Remainder, Review

(残り・復習)

Discrete Mathematics I

15th lecture, January 18, 2019

http://www.sw.it.aoyama.ac.jp/2018/Math1/lecture15.html

Martin J. Dürst

Today's Schedule

Remaining Schedule
Homewory and summary for last lecture
Mathematical Induction (continued)
Repetition
Outlook

Remaining Schedule

January 25, 11:10-12:35: Term final exam

Summary of Last Lecture

Digit sum and digital root are helpful when checking calculations using casting out nines
Proofs are the most important tool of Mathematics, and very useful for Information Technology
There are many different proof methods:
Deductive proof, inductive proof, proof by contradiction, proof by counterexample, proof about sets, proof by enumeration
Mathematical induction is an important proof method, consisting of two steps: Base and induction

Homework 1 from Last Lecture

Sorry, it was removed! :)

Homework 2 from Last Lecture

Sorry, it was removed! :)

Homework 3 from Last Lecture

Read the handout

[a small number of copies is still available in front of room O-529]

Proving Commutativity of Addition
from Peano Axioms

In lecture 2, we proved associativity of addition
(a + (b+c) = (a+b) + c) from the Peano Axioms.

This was a (very informal) proof by mathematical induction.

Here we prove commutativity of addition (a+b = b+a), also by mathematical induction.

Reminder: axioms of addition (expressed in predicate logic)

∀a∈ℕ⁺: a + 1 = s(a)
∀a, b∈ℕ⁺: a + s(b) = s(a + b)

Induction for Peano arithmetic (Peano axiom 5):P(1) ∧ (∀k∈ℕ⁺: P(k) → P(s(k)) ⇒ ∀n∈ℕ⁺: P(n)

For comparison: Inductive proof:
S(0) ∧ (∀k∈ℕ: S(k) → S(k+1)) ⇒ ∀n∈ℕ: S(n)

In other words, Peano axiom 5 states that mathematical induction works for natural numbers.

Lemma: Commutativity of Addition for 1

Lemma: 1+a = s(a) = a+1

(a lemma is a minor theorem, used as a stepping stone)

Proof:

Base case (a=1):
1+1 = 1+1[axiom of addition 1]
= s(1) [axiom of addition 1]
= 1+1
Inductive step: Proof of 1+s(k) = s(s(k)) = s(k)+1
1. Inductive assumption: For any k≧1, 1+k = s(k) = k+1
2. 1 + s(k) [axiom of addition 1]
  = 1 + (k+1) [associativity of addition]
  = (1+k) + 1 [assumption]
  = s(k) + 1 [axiom of addition 1]
  = s(s(k)) Q.E.D.

Proof of Main Theorem

Theorem: a+b = b+a

Method: Use induction over b.

[General remark: if there are two or more variables, try induction over one of them.]

Base case (b=1):
a+1 = 1+a [lemma on previous slide]
Inductive step: Proof of a+s(k) = s(k)+a
1. Inductive assumption: For any k≧1, a+k = k+a
2. a + s(k) [axiom of addition 1]
  = a + (k+1) [associativity of addition]
  = (a+k) + 1 [assumption]
  = (k+a) + 1 [associativity of addition]
  = k + (a+1) [lemma on previous slide]
  = k + (1+a) [associativity of addition]
  = (k+1) + a [axiom of addition 1]
  = s(k) + a Q.E.D.

Comment: The commutative law is simpler than the associative law. However, proving the commutative law was more difficult. We used the associative law four times to prove the commutative law.

Fibonacci Numbers

Number sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,...

Definition of Fibonacci function fib(n):

fib(0) = 0
fib(1) = 1
∀n>1: fib(n) = fib(n-1) + fib(n-2)

This definition directly leads to a very simple (but slow) recursive implementation.

Wide range of applications:

Number of rabbits after n months
Golden ratio
Number of sunflower seeds

Wide range of mathematical properties

Proof of a Property of Fibonacci Numbers

Property: fib(n+m) = fib(m) · fib(n+1) + fib(m-1) · fib(n) (∀n≥0, m≥1)

Proof by induction over n:

Base case (n=0):
fib(0+m) = fib(m) · fib(0+1) + fib(m-1) · fib(0)
fib(m) = fib(m) · 1 + fib(m-1) · 0
Inductive step: Proof of fib(k+1+m) = fib(m) · fib(k+2) + fib(m-1) · fib(k+1)
1. Inductive assumption: fib(k+p) = fib(p) · fib(k+1) + fib(p-1) · fib(k) (∀p≥1)
2. (Hint: Start with more difficult side)
  fib(m) · fib(k+2) + fib(m-1) · fib(k+1) [Definition of fib (k>0)]
  = fib(m) · (fib(k+1)+fib(k)) + fib(m-1) · fib(k+1) [arithmetic]
  = (fib(m)+ fib(m-1)) · fib(k+1) + fib(m) · fib(k) [Definition of fib (m>1)]
  = fib(m+1) · fib(k+1) + fib(m) · fib(k) [assumption, p = m+1]
  = fib(k+m+1) [arithmetic]
  = fib(k+1+m) Q.E.D.

Outlook

情報確率統計 (二年前期、楽先生)

情報数学 II: 情報理論、グラフ理論など (二年後期、大原先生)

データ構造とアルゴリズム (2年後期, Dürst)

情報総合プログラミング実習II: ウェブ技術 (3年前期, Dürst)

言語理論とコンパイラ (3年前期, Dürst)

卒業研究 (4年通年)

Repetition

Homework 4 from last lecture: Find a question regarding past examinations that you can ask in the next lecture.

Glossary

commutativity: 可換性 (注: 交換性ではない)
associativity: 結合性
(in)formal: (非)形式的
lemma: 補題 (補助定理)
golden ratio: 黄金比