proof by deduction)
• Inductive proof (proof by induction)
• Proof by contradiction
• Proof by counterexample
• Proof about sets
• Proof by enumeration

Proofs and Symbolic Logic

(S is the the theorem to be proven, expressed as a proposition or predicate)

• Deductive proof: (H ∧ (H→S)) ⇒ S, etc.
• Inductive proof: S(0) ∧ (∀k∈ℕ: S(k) → S(k+1)) ⇒ (∀n∈ℕ: S(n))
  
• Proof by contradiction: (¬S→S) ⇒ S

  Confirmation:

  S	¬S	¬S→S	(¬S→S) → S
  F	T	F	T
  T	F	T	T

• Proof by counterexample:
  (∃x:¬S(x)) ⇒ ¬∀x: S(x)
• Proof by enumeration (example: truth table)

Deduction and Induction

• Deduction: Conclude some specific fact from some general law
• Induction: Infer some general law from some sample observations
• Mathematical induction
  • Goal: Proof of some property about (almost) all parts of some structure
  • The integers are the most frequent "structure", but also tree structures,...
  • Mathematical induction is actually a kind of deduction, not induction
• Applications of mathematical induction in information technology:
  • Design and proof of properties of algorithms and data structures
  • Proofs about programs:
    • Loops
    • Recursion

The Two steps of Mathematical Induction

S(0) ∧ (∀k∈ℕ: S(k) → S(k+1)) ⇒ (∀n∈ℕ: S(n))

1. Base case (basis (step)): Proof of S(0)
2. Inductive step (induction, inductive case): Proof of ∀k∈ℕ: S(k) → S(k+1)
   
   1. (inductive) Assumption: clearly state S(k)
   2. Actual proof of inductive step
      Method: Formula manipulation so that the assumption can be used

Simple Example of Mathematical Induction

Look at the following equations:

 0 = (adding up no numbers results in 0)
 1 = 1
 4 = 1 + 3
 9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9

Express the general rule contained in the above additions as a hypothesis.

Prove the hypothesis using Mathematical induction.

Hypothesis

• The left side of the equations is n² (n≥0)
• The right side of the equations is the sum of the (consecutive) n smallest odd numbers (n≥0)
• Hypothesis: The sum of the n smallest odd numbers is n² for n≥0:
  ∀n≥0: n² = ∑ⁿ_i=1 2i-1

Proof

1. Basis:

Prove the property for n = 0: ∑⁰_i=1 2i-1 = 0 = 0²

2. Induction:

a. Inductive assumption: Assume that the property is true for some k≥0: k² = ∑^k_i=1 2i-1

b. Show that the property is true for k + 1:
(k+1)² = ∑^(k+1)_i=1 2i-1

[start with left side]
(k+1)² [expansion]
= k² + 2k + 1 [arithmetic]
= k² + 2k + 2 - 1 [arithmetic]
= k² + 2(k+1) - 1 [use assumption]
= ∑^k_i=12i-1 + 2(k+1) - 1 [property of ∑]
= ∑^k+1_i=1 2i-1

Both the basis and the induction are true. This proves the hypothesis. Q.E.D.

Variations of Mathematical Induction

• Start with S(1) or S(2),... instead of S(0)
  S(b) ∧ (∀k∈ℕ, k≥b: S(k) → S(k+1)) ⇒ (∀n∈ℕ, n≥b: S(n))
  We can interpret this as proving T(n-b) = S(n), so that we again start at 0
  
• In the proof of step 2 for S(k+1), use not only
  S(k), but some or all S(j) where 0≤ j≤k
  (this is called strong induction or complete induction)
  S(0) ∧ (∀k∈ℕ: (∀i (0≤i≤k): S(i)) → S(k+1)) ⇒ ∀n∈ℕ: S(n)
  
• Limit the proof to some subset of integers (examples: even numbers only, 2^m only)
  We can interpret this as proving T(n/2) or T(m = log₂ n)
• Proof not about integers, but about something that can be ordered using integers
• Branching tree structure or some other structure (e.g. (half) order, lattice,...): structural induction

Example of Structural Induction

• Theorem to be proved:
  In a binary tree with l leaves, the number of nodes is n = 2l-1
• Definition of binary tree:
  Directed graph with the following conditions:
  
  • No cycles
  • If there is a directed edge (arrow) from node a to node b, then a is the parent of b, and b is the child of a
    
  • All nodes except the root have a parent
  • There are two types of nodes:
    • Leaves, which have no children
    • Internal nodes, which have exactly two children

Proof of the Relation between the Number of Nodes and Leaves in a Binary Tree

We start with a very small tree consisting only of the root, and grow it step by step. We can create any shape of binary tree this way.

1. Base case: In a tree with only the root node, n=1 and l=1, therefore n = 2l-1 is correct.
2. Inductive step: Grow the tree one step by replacing a leaf with an internal node with two leaves.
   Denote the number of nodes before growing by n, the number of leaves before growing by l, the number of nodes after growth by n', and the number of leaves after growth by l'. We need to prove n' = 2l'-1.
   1. Inductive assumption: n = 2l-1
   2. In one growth step, the number of nodes increases by two: n'=n+2 (1)
      In one growth step, the number of leaves increases by two but is reduced by one: l'=l+2-1=l+1; l = l'-1 (2)
      n' [(1)]
      = n+2 [assumption]
      = 2l-1+2 [(2)]
      = 2(l'-1)-1+2 [arithmetic]
      = 2l'-1
      

Both the basis and the induction are true. This proves the hypothesis. Q.E.D.

Homework

(no need to submit)

1. Answer the question on the slide "Application of Congruence: Casting out Nines"
2. Find the problem in the following proof:

   Theorem: All n lines on a plane that are not parallel to each other will cross in a single point.

   Proof:

   1. Base case: Obviously true for n=2
   2. Induction:
      1. Assumption: k lines cross in a single point.
      2. For k+1 lines, both the first k lines and the last k lines will cross in a single point, and this point will have to be the same because it is shared by k-1 lines.
3. Read the handout
4. Find a question regarding past examinations that you can ask in the next lecture.

Glossary

digit

数字、桁

digit sum

数字和

digital root

数字根

casting out nines

九去法

(formal) language theory

言語理論

automata theory

オートマトン理論

proof

証明

to prove

証明する

data structure

データ構造

intuition

直感

test case

テスト・ケース

deductive proof

演繹的証明

inductive proof

帰納的証明

proof by contradiction

背理法

proof by counterexample

反例による証明

proof about sets

集合についての証明

proof by enumeration

列挙による「証明」

mathematical induction

数学的帰納法

structure

構造

loop

繰返し

base case

基底

inductive step

帰納

inductive assumption

(帰納の) 仮定

recursion

再帰

hypothesis

仮説

equation

方程式

consecutive

連続的な

odd (number)

奇数

inductive assumption

(帰納の) 仮定

strong induction

完全帰納法 (または累積帰納法)

structural induction

構造的帰納法

binary tree

二分木

node (of a tree/graph)

節

leaf (of a tree)

葉

directed graph

有効グラフ

cycle (of a graph)

閉路

parent (in a tree)

親

child (in a tree)

子

root (of a tree)

根

internal node

内部節

parallel

平行