Remainder, Review

(残り・復習)

Discrete Mathematics I

15th lecture, January 24, 2020

https://www.sw.it.aoyama.ac.jp/2019/Math1/lecture15.html

Martin J. Dürst

AGU

© 2006-20 Martin J. Dürst Aoyama Gakuin University

Today's Schedule

 

Remaining Schedule

January 31, 11:10-12:35: Term final exam

 

Summary of Last Lecture

 

Homework 1 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 2 ⇒ wrong

 

Homework 2 from Last Lecture

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:

 

Homework 3 from Last Lecture

Read the handout

[a small number of copies is still available in this lecture or 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)

  1. a∈ℕ+: a + 1 = s(a)
  2. 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:

  1. Base case (a=1):
    1+1 = 1+1[axiom of addition 1]
    = s(1) [axiom of addition 1]
    = 1+1
  2. 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.]

  1. Base case (b=1):
    a+1 = 1+a [lemma on previous slide]
  2. 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):

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

Wide range of applications:

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:

  1. 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
  2. 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