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

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))

Base case (basis (step)): Proof of S(0)
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.

Base case: In a tree with only the root node, n=1 and l=1, therefore n = 2l-1 is correct.
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)

Answer the question on the slide "Application of Congruence: Casting out Nines"
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.
Read the handout
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