# 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 # Today's Schedule

• Remaining Schedule
• Summary and homework for last lecture
• Mathematical Induction
• Fibonacci numbers
• Repetition
• Outlook

# Remaining Schedule

January 31, 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

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:

• Base case: Obviously true for n=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.

# Homework 3 from Last Lecture

[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):

• 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:

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)
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

データ構造とアルゴリズム (2年後期, Dürst)

# Repetition

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

commutativity

associativity

(in)formal
(非)形式的
lemma

golden ratio