Proof Methods
(証明の方法)
Discrete Mathematics I
14th lecture, Jan. 16, 2015
http://www.sw.it.aoyama.ac.jp/2014/Math1/lecture14.html
Martin J. Dürst
© 2006-15 Martin
J. Dürst Aoyama Gakuin
University
Today's Schedule
- Remaining Schedule
- Proof Methods
- Mathematical Induction
- Student Survey
Remaining Schedule
- January 23 (makeup class): 15th
lecture
- January 30, 11:10~12:35: Final exam
About makeup classes: The material in the makeup class is part of the final
exam. If you have another makeup class at the same time, please inform the
teacher as soon as possible.
補講について:
補講の内容は期末試験の対象。補講が別の補講とぶつかる場合には事前に申し出ること。
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 applying 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 A3FB is E
Application of Congruence: Casting out Nines
- The digital root of a number n in base b is equal
to n mod (b-1)
(dr(n) = n mod (b-1))
- If b=10, then b-1=9
- Reason: For n =
dk...d1d0
= dk×bk+
d(k-1)×b(k-1)+...+d1×b1+d0×b0,
because bm ≡(mod
(b-1)) 1m = 1, n
≡(mod (b-1)) dk+
d(k-1)+...+d1+d0
- This can be used for cross-checking the result of an arithmetic
operation:
a · b = c → dr(dr(a) ·
dr(b)) = dr(c)
a · b ≠ c ← dr(dr(a) ·
dr(b)) ≠ dr(c)
- Example: Only one of the two equations below is correct. Which one?
2485938 · 4962483 = 12336425064054
2354987 · 2498472 = 5883469079864
About the Handout
- From a (famous) book about (formal) language theory
Please ignore pieces about language theory and automata theory
- Selected because it is general and easily readable
- Sometimes, is uses a different mathematical "dialect"
- The contents is part of the final exam
Importance of Proofs
- Very important tool for Mathematics
(the goal is to prove as many useful and interesting theorems and
properties from very few axioms and definitions)
- For computer science and information technology:
- Proofs of properties of data structures
- Proofs of correctness or other properties of algorithms
- Proofs of correctness or other properties (e.g. speed) of a
program
- Proofs of correctness of program transformations
How Detailled Should a Proof Be?
- Intuition
- (Program) test
- Level of detail of a proof:
- Rough proof
Example: x + (y + 1) = (x + 1) +
y
- Detailled proof
Example: x + (y + 1) = (commutativity of
addition) x + (1 + y) =
=(associativity of addition) (x + 1) +
y
- How to express proofs:
- Mostly textual proof
- Proof using formulæ
- Automatically verified proof
Proof Methods
- Deductive proof (proof by deduction)
- Inductive proof (proof by induction)
- Proof by contradiction
- Proof by counterexample
- Proof about sets
- Proof by enumeration
Proofs and Symbolic Logic
(S is the proposition or predicate)
- Deductive proof: (H ∧
(H→S)) ⇒ S, etc.
- Inductive proof: S(0) ∧ (∀k∈ℕ:
S(k) → S(k+1)) ⇒
(∀n∈ℕ: S(n)), etc.
- Proof by contradiction: (¬S→S) ⇒
S
- Proof by counterexample: (∃x: ¬S(x)) ⇒ ¬∀x: S(x)
- Proof by enumeration: example: truth table
Deduction and Induction
- Deduction: Conclude some specific fact
from some general law
- Induction: Infer some general law from some sample observations
- Mathematical induction
- Goal: Proof of some property about (almost) all parts of some
structure
- The integers are the most frequent "structure", but also tree
structures,...
- Mathematical induction can actually be classified as some kind of
deduction, not induction
Applications of Mathematical Induction
Applications of mathematical induction in information technology:
- Design and proof of properties of algorithms and data structures
- Proofs about programs:
The Two steps of Mathematical Induction
S(0) ∧ (∀k∈ℕ: S(k) →
S(k+1)) ⇒ (∀n∈ℕ:
S(n))
- Base case (basis (step): proof of S(0)
- Inductive step (induction, inductive case): proof of
∀k∈ℕ: S(k) →
S(k+1)
- (inductive) Assumption: clearly stateS(k)
- Actual proof of inductive step
Method: Formula manipulation so that the inductive assumption cas 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
- The right side of the equations is m2
- The left side is the equations is the sum of m consecutive odd
numbers, starting with 1
- Hypothesis: ∀ n≥0:
∑ni=0 2i+1 =
(n+1)2
Proof
Basis:
Prove the property for n = 0:
∑0i=0 2i+1 = 1 =
12
Induction:
Inductive assumption: Assume that the hypothesis 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
hypothesis]
= ∑ki=0 (2i+1) +
2(k+1) + 1 [property of ∑]
= ∑(k+1)i=0 2i+1
Q.E.D.
Variations of Mathematical Induction
- Choice of S(1) or S(2),... instead of
S(0)
S(b) ∧ (∀k∈ℕ, k≥b:
S(k) → S(k+1)) ⇒
(∀n∈ℕ, n≥b:
S(n))
We can interpret this as proving T(n-b) =
S(n), so that we again start at 0
- In the proof of step 2 for S(k+1), use not only
S(k), but also some or all S(j)
where j≤k
(this is called strong induction or complete
induction)
S(0) ∧ (∀k∈ℕ: (∀i
(0≤i≤k): S(i)) →
S(k+1)) ⇒ ∀n∈ℕ:
S(n)
or (∀k∈ℕ: (∀i
(0≤i≤k): S(i)) →
S(k+1)) ⇒ ∀n∈ℕ:
S(n)
- Limit the proof to some subset of integers (examples: even numbers only,
2m only)
- Proof not about integers, but about something that can be ordered using
integers
- Branching tree structure or some other structure (e.g. (half) order)
Homework
(no need to submit)
- Find out the problem in the following proof:
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:
- Assumption: k lines cross in a single point.
- 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.
- Find a question regarding past examinations that you can ask in the next
lecture.
Student Survey
(授業改善のための学生アンケート)
お願い:
自由記述をできるだけ使って、具体的に書いてください
(例: 「英語をやめてほしい」のではなく、「Glossary を ...
に改善してください」)
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
- 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
- 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
- parallel
- 平行