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 6, 2023

https://www.sw.it.aoyama.ac.jp/2022/Math1/lecture14.html

Martin J. Dürst

© 2006-23 Martin J. Dürst Aoyama Gakuin University

Today's Schedule

• Remaining Schedule
• Homework and summary of last lecture
• Proof Methods
• Mathematical Induction
• Student Survey

Remaining Schedule

• January 6: this lecture
• January 13: 15th lecture
• January 27, 11:10-12:35: Term final exam (85 minutes)

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 are helpful when checking calculations using casting out nines

Homework from Last Lecture

• Create some addition problems in different bases, and then solve them (see lecture 2). Check your results using casting out nines (or b-1). Also, check your results by converting the numbers to decimal notation.
• Create some subtraction problems in different bases, and figure out how to do subtraction in bases other than 10. Check your results as above.
• Create some problems for remainder calculation using congruence, and solve them. Check your results using a calculator.
• Prepare for final exam

How to Prepare for Final Exam

• Use past exams for training and review
• Submit one question about past exams or lecture content as homework
• Use Q&A Forum for questions about lecture content
• Check Q&A Forum for answers on questions from others
• Ask questions now!

More Homework from Last Lecture

Only one of the two equations below is correct. Which one?

1346021 · 6292041 = 8469219318861

dr(1346021) · dr(6292041) ≡? dr(8469219318861)

8 · 6 ≡_{mod 9} 3 ⇒ maybe correct

3615137 · 8749019 = 31628902349603

dr(3615137) · dr(8749019) ≡? dr(31628902349603)

8 · 2 ≡_{mod 9} 7 ≠ 2 ⇒ wrong

Homework about Groups

• Homework due December 15, 2022
• Returning this homework after this lecture
• Some very nice results, but also some very bad results
• Advice to improve:
  • Read question carefully, and follow it:
    • Symmetric group of order 3 ⇒ group of size 3! (=6)
    • Use lexical order for the permutations. Use the notation introduced in this lecture.
    • Check all the conditions (axioms) for a group.
      (closure, identity, inverse, associativity)
  • If you don't understand something, consult a book or the web, or come to ask

Homework about Groups: Frequent Errors

• 3-element group instead of symmetric group of order 3 (size 6)
• Missing rows or columns (not a Caley table)
• Different row and column headings (not a group), or different order (not a Caley table)
• Results in table that are not part of group (not a group)
• Duplicate elements in same row or column (not a group)
• Identity element not in first row/column (not a Caley table)
• See examples on handout

Homework about Boolean Algebras: Frequent Errors

(Homework due December 22, 2022)

• {∅}: Empty set is {} or ∅, but not {∅}
• Sets written as sequences ({a, b, c} ≠ (a, b, c))
• Hasse diagrams do not use arrows
• Hasse diagrams have no horizontal lines
• Duplications:
  • {a, b, c, d}: 6 cases
  • {s, t, u, v}: 3 cases
  • (0110}: 5 cases
  • 11, 13, 17, 19: 3 cases

About the Handout

• Sections 1.2-1.4 of Introduction to Automata Theory, Languages, and Computation
  by John E. Hopcroft, Rajeev Motwani, and Jeffrey D. Ullman
• Please ignore pieces about language theory and automata theory
• Language theory is part of 3rd-year course Language Theory and Compilers
• Selected because it is general and easily readable
• Sometimes, uses a different mathematical "dialect"
• Contents is part of the final exam
• For this lecture only; do not distribute

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

Ways to Express a Proof

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

How Detailled Should a Proof Be?

• Intuition is not a proof
• (Program) test: Individual test cases are not enough for a proof
• Rough proof: Steps are grouped, general arithmetic knowledge is assumed
  Example (one step): x + y + 1 = 1 + x + y
• Detailled proof: Every small step is documented and justified
  Example: x + (y + 1) = ^{(commutativity of addition)}
  x + (1 + y) = ^{(associativity of addition)}
  (x + 1) + y = ^{(commutativity of addition)}
  

  (1 + x) + y

• Required details depend on topic and assumed knowledge

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 theorem to be proven, expressed as a proposition or predicate)

• Deductive proof: (H ∧ (H→S)) ⇒ S, etc.
• Inductive proof: see later slide
  
• 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 natural numbers 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

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
• Property to prove for all n≥0: S(n): 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:
We need to prove that (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.

Important: In exam, follow this structure, including final sentence.

The Two steps of Mathematical Induction

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

Mathematical induction in symbolic logic:
S(0) ∧ (∀k∈ℕ: S(k) → S(k+1)) ⇒ ∀n∈ℕ: S(n)

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 natural numbers (examples: even numbers only, 2^m only)
  We can interpret this as proving T(n/2) or T(m = log₂ n)
• Proof not about natural numbers, but about something that can be ordered using natural numbers
• Branching tree structure or other structure (e.g. (half) order, lattice,...): structural induction

Example of Structural Induction on Trees

• Theorem to be proven:
  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
• Theorem is helpful, shows that about half of nodes in a binary tree are leaves.

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 a binary tree of any shape..

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

1. Read the handout (Formal Proof, English or Japanese, some paper handouts available)
2. Find a question regarding past examinations or lecture content.
   Deadline: January 11, 2023 (Wednesday), 22:00.
   Where to submit: Moodle quiz
   Caution: Do not submit the same question as another student
   
3. Prove the closed formula for combinations (see lecture 7) by structural induction (no need to submit)
4. Find the problem in the following proof (no need to submit):

   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.

Student Survey

(授業改善のための学生アンケート)

WEB Survey

お願い: 自由記述に必ず良かった点、問題点を具体的に書きましょう

(悪い例: 発音が分かりにくい; 良い例: さ行が濁っているかどうか分かりにくい)

Glossary

(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