MAT246H1-S – LEC0201/9201 Concepts in Abstract Mathematics

POSITIONAL NUMERAL SYSTEM From the appendix of Chapter 4

February 9th, 2021

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Feb 9, 2021 1 / 9 Introduction

In our everyday life, we usually use a base ten positional notation (decimal numeral system).

It allows us to write all natural numbers using only 10 digits although ℕ is infinite. The idea is that the position of a digit changes its value:

590743 = 5 × 105 + 9 × 104 + 0 × 103 + 7 × 102 + 4 × 101 + 3 × 100

But it is not the only base that we can use: • Base 2 (binary) and base 16 (hexadecimal) are quite common nowadays in computer sciences. • The first known positional numeral system is the Babylonian one using abase60 (sexagesimal), circa 2000BC. • In our everyday life, we can observe the influence of bases 60 (1 hour is 60 minutes) or20 (in French 96 is litteraly pronounced 4 × 20 + 16).

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Feb 9, 2021 2 / 9 Positional numeral system with base 푏 – 1

Theorem Let 푏 ≥ 2 be an natural number. Then any natural number 푛 ∈ ℕ admits a unique expression

푛 = 푎 푏푘 ∑ 푘 푘≥0

where 푎푘 ∈ {0, 1, … , 푏 − 1} and 푎푘 = 0 for all but finitely many 푘 ≥ 0.

푟 We write 푎 푎 … 푎 푎 푏 for 푎 푏푘. 푟 푟−1 1 0 ∑ 푘 푘=0

Question 2 of PS1 was about the existence of the binary numeral system (i.e. when 푏 = 2).

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Feb 9, 2021 3 / 9 Positional numeral system with base 푏 – 2

Remark In order to pass from a base 10 expression to a base 푏 expression, we can perform successive Euclidean divisions by 푏.

Example: from base 10 to base 2

42 = 2 × 21 + 0 = 2 × (2 × 10 + 1) + 0 = 2 × (2 × (2 × 5 + 0) + 1) + 0 = 2 × (2 × (2 × (2 × 2 + 1) + 0) + 1) + 0 = 2 × (2 × (2 × (2 × (2 × 1 + 0) + 1) + 0) + 1) + 0 = 1 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20

10 2 Hence 42 = 101010 .

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Feb 9, 2021 4 / 9 Positional numeral system with base 푏 – 3

Proof. Existence. Let 푏 ≥ 2. We are going to prove by strong induction that for 푛 ≥ 0, there exist 푎 ∈ {0, 1, … , 푏 − 1} such that 푎 = 0 for all but finitely many 푘 ≥ 0 and 푛 = 푎 푏푘. 푘 푘 ∑ 푘 푘≥0 • Base case at 푛 = 0: 0 = 0푏푘. ∑ 푘≥0 • Induction step. Assume that 0, 1, … , 푛 admit an expression in base 푏, for some 푛 ≥ 0. By Euclidean division, 푛 + 1 = 푏푞 + 푟 where 푞, 푟 ∈ ℕ satisfy 0 ≤ 푟 < 푏. Note that if 푞 ≠ 0 then 푞 < 푏푞 ≤ 푏푞 + 푟 = 푛 + 1. Thus 0 ≤ 푞 ≤ 푛. Therefore, by the induction hypothesis, 푞 = 푎 푏푘 ∑ 푘 푘≥0

where 푎푘 ∈ {0, 1, … , 푏 − 1} and 푎푘 = 0 for all but finitely many 푘 ≥ 0. Hence, 푛 + 1 = 푏푞 + 푟 = 푎 푏푘+1 + 푟푏0 ∑ 푘 푘≥0 which ends the induction step.

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Feb 9, 2021 5 / 9 Positional numeral system with base 푏 – 3

Proof. Uniqueness. Assume that 푎 푏푘 = 푎′ 푏푘 where 푎 , 푎′ ∈ {0, 1, … , 푏 − 1} are zero for all but finitely many. ∑ 푘 ∑ 푘 푘 푘 푘≥0 푘≥0 ′ Assume by contradiction there exists 푘 ≥ 0 such that 푎푘 ≠ 푎푘. ′ Since {푘 ∈ ℕ ∶ 푎푘 ≠ 푎푘} is finite and non-empty, it admits a greatest element ℓ. ′ WLOG, we may assume that 푎ℓ < 푎 . Then ℓ ℓ 0 = 푎 푏푘 − 푎′ 푏푘 = (푎 − 푎′ )푏푘 = (푎 − 푎′ )푏푘 ∑ 푘 ∑ 푘 ∑ 푘 푘 ∑ 푘 푘 푘≥0 푘≥0 푘≥0 푘=0

ℓ−1 (푎′ − 푎 )푏ℓ = (푎 − 푎′ )푏푘 So that ℓ ℓ ∑ 푘 푘 푘=0

ℓ−1 ℓ−1 (푎′ − 푎 )푏ℓ ≤ |푎 − 푎′ |푏푘 ≤ (푏 − 1)푏푘 = 푏ℓ − 1 < 푏ℓ ≤ (푎′ − 푎 )푏ℓ Therefore ℓ ℓ ∑ 푘 푘 ∑ ℓ ℓ 푘=0 푘=0 Hence a contradiction. ■ Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Feb 9, 2021 6 / 9 Babylonian cuneiform numerals (circa 2000BC)

The first known positional numeral system is the Babylonian one (circa 2000BC) whose baseis 60 and whose digits are: 0 10 ⌋ 20 ⌋⌋ 30 ⌍ 40 ℭ 50 ℴ 1 ⁹ 11 ⌋⁹ 21 ⌋⌋⁹ 31 ⌍⁹ 41 ℭ⁹ 51 ℴ⁹ 2 ∫ 12 ⌋∫ 22 ⌋⌋∫ 32 ⌍∫ 42 ℭ∫ 52 ℴ∫ 3 ␈ 13 ⌋␈ 23 ⌋⌋␈ 33 ⌍␈ 43 ℭ␈ 53 ℴ␈ 4 ⃻ 14 ⌋⃻ 24 ⌋⌋⃻ 34 ⌍⃻ 44 ℭ⃻ 54 ℴ⃻ 5 ␊ 15 ⌋␊ 25 ⌋⌋␊ 35 ⌍␊ 45 ℭ␊ 55 ℴ␊ 6 ␋ 16 ⌋␋ 26 ⌋⌋␋ 36 ⌍␋ 46 ℭ␋ 56 ℴ␋ 7 ␌ 17 ⌋␌ 27 ⌋⌋␌ 37 ⌍␌ 47 ℭ␌ 57 ℴ␌ 8 ␍ 18 ⌋␍ 28 ⌋⌋␍ 38 ⌍␍ 48 ℭ␍ 58 ℴ␍ 9 ␎ 19 ⌋␎ 29 ⌋⌋␎ 39 ⌍␎ 49 ℭ␎ 59 ℴ␎

We want to write 13655 using Babylonian cuneiform numerals. We perform successive Euclidean divisions by 60 as follows: 13655 = 60 × 227 + 35 = 60 × (60 × 3 + 47) + 35 = 3 × 602 + 47 × 601 + 35 × 600. Hence it was written: ␈ ℭ␌ ⌍␊

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Feb 9, 2021 7 / 9 YBC 7289, clay tablet, between 1800BC and 1600BC.

Original picture from https://commons.wikimedia.org/wiki/File:YBC-7289-OBV-REV.jpg

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Feb 9, 2021 8 / 9 YBC 7289, clay tablet, between 1800BC and 1600BC.

It shows (extremely accurate) approximations of

24 51 10 √2 ≃ 1 + + + 60 602 603 and of 25 35 30√2 ≃ 42 + + 60 602 (diagonal of the square of side length 30, see the ⌍ above de square)

Original picture from https://commons.wikimedia.org/wiki/File:YBC-7289-OBV-REV.jpg

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