(整数の表現)

http://www.sw.it.aoyama.ac.jp/2016/Math1/lecture2.html

© 2005-15 Martin J. Dürst Aoyama Gakuin University

- Last week's homework, Moodle registration
- How to watch videos of this lecture
- History of numbers and numerals
- The historic origin of numbers: 1
- Creating natural numbers starting from 1
- The discovery of 0
- Positional notation (decimal, binary, ...)

- Mathematics is an important base for Information Technology
- Discrete Mathematics I mainly covers discrete mathematics (logic,...)
- Mathematics is a tool, a language, and a way of thinking
- English is very important for Information Technology
- English is best learned by diving in and practicing

- Create your account on http://moo.sw.it.aoyama.ac.jp
- Enroll for Discrete Mathematics I (deadline: September 22)
- Solve the Quiz Simple Arithmetic in English (290 attempts, 116 complete (all problems solved))
- Review decimal number representation, binary number representation, and
`n`-ary number representation based on high-school notes and Web resources - Check out/buy/lend a textbook or reference book

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(Georges Ifrah: The Universal History of Numbers, John Wiley & Sons, 1998)

- Humans have used many different representations of numbers throughout history
- The first number represented was 1
- Representations such as |, ||, |||,... are most frequent
- For bigger numbers, using groups of 10 is most frequent

(20 (French 80: quatre-vingt=4-20) and 60 (minutes, seconds) also exist

- Chinese numerals: 一、二、三、亖
- Roman numerals: I, II, III, IIII (or IV)
- (Arabic-)Indic numerals: ١, ٢, ٣ (used in Arabic)
- (European-)Arabic numerals: 1, 2, 3

- Chinese numerals: 十、廿、...
- Roman numerals: X, XX,...

Peano Axioms (Guiseppe Peano, 1858-1932)

- 1 is a natural number

(1∈ℕ) - If
`a`is a natural number, then`s`(`a`) is a natural number (`s`(`a`) is the*successor*of`a`)

(`a`∈ℕ ⇒`s`(`a`)∈ℕ) - There is no natural number
`x`so that`s`(`x`) = 1 - If two natural numbers are different, then their successors are
different

(`a`∈ℕ,`b`∈ℕ,`a`≠`b`⇒`s`(`a`) ≠`s`(`b`)) - If we can prove a property for 1,

and we can prove, for any natural number`a`, that if`a`has this property then`s`(`a`) also has this property,

then all natural numbers have this property.

(Nowadays, it is usual to start with 0 rather than with 1.)

(We will learn how to express axioms 3 and 5 as formulæ in the lesson about Predicate Logic)

1 | 1 |

2 | s(1) |

3 | s(s(1)) |

4 | s(s(s(1))) |

5 | s(s(s(s(1)))) |

6 | s(s(s(s(s(1))))) |

7 | s(s(s(s(s(s(1)))))) |

... | ... |

Axioms of addition:

`a`+ 1 = s(`a`)- If
`a`and`b`are natural numbers,`a`+ s(`b`) = s(`a`+`b`)

(`a`∈ℕ,`b`∈ℕ ⇒`a`+ s(`b`) = s(`a`+`b`))

Calculate 2 + 3 using Peano arithmetic:

(associative property)

A binary operation (represented by operator △) is associative if and only
if for all operands `a`, `b`, and `c`:

(`a`△`b`) △ `c` = `a` △
(`b`△`c`)

Examples:

- Addition of (natural) numbers is associative.

((`a`+`b`) +`c`=`a`+ (`b`+`c`)) - Multiplication of (natural) numbers is associative.

((`a`·`b`) ·`c`=`a`· (`b`·`c`)) - Multiplication of matrices is associative.

Counterexamples:

- Subtraction of (natural) numbers ((
`a`-`b`) -`c`≠`a`- (`b`-`c`)) - Exponentiation
((
`a`^{b})^{c}≠`a`^{(bc)})

What we want to prove:

Associative law for addition: (`d` + `e`) + `f` =
`d` + (`e` + `f`)

Let's prove this for all values of `f`.

- Let's distinguish two cases:
`f`=1 and`f`=`k`+1 - If
`f`= 1, then (`d`+`e`) + 1 = s(`d`+`e`) [by the 1st axiom of addition, with (`d`+`e`) for`a`]

=`d`+ s(`e`) [by the 2nd axiom of addition, with`d`for`a`and`e`for`b`]

=`d`+ (`e`+ 1) [by the 1st axiom of addition, backwards, with`e`for`a`] - Assuming that associativity holds for
`f`=`k`(i.e. (`d`+`e`) +`k`=`d`+ (`e`+`k`)), let's prove associativity for`f`=`k`+1, i.e let's prove

(`d`+`e`) +`k`=`d`+ (`e`+`k`) ⇒^{}(`d`+`e`) + (`k`+1) =`d`+ (`e`+ (`k`+1)):(

`d`+`e`) + (`k`+1) = (`d`+`e`) + s(`k`) [by the 1st axiom of addition, with`k`for`a`]

= s((`d`+`e`) +`k`) [by the 2nd axiom of addition, with (`d`+`e`) for a and`k`for`b`]

= s(`d`+ (`e`+`k`)) [using the assumption]

=`d`+`s`(`e`+`k`) [by the 2nd axiom of addition, backwards, with`d`for a and (`e`+`k`) for b]

=`d`+ ((`e`+`k`) + 1) [by the 1st axiom of addition, with (`e`+`k`) for`a`]

=`d`+ (`e`+ (`k`+ 1)) [using the case`f`=1, with`e`for`d`and`k`for`e`]

Q.E.D. [using the 5th Peano axiom, with`f`for`a`and the property (`d`+`e`) +`f`=`d`+ (`e`+`f`)]

- We have to be careful that we are only using the axioms, not any 'general knowledge'
- Because we have not yet established associativity, we always have to use parentheses.
- Once we have proved associativity, we can eliminate the parentheses.
- This proof uses mathematical induction.

- Peano Axiom 5 can be seen as the basis for mathematical induction.
- We will look at mathematical induction more closely later.

- Mathematics tries to start with very few facts or rules
- These are usually called
*axioms* - The axioms should be self-evident
- Other facts and rules (theorems) are deduced from the axioms using proofs
- The less axioms and the more theorems, the better (from a mathematical viewpoint)

- The latest (natural/integer) number discovered by humans
- Discovered around 800 A.D. in India
- Discovery spread West to Arabia and Europe, Easts to China and Japan
- 0 is very important for positional notations such as decimal, binary,...

Exponentiation (e.g. 2^{3}):

Two raised to the power of three is eight.

Two to the power of three is eight.

Two to the three (third) is eight.

The third power of two is eight.

Five to the power of four is six hundred twenty-five.

Three raised to the power of four is eight-one.

Modulo operation (remainder):

Twenty modulo six is two.

Twenty-five modulo seven is four.

Number representations before positional notation:

Chinese (Han) numerals: 二百五十六、二千十六

Roman numerals: CCLVI, MMXVI

Example of decimal notation: 256 = 2·10^{2} + 5·10^{1} +
6·10^{0}

Example containing 0: 206 = 2·10^{2} + 0·10^{1} +
6·10^{0}

Generalization:
`d`_{n}...`d`_{1}`d`_{0}
= `d`_{n}·10^{n}+...+`d`_{1}·10^{1}+`d`_{0}·10^{0}

Example with decimal point: 34.56 = 3·10^{1} + 4·10^{0} +
5·10^{-1} + 6·10^{-2}

(the base of a number is often given as a subscript)

1010011_{2} = 1·2^{6} + 0·2^{5} + 1·2^{4}
+ 0·2^{3} + 0·2^{2} + 1·2^{1} + 1·2^{0}
=

1·64 + 0·32 + 1·16 + 0·8 + 0·4 + 1·2 + 1·1 =

1·64 + 1·16 + 1·2 + 1·1 =

64 + 16 + 2 + 1 =

83

Calculate the sum of each of the digits multiplied by its positional weight.

The positional weight is a power of `b`, the zeroth power for the
rightmost digit.

The power increases by one when moving one position to the left.

`d`_{n}...`d`_{1}`d`_{0}
(in base `b`) =
`d`_{n}·`b`^{n}+...+`d`_{1}·`b`^{1}+`d`_{0}·`b`^{0}

Take the number to convert as the first quotient.

Repeatedly:

- Take the quotient of the previous division
- Divide that quotient by the base
`b` - Add the remainder of the division as a digit to the left of the result

dividend | divisor | quotient | remainder | digits of the result |
---|---|---|---|---|

23← | ||||

23 | 2 | 11← | 1↑ |
1 |

11 | 2 | 5← | 1↑ |
11 |

5 | 2 | 2← | 1↑ |
111 |

2 | 2 | 1← | 0↑ |
0111 |

1 | 0 | 1↑ |
10111 |

23 divided by 2 is 11 remainder 1

11 divided by 2 is 5 remainder 1

5 divided by 2 is 2 remainder 1

2 divided by 2 is 1 remainder 0

1 divided by 2 is 0 remainder 1

23 = 11·2^{1} + 1·2^{0}

= 5·2^{2} + 1·2^{1} + 1·2^{0}

= 2·2^{3} + 1·2^{2} + 1·2^{1} + 1·2^{0}

= 1·2^{4} + 0·2^{3} + 1·2^{2} + 1·2^{1} +
1·2^{0} = 10111

Using Horner's rule: 23 = (((1×2^{} + 0)×2^{} +
1)×2^{} + 1)×2^{} + 1

- It is possible to start from the most significant digit
- When starting with
`a`, first find`n`so that`b`^{n+1}>`a`≧`b`^{n} - Divide by
`b`^{n}, then by`b`^{n-1}, and so on

dividend | divisor | quotient | remainder | digits of the result |
---|---|---|---|---|

23← | ||||

23 | 16 | 1↓ |
7← | 1 |

7 | 8 | 0↓ |
7← | 10 |

7 | 4 | 1↓ |
3← | 101 |

3 | 2 | 1↓ |
1← | 1011 |

1 | 1 | 1↓ |
0← | 10111 |

- General method: Convert via base 10

Example: base 3 → base 10 → base 5 - Shortcut 1: If
`b`is a power of`c`(or the other way round), then convert the digits in groups

Example 1: base 3 → base 9 (9 is 3^{2}, therefore make groups of two digits and convert to a single digit)

Example 2: base 8 → base 2 (8 is 2^{3}, therefore convert each digit to a group of three digits) - Shortcut 2: If both
`b`and`c`are powers of`d`, then convert via base`d`

Example: base 4 → base 8

because 4 = 2^{2}and 8 = 2^{3},`d`= 2

therefore, convert base 4 → base 2 → base 8 (use shortcut 1 two times)

Convert 47623 (base 8) to base 4.

8 = 2^{3}, 4 = 2^{2}, therefore convert base 8 → base 2
→ base 4

47623_{8} →

4 | 7 | 6 | 2 | 3 | base 8 |

100 | 111 | 110 | 010 | 011 | convert each base-8 digit to three base-2 digits |

100111110010011_{2}

1 | 00 | 11 | 11 | 10 | 01 | 00 | 11 | split base 2 into groups of two digits |

1 | 0 | 3 | 3 | 2 | 1 | 0 | 3 | convert two base-2 digits to one base-4 digit |

→ 10332103_{4}

1AF = 1×16^{2} + A×16^{1} + F×16^{0} = 1×256 + 10×16 + 15×1 = 256 + 160
+ 15 = 431

digit (upper case) | digit (lower case) | value (decimal) |

A | a | 10 |

B | b | 11 |

C | c | 12 |

D | d | 13 |

E | e | 14 |

F | f | 15 |

base | name (adjective) and abbreviation | (reason for) use | constants in programming languages |
---|---|---|---|

2 | binary, bin | used widely in logic and circuits (hardware) | `0b101100` (Ruby,...) |

8 | octal, oct | shortened form of binary (rare these days) | `024570` (C and many others) |

10 | decimal, dec | for humans | `1234567` (all languages) |

16 | hexadecimal, hex | shortened form of binary, 1 byte (8bits) can be represented with two digits | `0xA3b5` (C and many others) |

10 | 2 | 16 |
---|---|---|

0 | 0000 | 0 |

1 | 0001 | 1 |

2 | 0010 | 2 |

3 | 0011 | 3 |

4 | 0100 | 4 |

5 | 0101 | 5 |

6 | 0110 | 6 |

7 | 0111 | 7 |

8 | 1000 | 8 |

9 | 1001 | 9 |

10 | 1010 | A |

11 | 1011 | B |

12 | 1100 | C |

13 | 1101 | D |

14 | 1110 | E |

15 | 1111 | F |

16 | 10000 | 10 |

n |
2^{n} |
in base 16 |
---|---|---|

0 | 1 | 1 |

1 | 2 | 2 |

2 | 4 | 4 |

3 | 8 | 8 |

4 | 16 | 10 |

5 | 32 | 20 |

6 | 64 | 40 |

7 | 128 | 80 |

8 | 256 | 100 |

9 | 512 | 200 |

10 | 1'024 ≈10^{3} (kilo) |
400 |

11 | 2'048 (the game) | 800 |

12 | 4'096 | 1000 |

16 | 65'536 | 1'0000 |

20 | 1'048'576 ≈ 10^{6} (mega) |
10'0000 |

30 | 1'073'741'824 ≈ 10^{9} (giga) |
4000'0000 |

40 | 1'099'511'627'776 ≈ 10^{12} (tera) |
100'0000'0000 |

Question: Why do computer scientist always think Christmas and Halloween are
the same?

(Hint: In the USA, Halloween is October 31st only)

Question: At what age do Information Technologists celebrate "Kanreki" (還暦)

- Solve Arithmetic
and Base Conversion (repeat until you get it 100% correct; deadline
**September 29, 22:00**) - Learn binary and hexadecimal numbers up to 16, and powers of 2 up to
2
^{12} - Try to find an answer to the joke questions (no need to submit)
- Use highschool texts or the Web to refresh your knowledge about propositions, logic, and functions

- Moodle で Arithmetic
and Base Conversion のクイズを解く
(満点まで繰り返す、
**締切: 9月29日 (木) 22:00**) - 冗談の問題の解答を考える (提出不要)
- 数学の「命題」、「論理演算」、「関数」について高校で学んだことを再確認し、ウェブで調べる

- number
- 数
- numeral
- 数字
- natural number
- 自然数
- discovery
- 発見
- origin
- 原点
- positional notation
- 位取り表現
- perfect score
- 満点
- confusion
- 混乱
- representation
- 表現
- exponentiation
- べき乗演算
- Modulo operation
- モジュロ演算
- remainder
- 剰余 (余り)
- decimal notation (decimal numeral system)
- 十進法
- Chinese numerals
- 漢数字
- Roman numerals
- ローマ数字
- discovery
- 発見
- axiom
- 公理
- Peano axioms
- ペアノの公理
- successor
- 後者
- formula (plural: formulæ)
- 式
- property
- 性質
- arithmetic
- 算術
- associative law (property)
- 結合律
- counterexample
- 反例
- operation
- 演算
- operator
- 演算子
- operand
- 被演算子
- binary operation
- 二項演算
- proof
- 証明
- prove
- 証明する
- Q.E.D. (quod erat demonstrandum)
- 証明終了
- parenthesis
- (丸・小) 括弧、複数 parentheses
- mathematical induction
- 数学的帰納法
- self-evident
- 自明
- remainder
- 余り、剰余
- subscript
- 下付き文字 (添え字)
- generalization
- 一般化
- decimal point
- 小数点
- base
- 基数
- base conversion
- 奇数変換
- positional weight
- (その桁の) 重み
- dividend (or numerator)
- 被除数、実 (株の配当という意味も)
- divisor (or denominator, modulus)
- 除数、法
- quotient
- 商 (割り算の結果)
- Horner's rule
- ホーナー法
- digit
- 数字
- shortcut
- 近道
- upper case
- 大文字
- lower case
- 小文字
- binary
- 二進数 (形容詞)
- octal
- 八進数 (形容詞)
- decimal
- 十進数 (形容詞)
- hexadecimal
- 十六進数 (形容詞)
- circuit
- 回路
- constant
- 定数
- joke
- 冗談
- submit
- 提出する
- proposition
- 命題
- function
- 関数