Remainder, Repetition

(残り・復習)

Discrete Mathematics I

15th lecture, Jan. 23, 2015

http://www.sw.it.aoyama.ac.jp/2014/Math1/lecture15.html

Martin J. Dürst

© 2006-15 Martin J. Dürst Aoyama Gakuin University

Today's Schedule

• Remaining Schedule
• Summary and homework for last lecture
• Mathematical Induction (continued)
• Repetition
• Lookout

Remaining Schedule

• January 30, 11:10～12:35: Final exam

Last Week's Homework 1

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.

[都合により削除]

Summary of Last Week's Lecture

• 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

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 with predicate logic)

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

Note: Induction for Peano arithmetic is:
S(1) ∧ (∀k∈ℕ⁺: S(k) → S(s(k)) ⇒ ∀n∈ℕ⁺: S(n)

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 [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. Induction hypothesis: 1+k = s(k) = k+1
   2. 1 + s(k) [axiom of addition 1]
      = 1 + (k+1) [associativity of addition]
      = (1+k) + 1 [hypothesis]
      = 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 from Lemma on previous slide
2. Inductive step: Proof of a+s(k) = s(k)+a
   1. Induction hypothesis: a+k = k+a
   2. a + s(k) [axiom of addition 1]
      = a + (k+1) [associativity of addition]
      = (a+k) + 1 [hypothesis]
      = (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, it was more difficult to prove the commutative law. Actually, we used the associative law to prove the commutative law.

Fibonacci Numbers

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

Definition of Fibonacci function fib(n):

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

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:

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. Induction hypothesis: 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) [hypothesis, p = m+1]
      = fib(k+m+1) [arithmetic]
      = fib(k+1+m) Q.E.D.

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.

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 numebr of nodes after growth by n', and the number of leaves after growth by l'. We need to prove n' = 2l'-1.
   1. Induction hypothesis: 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 [hypothesis]
      = 2l-1+2 [(2)]
      = 2(l'-1)-1+2 [arithmetic]
      = 2l'-1 Q.E.D.

Outlook

情報数学 II: 情報理論、グラフ理論など (二年前期、大原先生)

計算機実習 I (二年前期, Dürst)

データ構造とアルゴリズム (二年後期, Dürst)

言語理論とコンパイラ (三年前期, Dürst)

卒業研究 (四年通年)

Repetition

Questions, please!

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

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