Egyptian Fractions · 2018. 7. 3. · Egyptian Fractions Mario Antoniazzi Gaia Barella Paolo...

Egyptian Fractions

Mario Antoniazzi Gaia Barella Paolo Barisan Marco CarolloMarco Casagrande Alex Dal Cin Giusi Grotto Gioella Lorenzon

Antonio Lucchetta Klara Metaliu Azzurra PizzolottoSimona Sota Mattia Toffolon Marina Vitali

ISISS M. Casagrande

March 28, 2018 - Padova, Italy

Archimede Project (ISISS M. Casagrande) Egyptian Fractions March 28, 2018 - Padova 0 / 29

Problem

Egyptians used only fractions like1

n, with n a non-zero natural number;

such fraction will be called “unit fraction” (U.F.).

We tried to write irreducible proper fractions likea

b, with a,b ∈ N0, as a

sum of distinct unit fractions. This expansion will be called “Egyptianfraction”.

Example:

3

7=

1

4+

1

7+

1

28


Problem

Egyptians used only fractions like1

n, with n a non-zero natural number;

such fraction will be called “unit fraction” (U.F.).

We tried to write irreducible proper fractions likea

b, with a,b ∈ N0, as a

sum of distinct unit fractions. This expansion will be called “Egyptianfraction”.

Example:

3

7=

1

4+

1

7+

1

28


First questions:

Can we write every proper fraction as a sum of unit fractions?

Are there different methods of expanding a proper fraction?

Are there different types of expansions?

Are there infinite expansions for each proper fraction?

Is there a best expansion? If so, would it be the shortest one or theone that leads to a sum with the lowest denominators?


First questions:







First questions:







First questions:







First questions:







First questions:







Our ways:

1 Fibonacci’s Method

2 Golomb’s method

3 Method of practical numbers

4 Geometric method (our solution)

5 Comparison

6 The tree of fractions


Our ways:


2 Golomb’s method



5 Comparison



Our ways:


2 Golomb’s method



5 Comparison



Our ways:


2 Golomb’s method



5 Comparison



Our ways:


2 Golomb’s method



5 Comparison



Our ways:


2 Golomb’s method



5 Comparison



Fibonacci’s Method

Fibonacci’s algorithm

Subtracting the largest possible unit fraction that keeps a non-negative

difference from a proper irreducible fractiona

b

(a

b− 1

k≥ 0

), the

resulting fraction numerator is strictly less than the one of the initialfraction and the resulting fraction denominator is strictly greater than theone of the initial fraction.

The value of k

Leta

bbe an irreducible proper fraction, we get

a

b=

1

k+

A

Bwith k =

⌈b

a

⌉and A < a.

Example: Let’s consider the fraction3

5, k =

⌈5

3

⌉. So k =

⌈1.6⌉

= 2.

Then3

5− 1

k≥ 0 and we get

3

5=

1

2+

1

10






b

(a

b− 1

k≥ 0

), the


The value of k

Leta


a

b=

1

k+

A

Bwith k =

⌈b

a

⌉and A < a.


5, k =

⌈5

3

⌉. So k =

⌈1.6⌉

= 2.

Then3

5− 1

k≥ 0 and we get

3

5=

1

2+

1

10






b

(a

b− 1

k≥ 0

), the


The value of k

Leta


a

b=

1

k+

A

Bwith k =

⌈b

a

⌉and A < a.


5, k =

⌈5

3

⌉. So k =

⌈1.6⌉

= 2.

Then3

5− 1

k≥ 0 and we get

3

5=

1

2+

1

10



First Answers:

Every proper fraction can be expanded into distinct unit fractions(Fibonacci’s algorithm).

There are different ways to expand a proper fractionThere are infinite ways to expand a proper fraction

To give these last two answers we used the identity:

1

n=

1

n + 1+

1

n(n + 1)

Example:3

5=

1

2+

1

10but

1

2=

1

3+

1

6and

1

10=

1

11+

1

110so

3

5=

1

3+

1

6+

1

11+

1

110



First Answers:




1

n=

1

n + 1+

1

n(n + 1)

Example:3

5=

1

2+

1

10but

1

2=

1

3+

1

6and

1

10=

1

11+

1

110so

3

5=

1

3+

1

6+

1

11+

1

110



First Answers:




1

n=

1

n + 1+

1

n(n + 1)

Example:3

5=

1

2+

1

10but

1

2=

1

3+

1

6and

1

10=

1

11+

1

110so

3

5=

1

3+

1

6+

1

11+

1

110



First Answers:




1

n=

1

n + 1+

1

n(n + 1)

Example:3

5=

1

2+

1

10but

1

2=

1

3+

1

6and

1

10=

1

11+

1

110so

3

5=

1

3+

1

6+

1

11+

1

110



Merits and limits:

The merits of this method are:

It works with all irreducible proper fractions;

It can expand every proper fraction into an Egyptian fraction.

Its limits are:

The denominator of the resulting fractions increases as we keepsubtracting 1

k , it can grow quite huge;

It’s not always the shortest expansion.

Example:

With Fibonacci4

49=

1

13+

1

213+

1

67841+

1

9204734721

while4

49=

1

14+

1

98



Merits and limits:The merits of this method are:



Its limits are:




Example:

With Fibonacci4

49=

1

13+

1

213+

1

67841+

1

9204734721

while4

49=

1

14+

1

98






Its limits are:




Example:

With Fibonacci4

49=

1

13+

1

213+

1

67841+

1

9204734721

while4

49=

1

14+

1

98






Its limits are:




Example:

With Fibonacci4

49=

1

13+

1

213+

1

67841+

1

9204734721

while4

49=

1

14+

1

98


Golomb’s method

Given a proper fractiona

b, there exist A and B such as aA = bB + 1.

Thena

b=

aA

bA=

bB + 1

bA=

B

A+

1

bA

How can we find A and B?

Bezout’s theorem

For nonzero a and b in Z, there are x and y in Z, such thatgcd(a, b) = ax + by . In particular, when a and b are relatively prime,there are x and y in Z such that ax + by = 1.

Theorem for the Golomb’s Method

Let a < b be positive integers with a 6= 1 and with gcd(a, b) = 1, and

consider the fraction 0 <a

b< 1 then there exist a natural number

0 < A < b and a natural number B such that aA = bB + 1.


Golomb’s method



Thena

b=

aA

bA=

bB + 1

bA=

B

A+

1

bA


Bezout’s theorem








Golomb’s method



Thena

b=

aA

bA=

bB + 1

bA=

B

A+

1

bA


Bezout’s theorem








Golomb’s method



Thena

b=

aA

bA=

bB + 1

bA=

B

A+

1

bA


Bezout’s theorem








Golomb’s method



Thena

b=

aA

bA=

bB + 1

bA=

B

A+

1

bA


Bezout’s theorem








Golomb’s method

In order to find A and B we use the Euclidean algorithm.If A and B do not satisfy the condition of Golomb’s method, we have tocalculate A′ and B ′ such that A′ = A− kb, B ′ = B + ka with k ∈ Z.

Theorem

Given a, b ∈ N with a < b and A,B ∈ Z with aA + bB = 1 and A · B < 0,exist k ∈ Z such that, if A′ = A− kb and B ′ = B + ka then A′ > 0,B ′ < 0 and A′ < b.

We have to take k ∈ Z such thatA

b− 1 < k < −B

a.

A number with these properties exists and it is k =

⌊A

b

⌋


Golomb’s method

In order to find A and B we use the Euclidean algorithm.If A and B do not satisfy the condition of Golomb’s method, we have tocalculate A′ and B ′ such that A′ = A− kb, B ′ = B + ka with k ∈ Z.

Theorem

Given a, b ∈ N with a < b and A,B ∈ Z with aA + bB = 1 and A · B < 0,exist k ∈ Z such that, if A′ = A− kb and B ′ = B + ka then A′ > 0,B ′ < 0 and A′ < b.

We have to take k ∈ Z such thatA

b− 1 < k < −B

a.

A number with these properties exists and it is k =

⌊A

b

⌋


Golomb’s method

Leta

bbe an irreducible proper fraction;

Algorithm:

Find A e B of the Bezout’s Identity using the Euclidean algorithm;

If 0 < A < b and B < 0 then we have to go to the next step;

else we calculate A′ = A−⌊A

b

⌋· b and B ′ = B +

⌊A

b

⌋· a and we get

0 < A′ < b and B ′ < 0 and now go to the next step;

Use the Golomb’s method.


Golomb’s method

Leta


Algorithm:




b

⌋· b and B ′ = B +

⌊A

b

⌋· a and we get




Golomb’s method

Leta


Algorithm:




b

⌋· b and B ′ = B +

⌊A

b

⌋· a and we get




Golomb’s method

Leta


Algorithm:




b

⌋· b and B ′ = B +

⌊A

b

⌋· a and we get




Golomb’s method

Example:

Let’s consider the proper fraction38

121;

The Euclidean algorithm gives 38 · (−35) + 121 · (11) = 1.Then A = −35,B = 11 that is A < 0 and B > 0;

We calculate k =

⌊A

b

⌋=

⌊−35

121

⌋= b−0.2c = −1 therefore

A′ = A− (−1)b = −35 + 121 = 86

B ′ = B + (−1)a = 11− 38 = −27

so we get 0 < A′ < b and B ′ < 0;

With the Golomb’s method:a

b=

bB

bA+

1

bA⇒ 38

121=

27

86+

1

10406


Golomb’s method

Example:


121;


We calculate k =

⌊A

b

⌋=

⌊−35

121


A′ = A− (−1)b = −35 + 121 = 86

B ′ = B + (−1)a = 11− 38 = −27

so we get 0 < A′ < b and B ′ < 0;


b=

bB

bA+

1

bA⇒ 38

121=

27

86+

1

10406


Golomb’s method

Example:


121;


We calculate k =

⌊A

b

⌋=

⌊−35

121


A′ = A− (−1)b = −35 + 121 = 86

B ′ = B + (−1)a = 11− 38 = −27

so we get 0 < A′ < b and B ′ < 0;


b=

bB

bA+

1

bA⇒ 38

121=

27

86+

1

10406


Golomb’s method

Example:


121;


We calculate k =

⌊A

b

⌋=

⌊−35

121


A′ = A− (−1)b = −35 + 121 = 86

B ′ = B + (−1)a = 11− 38 = −27

so we get 0 < A′ < b and B ′ < 0;


b=

bB

bA+

1

bA⇒ 38

121=

27

86+

1

10406


Method of practical numbers

Definition

A practical number is a natural number N such that for all n < N, n ∈ N0

then n can be written as the sum of distinct divisors of N.

Theorem

The product of two practical numbers is a practical number.

Theorem

The subset of N of practical numbers is a countably infinite set i.e. withthe same cardinality of N (ℵ0).

Example with practical denominator: 12 is a practical number,

D12 = {1, 2, 3, 4, 6, 12} 7 = 4 + 2 + 1 (6 + 1, 4 + 3)

7

12=

4

12+

2

12+

1

12=

1

3+

1

6+

1

12



Definition



Theorem


Theorem



D12 = {1, 2, 3, 4, 6, 12} 7 = 4 + 2 + 1 (6 + 1, 4 + 3)

7

12=

4

12+

2

12+

1

12=

1

3+

1

6+

1

12



Definition



Theorem


Theorem



D12 = {1, 2, 3, 4, 6, 12} 7 = 4 + 2 + 1 (6 + 1, 4 + 3)

7

12=

4

12+

2

12+

1

12=

1

3+

1

6+

1

12



Definition



Theorem


Theorem



D12 = {1, 2, 3, 4, 6, 12} 7 = 4 + 2 + 1 (6 + 1, 4 + 3)

7

12=

4

12+

2

12+

1

12=

1

3+

1

6+

1

12



Definition



Theorem


Theorem



D12 = {1, 2, 3, 4, 6, 12} 7 = 4 + 2 + 1 (6 + 1, 4 + 3)

7

12=

4

12+

2

12+

1

12=

1

3+

1

6+

1

12



Definition



Theorem


Theorem



D12 = {1, 2, 3, 4, 6, 12} 7 = 4 + 2 + 1 (6 + 1, 4 + 3)

7

12=

4

12+

2

12+

1

12=

1

3+

1

6+

1

12



For not practical denominators b exists at least one number k thatmultiplied by b gives a practical denominator.

Theorem

The product of the first prime numbers is a practical number.

Theorem

The product of a practical number by one of his divisor is a practicalnumber.

Theorem

The product of powers of the first prime numbers is a practical number.

Example for not practical denominator: 10 is not a practical number,

10 = 2 · 5;7

10=

7

10· 3

3=

21

30=

15

30+

6

30=

1

2+

1

5




Theorem


Theorem


Theorem



10 = 2 · 5;7

10=

7

10· 3

3=

21

30=

15

30+

6

30=

1

2+

1

5




Theorem


Theorem


Theorem



10 = 2 · 5;7

10=

7

10· 3

3=

21

30=

15

30+

6

30=

1

2+

1

5




Theorem


Theorem


Theorem



10 = 2 · 5;7

10=

7

10· 3

3=

21

30=

15

30+

6

30=

1

2+

1

5




Theorem


Theorem


Theorem



10 = 2 · 5;7

10=

7

10· 3

3=

21

30=

15

30+

6

30=

1

2+

1

5




Theorem


Theorem


Theorem



10 = 2 · 5;7

10=

7

10· 3

3=

21

30=

15

30+

6

30=

1

2+

1

5



Comparison with other methods.

Theorem (A. Galletti - K.P.S. Bhaskara Rao)

The factorial of a natural number n is practical.

With our method

a

b=

7

10b = 2 · 5; ⇒ k = 3

Instead of using this theorem k would be:

9! = 362880

This theorem guarantees that one k always exists but it is far greater thanours.






With our method

a

b=

7

10b = 2 · 5; ⇒ k = 3


9! = 362880







With our method

a

b=

7

10b = 2 · 5; ⇒ k = 3


9! = 362880







With our method

a

b=

7

10b = 2 · 5; ⇒ k = 3


9! = 362880




Theorem (Stewart)

Let m be a practical number. If n is an integer such that1 ≤ n ≤ σ(m) + 1, then mn is a practical number, where σ(m) denotesthe sum of the positive divisors of m. In particular for 1 ≤ n ≤ 2m, mn ispractical.

So k would be the first practical number greater than b2 , for example:

a

b=

7

10k >

b

2= 5.5 ⇒ k = 6

This result is, in general, the main tool to construct practical numbers.In some cases, as in the proposed example, ours is preferable (k = 3).

Conclusion: All these variants of the method of practical numbers give anexpansion of the fraction, that confirms the existence but they donot find all the possible expansions.



Theorem (Stewart)



a

b=

7

10k >

b

2= 5.5 ⇒ k = 6





Theorem (Stewart)



a

b=

7

10k >

b

2= 5.5 ⇒ k = 6





Theorem (Stewart)



a

b=

7

10k >

b

2= 5.5 ⇒ k = 6





The Algorithm to expand a fraction with practical denominator:

we find all the n proper divisorsof the denominator,

we create a matrix whichdimension n x (2n − 1),

with a loop, we check for eachrow if the sum of only divisorswhose cell is equal to 1, is equalto the numerator,

if it is true the program willprint the expansion.

d1 d2 d3 d4

1 → 0 0 0 12 → 0 0 1 03 → 0 0 1 14 → 0 1 0 05 → 0 1 0 16 → 0 1 1 07 → 0 1 1 18 → 1 0 0 09 → 1 0 0 1

10 → 1 0 1 011 → 1 0 1 112 → 1 1 0 013 → 1 1 0 114 → 1 1 1 015 → 1 1 1 1



The Algorithm to expand a fraction with practical denominator:

we find all the n proper divisorsof the denominator,

we create a matrix whichdimension n x (2n − 1),

with a loop, we check for eachrow if the sum of only divisorswhose cell is equal to 1, is equalto the numerator,

if it is true the program willprint the expansion.

d1 d2 d3 d4

1 → 0 0 0 12 → 0 0 1 03 → 0 0 1 14 → 0 1 0 05 → 0 1 0 16 → 0 1 1 07 → 0 1 1 18 → 1 0 0 09 → 1 0 0 1

10 → 1 0 1 011 → 1 0 1 112 → 1 1 0 013 → 1 1 0 114 → 1 1 1 015 → 1 1 1 1


Geometric method, our solution

Two unit fractions.Given a proper fraction a

b with (a, b) = 1 we want to find two natural

numbers, x and y such that kakb = 1

x + 1y for k ∈ Z that is to solve{

x + y = ka

xy = kb

We decide to analyse the problem from ageometric point of view:{x + y = ka −→ sheaf of straight lines

xy = kb −→ sheaf of hyperbola







x + y = ka

xy = kb









x + y = ka

xy = kb









x + y = ka

xy = kb





Theorem: two unit fractions.

Let ab be an irreducible proper fraction, then it can be expanded into two

distinct unit fractions 1x + 1

y , with 0 < x < y , if

b

a< x <

2b

a∧ y =

bx

ax − b

Example: Let’s consider ab = 5

18 that we want to write as 518 = 1

x + 1y . The

natural numbers x such that ba < x < 2b

a that is 185 < x < 36

5 are:

x = 4→ y = bxax−b = 36, acceptable

x = 5→ y 6∈ N0


x = 7→ y 6∈ N0

So:5

18=

1

4+

1

36=

1

6+

1

9






y , with 0 < x < y , if

b

a< x <

2b

a∧ y =

bx

ax − b



x + 1y .

The


a that is 185 < x < 36

5 are:


x = 5→ y 6∈ N0


x = 7→ y 6∈ N0

So:5

18=

1

4+

1

36=

1

6+

1

9






y , with 0 < x < y , if

b

a< x <

2b

a∧ y =

bx

ax − b



x + 1y . The


a that is 185 < x < 36

5 are:


x = 5→ y 6∈ N0


x = 7→ y 6∈ N0

So:5

18=

1

4+

1

36=

1

6+

1

9






y , with 0 < x < y , if

b

a< x <

2b

a∧ y =

bx

ax − b



x + 1y . The


a that is 185 < x < 36

5 are:


x = 5→ y 6∈ N0


x = 7→ y 6∈ N0

So:5

18=

1

4+

1

36=

1

6+

1

9



Three unit fractions.We tried to retrace an analogous procedure for the expansion of a fractionin a sum of three unit fractions.

ka

kb=

1

x+

1

y+

1

zthat is

{xy + xz + yz = ka

xyz = kb

we determine the geometric locus of the solutions:

z =bxy

axy − b(x + y)

We choose to impose 0 < x < y < z and so

x < y ∧ y >bx

ax − b

y <bx

ax − 2b∧ y <

2bx

ax − b




ka

kb=

1

x+

1

y+

1

zthat is

{xy + xz + yz = ka

xyz = kb


z =bxy

axy − b(x + y)


x < y ∧ y >bx

ax − b

y <bx

ax − 2b∧ y <

2bx

ax − b




ka

kb=

1

x+

1

y+

1

zthat is

{xy + xz + yz = ka

xyz = kb


z =bxy

axy − b(x + y)


x < y ∧ y >bx

ax − b

y <bx

ax − 2b∧ y <

2bx

ax − b




ka

kb=

1

x+

1

y+

1

zthat is

{xy + xz + yz = ka

xyz = kb


z =bxy

axy − b(x + y)


x < y ∧ y >bx

ax − b

y <bx

ax − 2b∧ y <

2bx

ax − b



Theorem: tree unit fractions.

Let ab be an irreducible proper fraction, then it can be expanded into three


y + 1z with 0 < x < y < z , if

b

a< x <

3b

a∧ bx

ax − b< y <

2bx

ax − b∧ z =

bxy

axy − b(x + y)



x + 1y + 1

z .

The natural numbers x such that ba < x < 3b

a that is 54 < x < 15

4 are:

x = 2 −→ 103 < y < 20

3

→ y = 4→ z = bxyaxy−b(x+y) = 20, acceptable


→ y = 6→ z 6∈ N0

x = 3 −→ 157 < y < 30

7

→ y = 3→ z 6∈ N0

→ y = 4→ z 6∈ N0

So:4

5=

1

2+

1

4+

1

20=

1

2+

1

5+

1

10






y + 1z with 0 < x < y < z , if

b

a< x <

3b

a∧ bx

ax − b< y <

2bx

ax − b∧ z =

bxy

axy − b(x + y)



x + 1y + 1

z .


a that is 54 < x < 15

4 are:

x = 2 −→ 103 < y < 20

3



→ y = 6→ z 6∈ N0

x = 3 −→ 157 < y < 30

7

→ y = 3→ z 6∈ N0

→ y = 4→ z 6∈ N0

So:4

5=

1

2+

1

4+

1

20=

1

2+

1

5+

1

10






y + 1z with 0 < x < y < z , if

b

a< x <

3b

a∧ bx

ax − b< y <

2bx

ax − b∧ z =

bxy

axy − b(x + y)



x + 1y + 1

z .


a that is 54 < x < 15

4 are:

x = 2 −→ 103 < y < 20

3



→ y = 6→ z 6∈ N0

x = 3 −→ 157 < y < 30

7

→ y = 3→ z 6∈ N0

→ y = 4→ z 6∈ N0

So:4

5=

1

2+

1

4+

1

20=

1

2+

1

5+

1

10






y + 1z with 0 < x < y < z , if

b

a< x <

3b

a∧ bx

ax − b< y <

2bx

ax − b∧ z =

bxy

axy − b(x + y)



x + 1y + 1

z .


a that is 54 < x < 15

4 are:

x = 2 −→ 103 < y < 20

3



→ y = 6→ z 6∈ N0

x = 3 −→ 157 < y < 30

7

→ y = 3→ z 6∈ N0

→ y = 4→ z 6∈ N0

So:4

5=

1

2+

1

4+

1

20=

1

2+

1

5+

1

10






y + 1z with 0 < x < y < z , if

b

a< x <

3b

a∧ bx

ax − b< y <

2bx

ax − b∧ z =

bxy

axy − b(x + y)



x + 1y + 1

z .


a that is 54 < x < 15

4 are:

x = 2 −→ 103 < y < 20

3



→ y = 6→ z 6∈ N0

x = 3 −→ 157 < y < 30

7

→ y = 3→ z 6∈ N0

→ y = 4→ z 6∈ N0

So:4

5=

1

2+

1

4+

1

20=

1

2+

1

5+

1

10






y + 1z with 0 < x < y < z , if

b

a< x <

3b

a∧ bx

ax − b< y <

2bx

ax − b∧ z =

bxy

axy − b(x + y)



x + 1y + 1

z .


a that is 54 < x < 15

4 are:

x = 2 −→ 103 < y < 20

3



→ y = 6→ z 6∈ N0

x = 3 −→ 157 < y < 30

7

→ y = 3→ z 6∈ N0

→ y = 4→ z 6∈ N0

So:4

5=

1

2+

1

4+

1

20=

1

2+

1

5+

1

10






y + 1z with 0 < x < y < z , if

b

a< x <

3b

a∧ bx

ax − b< y <

2bx

ax − b∧ z =

bxy

axy − b(x + y)



x + 1y + 1

z .


a that is 54 < x < 15

4 are:

x = 2 −→ 103 < y < 20

3



→ y = 6→ z 6∈ N0

x = 3 −→ 157 < y < 30

7

→ y = 3→ z 6∈ N0

→ y = 4→ z 6∈ N0

So:4

5=

1

2+

1

4+

1

20=

1

2+

1

5+

1

10



Theorem: general case.

Leta

bbe an irreducible proper fraction, then it can be expanded as sum of

n distinct unit fractions1

x1+ ...+

1

xn, with x0 < x1 < ... < xn, (x0 = 0), if,

for j = 1, ...n − 1

max

{xj−1,

b

a− b∑j−1

i=11xi

}< xj <

(n + 1− j)b

a− b∑j−1

i=11xi

, xn =b

a− b∑n−1

i=11xi

Observation. This method finds the expansions only given a number offractions of the sum but has the advantage that it finds all possibleexpansions in this case.



Theorem: general case.

Leta

bbe an irreducible proper fraction, then it can be expanded as sum of

n distinct unit fractions1

x1+ ...+

1

xn, with x0 < x1 < ... < xn, (x0 = 0), if,

for j = 1, ...n − 1

max

{xj−1,

b

a− b∑j−1

i=11xi

}< xj <

(n + 1− j)b

a− b∑j−1

i=11xi

, xn =b

a− b∑n−1

i=11xi

Observation. This method finds the expansions only given a number offractions of the sum but has the advantage that it finds all possibleexpansions in this case.


Comparison

MethodNo. ofexpan-sions

DenominatorsNo. of unit

fractionsExample: a

b = 79

Fibonacci 1 Huge growth Less than orequal to a

79 = 1

2 + 14 + 1

36

Golomb 1 Less thanb(b − 1)

Less than orequal to thenumerator a

79 = 1

2 + 16 + 1

12 + 136

Practicalnumbers

At least1

Less than orequal to the

denominator bor kb

Less than orequal to thenumber ofdivisor of b

or kb

79 = 1

2 + 16 + 1

9

GeometricAll theexpan-sions

Unforeseeable Fixed79 = 1

2 + 14 + 1

36

79 = 1

2 + 16 + 1

9


Comparison

MethodNo. ofexpan-sions

DenominatorsNo. of unit

fractionsExample: a

b = 79

Fibonacci 1 Huge growth Less than orequal to a

79 = 1

2 + 14 + 1

36

Golomb 1 Less thanb(b − 1)

Less than orequal to thenumerator a

79 = 1

2 + 16 + 1

12 + 136

Practicalnumbers

At least1

Less than orequal to the

denominator bor kb

Less than orequal to thenumber ofdivisor of b

or kb

79 = 1

2 + 16 + 1

9

GeometricAll theexpan-sions

Unforeseeable Fixed79 = 1

2 + 14 + 1

36

79 = 1

2 + 16 + 1

9


The tree of fractions

From the identity:1

n=

1

n + 1+

1

n(n + 1)we can build this tree:

1

2

1

3

1

6

1

4

1

12

1

42

1

7

1

5

1

20

1

13

1

156

1

1806

1

43

1

56

1

8... ... ... ... ... ... ... ... ... ...



From the identity:1

n=

1

n + 1+

1


1

2

1

3

1

6

1

4

1

12

1

42

1

7

1

5

1

20

1

13

1

156

1

1806

1

43

1

56

1

8... ... ... ... ... ... ... ... ... ...



From the identity:1

n=

1

n + 1+

1


1

2

1

3

1

6

1

4

1

12

1

42

1

7

1

5

1

20

1

13

1

156

1

1806

1

43

1

56

1

8... ... ... ... ... ... ... ... ... ...



Given r the number of the row and c the number of the column we canfind the denominator A(r , c) of the unit fraction which is in the positionr , c :

Formula

For all r , c ∈ N0 with 1 ≤ c ≤ 2r−1

A (r , c) =

{2 if r = c = 1[A(r − 1,

⌈c2

⌉)+ 1]·[A(r − 1,

⌈c2

⌉)]pif r > 1

with

p = 1−{c +

1

2· sgn(r − 2) ·

[sgn

(c − 2r−2 − 1

2

)+ 1

]}mod 2



Given r the number of the row and c the number of the column we canfind the denominator A(r , c) of the unit fraction which is in the positionr , c :

Formula

For all r , c ∈ N0 with 1 ≤ c ≤ 2r−1

A (r , c) =

{2 if r = c = 1[A(r − 1,

⌈c2

⌉)+ 1]·[A(r − 1,

⌈c2

⌉)]pif r > 1

with

p = 1−{c +

1

2· sgn(r − 2) ·

[sgn

(c − 2r−2 − 1

2

)+ 1

]}mod 2



How did we find it?

The problems are:

the recursion of the formula;

the need to operate differently with the unit fractions1

n + 1and

1

n(n + 1);

the symmetry of the tree;

the non-symmetry in the second row of the tree.



How did we find it?

The problems are:



n + 1and

1

n(n + 1);





How did we find it?

The problems are:



n + 1and

1

n(n + 1);





How did we find it?

The problems are:



n + 1and

1

n(n + 1);





How did we find it?

The problems are:



n + 1and

1

n(n + 1);





How did we find it?

The problems are:



n + 1and

1

n(n + 1);





The recursion of the formula:

Each element of the tree is obtained from an element of the previous row⇒ r − 1. The two-by-two elements are obtained from the one in theprevious row in the middle of them ⇒

⌈c2

⌉that is we used the ceiling

function.It is necessary to start this process therefore we fixed at first A(1, 1) = 2and then A(r , s) must be a function of A

(r − 1,

⌈c2

⌉)so

A (r , c) =

{2 if r = c = 1

f[A(r − 1,

⌈c2

⌉)]if r > 1



The recursion of the formula:Each element of the tree is obtained from an element of the previous row⇒ r − 1. The two-by-two elements are obtained from the one in theprevious row in the middle of them ⇒

⌈c2


function.

It is necessary to start this process therefore we fixed at first A(1, 1) = 2and then A(r , s) must be a function of A

(r − 1,

⌈c2

⌉)so

A (r , c) =

{2 if r = c = 1

f[A(r − 1,

⌈c2

⌉)]if r > 1




⌈c2



(r − 1,

⌈c2

⌉)so

A (r , c) =

{2 if r = c = 1

f[A(r − 1,

⌈c2

⌉)]if r > 1




⌈c2



(r − 1,

⌈c2

⌉)so

A (r , c) =

{2 if r = c = 1

f[A(r − 1,

⌈c2

⌉)]if r > 1



The need to operate differently with the unit fractions1

n + 1and

1

n(n + 1)

For every pair of elements of the tree the first one is obtained from theelement that is in the row r − 1 and in the column

⌈c2

⌉with the formula

n + 1 while the second with the formula n(n + 1). Thus is the factor n inone case there must be and in the other no. Therefore we put an exponentp in this factor which becomes 0 or 1. In order to do this we build theformula:

A (r , c) =[A(r − 1,

⌈c2

⌉)+ 1]·[A(r − 1,

⌈c2

⌉)]pwith p = 1− c mod 2where ”mod” is the modulo operation.




n + 1and

1

n(n + 1)


⌈c2

⌉with the formula


A (r , c) =[A(r − 1,

⌈c2

⌉)+ 1]·[A(r − 1,

⌈c2





n + 1and

1

n(n + 1)


⌈c2

⌉with the formula


A (r , c) =[A(r − 1,

⌈c2

⌉)+ 1]·[A(r − 1,

⌈c2




The symmetry of the tree.

In order to build a symmetrical tree, that is with the greatest fractions onthe left for the first 2r−2 elements and on the right for the element from2r−2 + 1 to 2r−1, we have to change the exponent p before and after theelement 2r−2. We want the following exponents:

...0 1 1 0

0 1 0 1 1 0 1 00 1 0 1 0 1 1 0 1 0 1 0

we used the fuction:

x

|x |=

{−1 if x < 0

+1 if x > 0⇒ 1

2·(

x

|x |+ 1

)=

{0 if x < 0

+1 if x > 0

and we get: p = 1−[c + 1

2 ·(

c−2r−2− 12

|c−2r−2− 12 |

+ 1

)]mod 2



The symmetry of the tree.In order to build a symmetrical tree, that is with the greatest fractions onthe left for the first 2r−2 elements and on the right for the element from2r−2 + 1 to 2r−1, we have to change the exponent p before and after theelement 2r−2.

We want the following exponents:...

0 1 1 00 1 0 1 1 0 1 0

0 1 0 1 0 1 1 0 1 0 1 0


x

|x |=

{−1 if x < 0

+1 if x > 0⇒ 1

2·(

x

|x |+ 1

)=

{0 if x < 0

+1 if x > 0


2 ·(

c−2r−2− 12

|c−2r−2− 12 |

+ 1

)]mod 2



The symmetry of the tree.In order to build a symmetrical tree, that is with the greatest fractions onthe left for the first 2r−2 elements and on the right for the element from2r−2 + 1 to 2r−1, we have to change the exponent p before and after theelement 2r−2. We want the following exponents:

...0 1 1 0

0 1 0 1 1 0 1 00 1 0 1 0 1 1 0 1 0 1 0


x

|x |=

{−1 if x < 0

+1 if x > 0⇒ 1

2·(

x

|x |+ 1

)=

{0 if x < 0

+1 if x > 0


2 ·(

c−2r−2− 12

|c−2r−2− 12 |

+ 1

)]mod 2



The symmetry of the tree.In order to build a symmetrical tree, that is with the greatest fractions onthe left for the first 2r−2 elements and on the right for the element from2r−2 + 1 to 2r−1, we have to change the exponent p before and after theelement 2r−2. We want the following exponents:

...0 1 1 0

0 1 0 1 1 0 1 00 1 0 1 0 1 1 0 1 0 1 0


x

|x |=

{−1 if x < 0

+1 if x > 0⇒ 1

2·(

x

|x |+ 1

)=

{0 if x < 0

+1 if x > 0


2 ·(

c−2r−2− 12

|c−2r−2− 12 |

+ 1

)]mod 2



The non-symmetry in the second row of the tree:

The exponents in the second row are 0 and 1 hence they are notsymmetric.

0 10 1 1 0

0 1 0 1 1 0 1 0

Therefore we need a function that changes the exponents only from thesecond row onwards. We use the sign function:

sgn (x) =

−1 if x < 0

0 if x = 0

+1 if x > 0

⇒ sgn (r − 2) =

−1 if r < 2(it never happens)

0 if r = 2

+1 if r > 2

So the exponent becomes:

p = 1−{c + 1

2 · sgn(r − 2) ·[

sgn

(c − 2r−2 − 1

2

)+ 1

]}mod 2



The non-symmetry in the second row of the tree:The exponents in the second row are 0 and 1 hence they are notsymmetric.

0 10 1 1 0

0 1 0 1 1 0 1 0


sgn (x) =

−1 if x < 0

0 if x = 0

+1 if x > 0

⇒ sgn (r − 2) =


0 if r = 2

+1 if r > 2


p = 1−{c + 1

2 · sgn(r − 2) ·[

sgn

(c − 2r−2 − 1

2

)+ 1

]}mod 2




0 10 1 1 0

0 1 0 1 1 0 1 0

Therefore we need a function that changes the exponents only from thesecond row onwards.

We use the sign function:

sgn (x) =

−1 if x < 0

0 if x = 0

+1 if x > 0

⇒ sgn (r − 2) =


0 if r = 2

+1 if r > 2


p = 1−{c + 1

2 · sgn(r − 2) ·[

sgn

(c − 2r−2 − 1

2

)+ 1

]}mod 2




0 10 1 1 0

0 1 0 1 1 0 1 0


sgn (x) =

−1 if x < 0

0 if x = 0

+1 if x > 0

⇒ sgn (r − 2) =


0 if r = 2

+1 if r > 2


p = 1−{c + 1

2 · sgn(r − 2) ·[

sgn

(c − 2r−2 − 1

2

)+ 1

]}mod 2




0 10 1 1 0

0 1 0 1 1 0 1 0


sgn (x) =

−1 if x < 0

0 if x = 0

+1 if x > 0

⇒ sgn (r − 2) =


0 if r = 2

+1 if r > 2


p = 1−{c + 1

2 · sgn(r − 2) ·[

sgn

(c − 2r−2 − 1

2

)+ 1

]}mod 2




0 10 1 1 0

0 1 0 1 1 0 1 0


sgn (x) =

−1 if x < 0

0 if x = 0

+1 if x > 0

⇒ sgn (r − 2) =


0 if r = 2

+1 if r > 2


p = 1−{c + 1

2 · sgn(r − 2) ·[

sgn

(c − 2r−2 − 1

2

)+ 1

]}mod 2



The expansion of the natural numbers.

1

2

1

3+

1

6

1

4+

1

12+

1

42+

1

7

1

5+

1

20+

1

13+

1

156+

1

1806+

1

43+

1

56+

1

8... ... ... ... ... ... ... ... ...

+1

2= 1

+1

2= 1

+1

2= 1

+1

2= 1

n

the sum of each row yields 12 ;

the sum of different parts of the tree yields 12 ;

the sum of two rows yields 1;

the sum of 2n rows, with n ∈ N0, yields n.⇒ all the natural numbers can be expanded into distinct U.F.;⇒ also the improper fractions can be expanded into distinct U.F.



The expansion of the natural numbers.1

2

1

3+

1

6

1

4+

1

12+

1

42+

1

7

1

5+

1

20+

1

13+

1

156+

1

1806+

1

43+

1

56+

1

8... ... ... ... ... ... ... ... ...

+1

2= 1

+1

2= 1

+1

2= 1

+1

2= 1

n








2

1

3+

1

6

1

4+

1

12+

1

42+

1

7

1

5+

1

20+

1

13+

1

156+

1

1806+

1

43+

1

56+

1

8... ... ... ... ... ... ... ... ...

+1

2= 1

+1

2= 1

+1

2= 1

+1

2= 1

n








2

1

3+

1

6

1

4+

1

12+

1

42+

1

7

1

5+

1

20+

1

13+

1

156+

1

1806+

1

43+

1

56+

1

8... ... ... ... ... ... ... ... ...

+1

2= 1

+1

2= 1

+1

2= 1

+1

2= 1

n








2

1

3+

1

6

1

4+

1

12+

1

42+

1

7

1

5+

1

20+

1

13+

1

156+

1

1806+

1

43+

1

56+

1

8... ... ... ... ... ... ... ... ...

+1

2= 1

+1

2= 1

+1

2= 1

+1

2= 1

n








2

1

3+

1

6

1

4+

1

12+

1

42+

1

7

1

5+

1

20+

1

13+

1

156+

1

1806+

1

43+

1

56+

1

8... ... ... ... ... ... ... ... ...

+1

2= 1

+1

2= 1

+1

2= 1

+1

2= 1

n








2

1

3+

1

6

1

4+

1

12+

1

42+

1

7

1

5+

1

20+

1

13+

1

156+

1

1806+

1

43+

1

56+

1

8... ... ... ... ... ... ... ... ...

+1

2= 1

+1

2= 1

+1

2= 1

+1

2= 1

n




the sum of 2n rows, with n ∈ N0, yields n.

⇒ all the natural numbers can be expanded into distinct U.F.;⇒ also the improper fractions can be expanded into distinct U.F.




2

1

3+

1

6

1

4+

1

12+

1

42+

1

7

1

5+

1

20+

1

13+

1

156+

1

1806+

1

43+

1

56+

1

8... ... ... ... ... ... ... ... ...

+1

2= 1

+1

2= 1

+1

2= 1

+1

2= 1

n




the sum of 2n rows, with n ∈ N0, yields n.⇒ all the natural numbers can be expanded into distinct U.F.;

⇒ also the improper fractions can be expanded into distinct U.F.




2

1

3+

1

6

1

4+

1

12+

1

42+

1

7

1

5+

1

20+

1

13+

1

156+

1

1806+

1

43+

1

56+

1

8... ... ... ... ... ... ... ... ...

+1

2= 1

+1

2= 1

+1

2= 1

+1

2= 1

n






Thank you for your attention!


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