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Unary, Binary, and Beyond. X n – 1 . 1 + X 1 + X 2 + X 3 + … + X n-2 + X n-1 =. X - 1. We will soon need this fundamental sum: The Geometric Series. A Frequently Arising Calculation. ( X - 1 ) ( 1 + X 1 + X 2 + X 3 + … + X n-2 + X n-1 )

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slide2

Xn – 1

1 + X1 + X2 + X 3 + … + Xn-2 + Xn-1 =

X- 1

We will soon need this fundamental sum:

The Geometric Series

a frequently arising calculation
A Frequently Arising Calculation
  • (X-1) ( 1 + X1 + X2 + X 3 + … + Xn-2 + Xn-1 )
  • = X1 + X2 + X 3 + … + Xn-1 + Xn
  • - 1 - X1 - X2 - X 3 - … - Xn-2 – Xn-1
  • = - 1 + Xn
  • = Xn - 1
the geometric series
The Geometric Series
  • (X-1) ( 1 + X1 + X2 + X 3 + … + Xn-2 + Xn-1 ) = Xn - 1

Xn – 1

1 + X1 + X2 + X 3 + … + Xn-2 + Xn-1 =

X- 1

when X1

n th triangular number
nth Triangular Number
  • n = 1 + 2 + 3 + . . . + n-1 + n
  • = n(n+1)/2
n th square number
nth Square Number
  • n = 1 + 3 + … + 2n-1
  • = Sum of first n odd numbers
n th square number8
nth Square Number
  • n = n + n-1
  • = n2
slide9

(n)2 = + + . . . +

n

(n)2 =(n-1)2 +

n

slide10

Today:

we will define sequences of symbols that give us a unary representation of the Natural numbers.

strings of symbols
Strings Of Symbols.
  • We take the idea of symbol and sequence of symbols as primitive.
  • Let  be any fixed finite set of symbols.
  •  is called an alphabet, or aset of symbols.
  • Examples:  = {0,1,2,3,4}
  • = {a,b,c,d, …, z}
  • = all typewriter symbols.
  •  = { a , b , c , d , …, z }
strings over the alphabet
Strings over the alphabet .
  • A string is a sequence of symbols from .
  • Let s and t be strings. Then st denotes the concatenation of s and t; i.e., the string obtained by the string s followed by the string t.
  • Now define + by these inductive rules:
  • x 2 ) x 2+
  • s,t 2+) st 2+
slide14

Intuitively, + is the set of all finite strings that we can make using (at least one) letters from .

the set
The set *
  • Define  be the empty string.
  • I.e., XY= XY for all strings X and Y.
  • is also called the string of length 0.

Define 0 = {  }

  • Define * = +[ {}
slide16

Intuitively, * is the set of all finite strings that we can make using letters from , including the empty string.

the natural numbers
The Natural Numbers
  •  = { 0, 1, 2, 3, . . .}

Notice that we include 0 as a Natural number.

god made the natural numbers everything else is the work of man
“God Made The Natural Numbers. Everything Else Is The Work Of Man.”

Kronecker

  •  = { 0, 1, 2, 3, . . .}
peano unary pu
Peano Unary (PU)
  • 0 0
  • 1 S0
  • 2 SS0
  • 3 SSS0
  • 4 SSSS0
  • 5 SSSSS0
  • 6 SSSSSS0

Giuseppe Peano [1889]

each number is a sequence of symbols in s 0
Each number is a sequence of symbols in {S, 0}+
  • 0 0
  • 1 S0
  • 2 SS0
  • 3 SSS0
  • 4 SSSS0
  • 5 SSSSS0
  • 6 SSSSSS0
peano unary representation of natural number
Peano Unary Representation of Natural Number
  •  = { 0, S0, SS0, SSS0, . . .}

0 is a natural number called zero.

Set notation: 02

If X is a natural number, then SX is a natural number called successor of X.

Set notation: X2)SX2

inductive definition of
Inductive Definition of “+”
  •  = { 0, S0, SS0, SSS0, . . .}

Inductive definition of addition (+):

X, Y2)

X + 0 = X

X + SY = S(X + Y)

proving 1 1 2
Proving “1+1=2”
  • Claim: S0 + S0 = SS0
  • Proof:
  • S0 + S0 = S(S0 + 0) =
  • S(S0) =
  • SS0

X, Y 2  )

X + 0 = X

X + SY = S(X + Y)

inductive definition of26
Inductive Definition of “*”
  •  = { 0, S0, SS0, SSS0, . . .}

Inductive definition of times (*):

X, Y 2  )

X * 0 = 0

X * SY = (X * Y) + X

inductive definition of27
Inductive Definition of “^”
  •  = { 0, S0, SS0, SSS0, . . .}

Inductive definition of raised to the (^):

X, Y 2  )

X ^ 0 = 1 (“X0 = 1”)

X ^ SY = (X ^ Y) * X (“XSY = XY * X”)

0 s0 ss0 sss0
 = { 0, S0, SS0, SSS0, . . .}
  • Defining < for :
  • 8x,y2
  • “x > y” is TRUE  “y < x” is FALSE
  • “x > y” is TRUE  “y > x” is FALSE
  • “x+1 > 0” is TRUE
  • “x+1 > y+1” is TRUE ) “x > y” is TRUE
slide29

Shortcut:

We will write 2 or 7 instead of SS0 or SSSSSSS0.

Red numbers are Peano numbers.

0 1 2 3
 = { 0, 1, 2, 3, . . .}
  • Defining a partial minus for :
  • 8x,y2:
  • x - 0 = x
  • x > y)
  • (x+1) – (y+1) = x - y
a a div b b a mod b
a = [a DIV b]*b + [a MOD b]
  • Defining DIV and MOD for :
  • 8 a,b 2 
  • a<b )
  • a DIV b = 0
  • a¸b>0 )
  • a DIV b = 1 + (a-b) DIV b
  • a MOD b = a – [b*(a DIV b)]

The maximum number of times b goes into a without going over.

The remainder when a is divided by b.

45 45 div 10 10 45 mod 10 4 10 5
45 = [45 DIV 10]*10 + [ 45 MOD 10]= 4*10 + 5
  • Defining DIV and MOD for :
  • 8 a,b 2 
  • a<b )
  • a DIV b = 0
  • a¸b>0 )
  • a DIV b = 1 + (a-b) DIV b
  • a MOD b = a – [b*(a DIV b)]
slide33

We have defined the Peano representation of the Naturals, with a notion of +, *, ^, <, -, DIV, and MOD.

slide34

PU takes size n+1 to represent the number n.

Higher bases are much more compact.

slide35

1000000 in Peano Unary takes one million one symbols to write.

1000000 only takes 7 symbols to write in base 10.

slide37

Let = {0, 1,2,3,4,5,6,7,8,9} be our symbol alphabet.

Any string in + will be called a decimal number.

slide38

Let X be a decimal number. Let’s define the length of X inductively:

length() = 0

X= aY, a2, Y2* ) length(X) = S(length(Y))

slide39

Let X be decimal. Let n (unary) = length(X).

For each unary i<n, we want to be able to talk about the ith symbol of X.

slide40

The ith symbol of X.

Defining rule:

X = PaS where P,S2* and a2. i=length(S).)a is the ith symbol of X.

slide41

For any string X, we can define its length n2 PU. For all i2 PU s.t. i<n, we can define the ith symbol ai.

X = an-1 an-2 …a0

slide42

We define a base conversion function, Base10 to 1 (X), to convert any decimal X to its unary representation.

initial cases length x 1
Initial Cases, length(X)=1
  • Base10 to 1 (0) = 0
  • Base10 to 1 (1) = S0
  • Base10 to 1 (2) = SS0
  • Base10 to 1 (3) = SSS0
  • ……
  • Base10 to 1 (9) = SSSSSSSSS0
suppose n length x s0
Suppose n=length(X)>S0
  • For all i<n, let ai be the ith symbol of X. Hence, X = an-1 an-2 …a0 .
  • Define Base10 to 1(X) =
  • i<n Base10 to 1(ai)*10i
  • where +, *, and ^ are defined over PU
example x 238
Example X= 238
  • Base10 to 1 (238) =
  • 2*100 + 3*10 + 8
slide46

Base10 to 1 (an-1 an-2 … a0) =

an-1*10n-1 + an-2 * 10n-2 +…+ a0 100

slide47

Now we want to go back the other way.

We want to convert PU to decimal.

slide48

No Leading Zero:

Let NLZ be the set of decimal numbers with no leading 0 (leftmost symbol  0), or the decimal number 0.

slide49

We define Base1 to 10 from PU to NLZ

It will turn out to be the inverse function to Base10 to 1.

one digit cases
One digit cases.
  • Base1 to 10(0) = “0”
  • Base1 to 10(S0) = “1”
  • Base1 to 10(SSo) =“2”
  • ….
  • Base1 to 10(SSSSSSSSS0) = “9”
suppose x 9
Suppose X > 9
  • Let n be the smallest unary number such that 10n > X, 10n-1· X.
  • Let d = X DIV 10n-1 [Notice that 1· d · 9]
  • Let Y = X MOD 10n-1
  • Define Base1 to 10(X) 2 NLZ to be
  • Base1 to 10 (d) Base1 to 10(Y)
slide52

For each n2 PU define its decimal representation to beBase1 to 10 (n).

Base1 to 10 goes from PU to NLZ.

slide53

Let’s verify that Base1 to 10 and Base10 to 1 really are inverse functions.

We need to show X2 NLZ )

Base1 to 10 (Base10 to 1 (X)) = X.

clear when x is a single digit
Clear when X is a single digit.
  • Base1 to 10 (Base10 to 1 (0) ) = 0
  • Base1 to 10 (Base10 to 1 (1) ) = 1
  • Base1 to 10 (Base10 to 1 (2) ) = 2
  • Base1 to 10 (Base10 to 1 (3) ) = 3
  • Base1 to 10 (Base10 to 1 (4) ) = 4
  • Base1 to 10 (Base10 to 1 (5) ) = 5
  • Base1 to 10 (Base10 to 1 (6) ) = 6
  • Base1 to 10 (Base10 to 1 (7) ) = 7
  • Base1 to 10 (Base10 to 1 (8) ) = 8
  • Base1 to 10 (Base10 to 1 (9) ) = 9
slide55

Let X be a shortest counter-example to the statement:

X2 NLZ )

Base1 to 10 (Base10 to 1(Z)) =Z

slide56

For all Y shorter than X:

Y2 NLZ )

Base1 to 10 (Base10 to 1(Y)) = Y

x 2 nlz n length z 1
X2 NLZ, n=length(Z) > 1
  • Suppose X=an-1 an-2 … a0
  • Base10 to 1(X) = Y =
  • i<n Base10 to 1(ai)*10i
  • Base1 to 10 (Y) = ?
  • n is smallest PU s.t. 10n-1· Y < 10n
  • Calculate d = Y DIV 10n-1, Z= Y MOD 10n-1
  • d= an-1, Z= i<n-1 Base10 to 1(ai)*10i
  • Output an-1 Base1 to 10 (Z) [Z shorter than X]
  • = an-1 an-2 … a1 a0
  • Contradiction of X being a counter-example.
slide58

No counter-example means that

For all Y2 NLZ,

Base1 to 10 (Base10 to 1 (Y)) = Y

slide59

So we can move back and forth between representing a number in PU or in NLZ.

Each string in PU corresponds to one and only one string in NLZ.

base x notation
Base X Notation
  • Let  be an alphabet of size X. A base X digit is any element of .
  • Let S = an-1, an-2, …, a0 be a sequence of base X digits.
  • Let BaseX to 1(S) = an-1 Xn-1 + … a2X2 + a1X + a0
  • S is called the base X representation of the number BaseX to 1(S).
s a n 1 a n 2 a 0 represents the number a n 1 x n 1 a n 2 x n 2 a 0 x 0
S = an-1, an-2, …, a0 represents the number: an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0
  • Base 2 [Binary Notation]
  • 101 represents 1 (2)2 + 0 (21) + 1 (20)
  • Base 7
  • 015 represents 0 (7)2 + 1 (71) + 5 (70)
s a n 1 a n 2 a 0 represents the number a n 1 x n 1 a n 2 x n 2 a 0 x 062
S = an-1, an-2, …, a0 represents the number: an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0
  • Base 2 [Binary Notation]
  • 101 represents 1 (2)2 + 0 (21) + 1 (20)
  • =
  • Base 7
  • 015 represents 0 (7)2 + 1 (71) + 5 (70)
  • =
bases in different cultures
Bases In Different Cultures
  • Sumerian-Babylonian: 10, 60, 360
  • Egyptians: 3, 7, 10, 60
  • Africans: 5, 10
  • French: 10, 20
  • English: 10, 12,20
biggest n digit number in base x would be x 1 x n 1 x 1 x n 2 x 1 x 0
Biggest n “digit” number in base X would be:(X-1)Xn-1 + (X-1)Xn-2 +…+ (X-1)X0
  • Base 2
  • 111 represents 1 (2)2 + 1 (21) + 1 (20)
  • Base 7
  • 666 represents 6 (7)2 + 6 (71) + 6 (70)
biggest n digit number in base x would be x 1 x n 1 x 1 x n 2 x 1 x 065
Biggest n “digit” number in base X would be:(X-1)Xn-1 + (X-1)Xn-2 +…+ (X-1)X0
  • Base 2
  • 111 represents 1 (2)2 + 1 (21) + 1 (20)
  • 111 + 1 = 1000 represents 23
  • Thus, 111 represents 23 – 1
  • Base 7
  • 666 represents 6 (7)2 + 6 (71) + 6 (70)
  • 666 + 1 = 1000 represents 73
  • Thus, 666 represents 73 - 1
slide66

Biggest n “digit” number in base X would be:S= (X-1)Xn-1 + (X-1)Xn-2 +…+ (X-1)X0\Add 1 to get: Xn + 0 Xn-1 + … + 0 X0Thus, S = Xn - 1

  • Base 2
  • 111 represents 1 (2)2 + 1 (21) + 1 (20)
  • 111 + 1 = 1000 represents 23
  • Thus, 111 represents 23 – 1
  • Base 7
  • 666 represents 6 (7)2 + 6 (71) + 6 (70)
  • 666 + 1 = 1000 represents 73
  • Thus, 666 represents 73 - 1
slide67

Xn – 1

1 + X1 + X2 + X 3 + … + Xn-2 + Xn-1 =

X- 1

Recall the

Geometric Series…

x 1 x n 1 x 1 x n 2 x 1 x 0 x n 1
(X-1)Xn-1 + (X-1)Xn-2 +…+ (X-1)X0 = Xn - 1
  • Proof:
  • Factoring out (X-1), we obtain:
  • (X-1) [Xn-1 + Xn-2 + ….. + X + 1] =
  • (X-1) [(Xn – 1)/(X-1)] =
  • Xn – 1
the highest n digit number in base x
The highest n digit number in base X.
  • (X-1) ( 1 + X1 + X2 + X3 + … + Xn-1 ) = Xn - 1

Base X. Let S be the sequence of n digits each of which is X-1. S is the largest number of length n that can be represented in base X.

S = Xn - 1

slide70

Each of the numbers from 0 to Xn-1 is uniquely represented by an n-digit number in base X.

We could prove this using our previous method, but let’s do it another way.

plus minus base x
Plus/Minus Base X
  • An plus/minusbase X digit is any integer –X < a < X
  • Let S = an-1, an-2, …, a0 be a sequence of plus/minus base X digits. S is said to be the plus/minus base X representation of the number:
  • an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0
slide74

No. Consider plus/minus binary.

5 = 0 1 0 1 = 1 0 -1 -1

slide75

Humm..

So what is it good for?

0 has a unique n digit plus minus base x representation as all 0 s
0 has a unique n-digit plus/minus base X representation as all 0’s.
  • Suppose an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0 = S =0, where there is some highest k such that ak 0. Wl.o.g. assume ak > 0.
  • Because Xk > (X-1)(1 + X1 + X2 + X 3 + … + Xk-1)
  • no sequence of digits from ak-1 to a0 can represent a number as small as –Xk
  • Hence S  0. Contradiction.
each of the numbers from 0 to x n 1 is uniquely represented by an n digit number in base x
Each of the numbers from 0 to Xn-1 is uniquely represented by an n-digit number in base X.
  • We already know that n–digits will represent something between 0 and Xn – 1.
  • Suppose two distinct sequences represent the same number:
  • an-1 Xn-1 + an-2 Xn-2 + . . . + a0 X0 =
  • bn-1 Xn-1 + bn-2 Xn-2 + . . . + b0 X0
  • The difference of the two would be an plus/minus base X representation of 0, but it would have a non-zero digit. Contradiction.
each of the numbers from 0 to x n 1 is uniquely represented by an n digit number in base x79
Each of the numbers from 0 to Xn-1 is uniquely represented by an n-digit number in base X.
  • n digits represent up to Xn – 1
  • n-1 digits represents up to Xn-1 - 1
  • Let k be a number: Xn-1 ≤ k ≤ Xn - 1
  • So k can be represented with n digits.
  • For all k:  logxk  = n – 1
  • So k uses  logxk  + 1 digits.
slide80
Fundamental Theorem For Base X:Each of the numbers from 0 to Xn-1 is uniquely represented by an n-digit number in base X.
  • k uses  logxk  + 1 digits in base X.
slide81

n has length n in unary, but has length  log2n  + 1 in binary.

Unary is exponentially longer than binary.

egyptian multiplication
Egyptian Multiplication

The Egyptians used decimal numbers but multiplied and divided in binary

egyptian multiplication a times b by repeated doubling
Egyptian Multiplication a times bby repeated doubling
  • b has some n-bit representation: bn..b0
  • Starting with a, repeatedly double largest so far to obtain: a, 2a, 4a, …., 2na
  • Sum together all 2ka where bk = 1
egyptian multiplication 15 times 5 by repeated doubling
Egyptian Multiplication 15 times 5by repeated doubling
  • 5 has some 3-bit representation: 101
  • Starting with 15, repeatedly double largest so far to obtain: 15, 30, 60
  • Sum together all 2k(15) where bk = 1:
  • 15 + 60 = 75
why does that work
Why does that work?
  • b = b020 + b121 + b222 + … + bn2n
  • ab = b020a + b121a + b222a + … + bn2na
  • If bk is 1 then 2ka is in the sum.
  • Otherwise that term will be 0.
egyptian base conversion
Egyptian Base Conversion
  • Output stream will print right to left.
  • Input X.
  • Repeat until X=0
  • {
  • If X is even then Output O;
  • Otherwise {X:=X-1; Output 1}
  • X:=X/2
  • }
egyptian base conversion89
Egyptian Base Conversion
  • Output stream will print right to left.
  • Input X.
  • Repeat until X=0
  • {
  • If X is even then Output O;
  • Otherwise Output 1
  • X:= X/2
  • }
start the algorithm
Start the algorithm

010101

  • Repeat until X=0
  • { If X is even then Output O; Otherwise Output 1;
  • X:= X/2
  • }

1

start the algorithm91
Start the algorithm

01010

  • Repeat until X=0
  • { If X is even then Output O; Otherwise Output 1;
  • X:= X/2
  • }

1

start the algorithm92
Start the algorithm

01010

  • Repeat until X=0
  • { If X is even then Output O; Otherwise Output 1;
  • X:= X/2
  • }

01

start the algorithm93
Start the algorithm

0101

  • Repeat until X=0
  • { If X is even then Output O; Otherwise Output 1;
  • X:= X/2
  • }

01

start the algorithm94
Start the algorithm

0101

  • Repeat until X=0
  • { If X is even then Output O; Otherwise Output 1;
  • X:= X/2
  • }

101

start the algorithm95
Start the algorithm

010

  • Repeat until X=0
  • { If X is even then Output O; Otherwise Output 1;
  • X:= X/2
  • }

101

and keep going until 0
And Keep Going until 0

0

  • Repeat until X=0
  • { If X is even then Output O; Otherwise Output 1;
  • X:= X/2
  • }

010101

slide97

Sometimes the Egyptian combined the base conversion by halving and the multiplication by doubling into one algorithm

rhind papyrus 1650 bc 70 13
70

140

280

560

* 70

6

* 350

1 * 910

Rhind Papyrus (1650 BC)70*13
rhind papyrus 1650 bc 70 1399
70

140

280

560

* 70

6

* 350

1 * 910

Rhind Papyrus (1650 BC)70*13

Binary for 13 is 1101 = 23 + 22 + 20

70*13 = 70*23 + 70*22 + 70*20

rhind papyrus 1650 bc
17

34

68

136

1

2 *

4

8 *

Rhind Papyrus (1650 BC)

184 48 14

rhind papyrus 1650 bc101
17

34

68

136

1

2 *

4

8 *

Rhind Papyrus (1650 BC)

184 48 14

184 = 17*8 + 17*2 + 14

184/17 = 10 with remainder 14

slide102

This method is called “Egyptian Multiplication/Division” or “Russian Peasant Multiplication/Division”.

standard binary multiplication egyptian multiplication
Standard Binary Multiplication= Egyptian Multiplication

* * * * * * * *

X

101

* * * * * * * *

* * * * * * * *

* * * * * * * * * * *

egyptian base 3
Egyptian Base 3
  • We have defined
  • Base 3: Each digit can be 0, 1, or 2
  • Plus/Minus Base 3 uses -2, -1, 0, 1, 2
  • Here is a new one:
  • Egyptian Base 3 uses -1, 0, 1
  • Example: 1 -1 -1 = 9 - 3 - 1 = 5
slide106

Unique Representation Theorem for Egyptian Base 3No integer has 2 distinct, n-digit, Egyptian base-3 representations. We can represent all integers from -(3n-1)/2 to (3n-1)/2

  • Proof; If so, their difference would be a non-trivial plus/minus base 3 representation of 0. Contradiction.
  • Highest number = 1111…1 = (3n-1)/2
  • Lowest number = -1-1-1-1…-1 = -(3n-1)/2
slide107

Unique Representation Theorem for Egyptian Base 3No integer has 2 distinct, n-digit, Egyptian base-3 representations. We can represent all integers from -(3n-1)/2 to (3n-1)/2

  • 3n, n-digit, base 3 representations of the numbers from 0 to 3n – 1
  • Subtract 111111…111 = (3n – 1)/2 from each to get an Egyptian base 3 representation of all the numbers from -(3n-1)/2 to (3n-1)/2.
slide108

How could this be Egyptian? Historically, negative numbers first appear in the writings of the Hindu mathematician Brahmagupta (628 AD).

references
References
  • The Book Of Numbers, by J. Conway and R. Guy
  • History of Mathematics, Histories of Problems, by The Inter-IREM Commission