Mathematical Induction and Other Proof Methods

(数学的帰納法などの証明方法)

Discrete Mathematics I

14th lecture, Jan. 13, 2016

http://www.sw.it.aoyama.ac.jp/2016/Math1/lecture14.html

Martin J. Dürst

AGU

© 2006-16 Martin J. Dürst Aoyama Gakuin University

Today's Schedule

Remaining Schedule

 

Questions about Final Exam

 

Remainder Calculation: More Examples

318 mod 7 = ?

318 = (33)6 = 276 (mod 7) (-1)6 = 1

4110 mod 37 = ?

4110 (mod 37) 410 = 220 = 324 (mod 37) (-5)4 = 252 (mod 37) (-12)2 = (6 · 2)2 = 36 · 4 (mod 37) (-1) · 4 = -4 (mod 37) 33

 

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

 

About the Handout

 

Importance of Proofs

 

How Detailled Should a Proof Be?

 

Proof Methods

 

Proofs and Symbolic Logic

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

 

Deduction and Induction

 

Applications of Mathematical Induction

Applications of mathematical induction in information technology:

 

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:

1 = 1

1 + 3 = 4

1 + 3 + 5 = 9

1 + 3 + 5 + 7 = 16

1 + 3 + 5 + 7 + 9 = 25

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

Prove the hypothesis using Mathematical induction.

 

Hypothesis

Proof

Basis:

Prove the property for n = 0: ∑0i=0 2i+1 = 1 = 12

Induction:

Inductive assumption: Assume that the property is true for some k≥0: ∑ki=0 2i+1 = (k+1)2

Show that the property is true for k + 1:
(k+1)i=0 2i+1 = (k+2)2

[start with right side]
(k+2)2 [expansion]
= k2 + 4k + 4 [arithmetic]
= k2 + 2k +1 + 2k + 3 [arithmetic]
= (k+1)2 + 2(k+1) + 1 [use assumption]
= ∑ki=0 (2i+1) + 2(k+1) + 1 [property of ∑]
= ∑(k+1)i=0 2i+1 Q.E.D.

 

Variations of Mathematical Induction

 

Homework

(no need to submit)

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

 

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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)
parallel
平行