Mumbai, January 5, 2008 Early history of irrational and transcendental numbers

(1)

Early history of irrational and transcendental numbers

Michel Waldschmidt

http://www.math.jussieu.fr/∼miw/

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The transcendence proofs for constants of analysis are essentially all based on the seminal work by Ch. Hermite : his proof of the transcendence of the number e in 1873 is the prototype of the methods which have been subsequently developed. We first show how the founding paper by

Hermite was influenced by earlier authors (Lambert, Euler,

Fourier, Liouville), next we explain how his arguments have

been expanded in several directions : Pad´ e approximants,

interpolation series, auxiliary functions.

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Numbers = real or complex numbers R, C.

Natural integers : N = {0, 1, 2, . . .}.

Rational integers : Z = {0, ±1, ±2, . . .}.

Rational numbers :

a/b with a and b rational integers, b > 0.

Irreducible representation :

p/q with p and q in Z, q > 0 and gcd(p, q) = 1.

Irrational number : a real (or complex) number which is

not rational.

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Numbers = real or complex numbers R, C.

Natural integers : N = {0, 1, 2, . . .}.

Rational integers : Z = {0, ±1, ±2, . . .}.

Rational numbers :

a/b with a and b rational integers, b > 0.

Irreducible representation :

p/q with p and q in Z, q > 0 and gcd(p, q) = 1.

Irrational number : a real (or complex) number which is

not rational.

(5)

Numbers = real or complex numbers R, C.

Natural integers : N = {0, 1, 2, . . .}.

Rational integers : Z = {0, ±1, ±2, . . .}.

Rational numbers :

a/b with a and b rational integers, b > 0.

Irreducible representation :

p/q with p and q in Z, q > 0 and gcd(p, q) = 1.

Irrational number : a real (or complex) number which is

not rational.

(6)

Numbers = real or complex numbers R, C.

Natural integers : N = {0, 1, 2, . . .}.

Rational integers : Z = {0, ±1, ±2, . . .}.

Rational numbers :

a/b with a and b rational integers, b > 0.

Irreducible representation :

p/q with p and q in Z, q > 0 and gcd(p, q) = 1.

Irrational number : a real (or complex) number which is

not rational.

(7)

Numbers = real or complex numbers R, C.

Natural integers : N = {0, 1, 2, . . .}.

Rational integers : Z = {0, ±1, ±2, . . .}.

Rational numbers :

a/b with a and b rational integers, b > 0.

Irreducible representation :

p/q with p and q in Z, q > 0 and gcd(p, q) = 1.

Irrational number : a real (or complex) number which is

not rational.

(8)

Numbers = real or complex numbers R, C.

Natural integers : N = {0, 1, 2, . . .}.

Rational integers : Z = {0, ±1, ±2, . . .}.

Rational numbers :

a/b with a and b rational integers, b > 0.

Irreducible representation :

p/q with p and q in Z, q > 0 and gcd(p, q) = 1.

Irrational number : a real (or complex) number which is

not rational.

(9)

Numbers = real or complex numbers R, C.

Natural integers : N = {0, 1, 2, . . .}.

Rational integers : Z = {0, ±1, ±2, . . .}.

Rational numbers :

a/b with a and b rational integers, b > 0.

Irreducible representation :

p/q with p and q in Z, q > 0 and gcd(p, q) = 1.

Irrational number : a real (or complex) number which is

not rational.

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Sums and products of rational numbers are rational numbers :

a b + c

d = ad + bc

bd , a

b · c d = ac

bd ·

The set of rational numbers is a field.

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Sums and products of rational numbers are rational numbers :

a b + c

d = ad + bc

bd , a

b · c d = ac

bd ·

The set of rational numbers is a field.

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The sum of two irrational numbers may be rational or irrational. For instance if x is irrational then −x and 2x are irrational, while x − x = 0 is rational

The product of two irrational numbers may be rational or irrational. For instance if x is irrational then 1/x and √

x are irrational, the product √

x · √

x = x is irrational, while x/x = 1 is rational.

The sum of an rational number and an irrational number is irrational. This is a consequence of the fact that the sum of two rational numbers is rational.

The product of a non–zero rational number and an

irrational number is irrational. This is a consequence of the

fact that the product of two rational numbers is rational.

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irrational. For instance if x is irrational then −x and 2x are irrational, while x − x = 0 is rational

The product of two irrational numbers may be rational or irrational. For instance if x is irrational then 1/x and √

x are irrational, the product √

x · √

x = x is irrational, while x/x = 1 is rational.

The sum of an rational number and an irrational number is irrational. This is a consequence of the fact that the sum of two rational numbers is rational.

The product of a non–zero rational number and an

irrational number is irrational. This is a consequence of the

fact that the product of two rational numbers is rational.

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The sum of two irrational numbers may be rational or irrational. For instance if x is irrational then −x and 2x are irrational, while x − x = 0 is rational

The product of two irrational numbers may be rational or irrational. For instance if x is irrational then 1/x and √

x are irrational, the product √

x · √

x = x is irrational, while x/x = 1 is rational.

The sum of an rational number and an irrational number is irrational. This is a consequence of the fact that the sum of two rational numbers is rational.

The product of a non–zero rational number and an

irrational number is irrational. This is a consequence of the

fact that the product of two rational numbers is rational.

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irrational. For instance if x is irrational then −x and 2x are irrational, while x − x = 0 is rational

The product of two irrational numbers may be rational or irrational. For instance if x is irrational then 1/x and √

x are irrational, the product √

x · √

x = x is irrational, while x/x = 1 is rational.

The sum of an rational number and an irrational number is irrational. This is a consequence of the fact that the sum of two rational numbers is rational.

The product of a non–zero rational number and an

irrational number is irrational. This is a consequence of the

fact that the product of two rational numbers is rational.

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Main question : Is the sum of a convergent series of rational numbers a rational or an irrational number ? Answer : It may be rational or irrational !

Example of a rational sum (geometric series) : 2 = 1 + 1

2 + 1 2

²

+ 1

2

³

+ 1

2

⁴

+ · · · =

∞

X

n=0

1 2

ⁿ

· Example of an irrational sum :

e = 1 + 1 1 + 1

1 · 2 + 1

1 · 2 · 3 + 1

1 · 2 · 3 · 4 + · · · =

∞

X

n=0

1 n! ·

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rational numbers a rational or an irrational number ? Answer : It may be rational or irrational !

Example of a rational sum (geometric series) : 2 = 1 + 1

2 + 1 2

²

+ 1

2

³

+ 1

2

⁴

+ · · · =

∞

X

n=0

1 2

ⁿ

· Example of an irrational sum :

e = 1 + 1 1 + 1

1 · 2 + 1

1 · 2 · 3 + 1

1 · 2 · 3 · 4 + · · · =

∞

X

n=0

1 n! ·

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Main question : Is the sum of a convergent series of rational numbers a rational or an irrational number ? Answer : It may be rational or irrational !

Example of a rational sum (geometric series) : 2 = 1 + 1

2 + 1 2

²

+ 1

2

³

+ 1

2

⁴

+ · · · =

∞

X

n=0

1 2

ⁿ

· Example of an irrational sum :

e = 1 + 1 1 + 1

1 · 2 + 1

1 · 2 · 3 + 1

1 · 2 · 3 · 4 + · · · =

∞

X

n=0

1 n! ·

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rational numbers a rational or an irrational number ? Answer : It may be rational or irrational !

Example of a rational sum (geometric series) : 2 = 1 + 1

2 + 1 2

²

+ 1

2

³

+ 1

2

⁴

+ · · · =

∞

X

n=0

1 2

ⁿ

· Example of an irrational sum :

e = 1 + 1 1 + 1

1 · 2 + 1

1 · 2 · 3 + 1

1 · 2 · 3 · 4 + · · · =

∞

X

n=0

1 n! ·

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Algebraic number : a complex number which is root of a non-zero polynomial with rational coefficients.

Examples :

rational numbers : a/b, root of bX − a.

√ 2, root of X

²

− 2.

i, root of X

²

+ 1.

The sum and the product of algebraic numbers are

algebraic numbers. The set of complex algebraic numbers is a field, the algebraic closure of Q in C.

A transcendental number is a complex number which is not

algebraic.

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Algebraic number : a complex number which is root of a non-zero polynomial with rational coefficients.

Examples :

rational numbers : a/b, root of bX − a.

√ 2, root of X

²

− 2.

i, root of X

²

+ 1.

The sum and the product of algebraic numbers are

algebraic numbers. The set of complex algebraic numbers is a field, the algebraic closure of Q in C.

A transcendental number is a complex number which is not

algebraic.

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Algebraic number : a complex number which is root of a non-zero polynomial with rational coefficients.

Examples :

rational numbers : a/b, root of bX − a.

√ 2, root of X

²

− 2.

i, root of X

²

+ 1.

The sum and the product of algebraic numbers are

algebraic numbers. The set of complex algebraic numbers is a field, the algebraic closure of Q in C.

A transcendental number is a complex number which is not

algebraic.

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Algebraic number : a complex number which is root of a non-zero polynomial with rational coefficients.

Examples :

rational numbers : a/b, root of bX − a.

√ 2, root of X

²

− 2.

i, root of X

²

+ 1.

The sum and the product of algebraic numbers are

algebraic numbers. The set of complex algebraic numbers is a field, the algebraic closure of Q in C.

A transcendental number is a complex number which is not

algebraic.

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Algebraic number : a complex number which is root of a non-zero polynomial with rational coefficients.

Examples :

rational numbers : a/b, root of bX − a.

√ 2, root of X

²

− 2.

i, root of X

²

+ 1.

The sum and the product of algebraic numbers are

algebraic numbers. The set of complex algebraic numbers is a field, the algebraic closure of Q in C.

A transcendental number is a complex number which is not

algebraic.

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algebraic numbers.

For instance a polynomial with rational coefficients vanishing at √

2 + √ 3 is (X − √

2 − √

3)(X − √ 2 + √

3)(X + √ 2 − √

3)(X + √ 2 + √

3).

In general if

m

Y

i=1

(X − α

i

) and

n

Y

j=1

(X − β

j

) have rational coefficients then

m

Y

i=1 n

Y

j=1

(X − α

_i

− β

_j

) and

m

Y

i=1 n

Y

j=1

(X − α

_i

β

_j

)

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Sums and products of algebraic numbers are algebraic numbers.

For instance a polynomial with rational coefficients vanishing at √

2 + √ 3 is (X − √

2 − √

3)(X − √ 2 + √

3)(X + √ 2 − √

3)(X + √ 2 + √

3).

In general if

m

Y

i=1

(X − α

i

) and

n

Y

j=1

(X − β

j

) have rational coefficients then

m

Y

i=1 n

Y

j=1

(X − α

_i

− β

_j

) and

m

Y

i=1 n

Y

j=1

(X − α

_i

β

_j

)

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algebraic numbers.

For instance a polynomial with rational coefficients vanishing at √

2 + √ 3 is (X − √

2 − √

3)(X − √ 2 + √

3)(X + √ 2 − √

3)(X + √ 2 + √

3).

In general if

m

Y

i=1

(X − α

i

) and

n

Y

j=1

(X − β

j

) have rational coefficients then

m

Y

i=1 n

Y

j=1

(X − α

_i

− β

_j

) and

m

Y

i=1 n

Y

j=1

(X − α

_i

β

_j

)

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For instance a polynomial with rational coefficients vanishing at √

2 + √

³

5 is (X − √

2 − √

³

5)(X − √

2 − j √

³

5)(X − √

2 − j

²

√

³

5) (X + √

2 − √

³

5)(X + √

2 − j √

³

5)(X + √

2 − j

²

√

³

5), where j is root of X

²

+ X + 1, so that

X

³

− 5 = (X − √

³

5)(X − j √

³

5)(X − j

²

√

³

5).

The set of complex algebraic numbers is a field.

Another proof of this fact is the following : a complex

number α is algebraic if and only if the vector space spanned

by 1, α, α

²

, α

³

. . . over the rational has finite dimension.

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vanishing at √ 2 + √

³

5 is (X − √

2 − √

³

5)(X − √

2 − j √

³

5)(X − √

2 − j

²

√

³

5) (X + √

2 − √

³

5)(X + √

2 − j √

³

5)(X + √

2 − j

²

√

³

5), where j is root of X

²

+ X + 1, so that

X

³

− 5 = (X − √

³

5)(X − j √

³

5)(X − j

²

√

³

5).

The set of complex algebraic numbers is a field.

Another proof of this fact is the following : a complex

number α is algebraic if and only if the vector space spanned

by 1, α, α

²

, α

³

. . . over the rational has finite dimension.

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For instance a polynomial with rational coefficients vanishing at √

2 + √

³

5 is (X − √

2 − √

³

5)(X − √

2 − j √

³

5)(X − √

2 − j

²

√

³

5) (X + √

2 − √

³

5)(X + √

2 − j √

³

5)(X + √

2 − j

²

√

³

5), where j is root of X

²

+ X + 1, so that

X

³

− 5 = (X − √

³

5)(X − j √

³

5)(X − j

²

√

³

5).

The set of complex algebraic numbers is a field.

Another proof of this fact is the following : a complex

number α is algebraic if and only if the vector space spanned

by 1, α, α

²

, α

³

. . . over the rational has finite dimension.

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or transcendental. For instance if x is transcendental then

−x and 2x are transcendental, while x − x = 0 is algebraic.

The product of two transcendental numbers may be algebraic or transcendental. For instance if x is

transcendental then 1/x and x

²

are transcendental, while x/x = 1 is algebraic.

The sum of an algebraic number and a transcendental number is transcendental. This is a consequence of the fact that the sum of two algebraic numbers is algebraic.

The product of a non–zero algebraic number and a transcendental number is transcendental. This is a

consequence of the fact that the product of two algebraic

numbers is algebraic.

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The sum of two transcendental numbers may be algebraic or transcendental. For instance if x is transcendental then

−x and 2x are transcendental, while x − x = 0 is algebraic.

The product of two transcendental numbers may be algebraic or transcendental. For instance if x is

transcendental then 1/x and x

²

are transcendental, while x/x = 1 is algebraic.

The sum of an algebraic number and a transcendental number is transcendental. This is a consequence of the fact that the sum of two algebraic numbers is algebraic.

The product of a non–zero algebraic number and a transcendental number is transcendental. This is a

consequence of the fact that the product of two algebraic

numbers is algebraic.

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or transcendental. For instance if x is transcendental then

−x and 2x are transcendental, while x − x = 0 is algebraic.

The product of two transcendental numbers may be algebraic or transcendental. For instance if x is

transcendental then 1/x and x

²

are transcendental, while x/x = 1 is algebraic.

The sum of an algebraic number and a transcendental number is transcendental. This is a consequence of the fact that the sum of two algebraic numbers is algebraic.

The product of a non–zero algebraic number and a transcendental number is transcendental. This is a

consequence of the fact that the product of two algebraic

numbers is algebraic.

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The sum of two transcendental numbers may be algebraic or transcendental. For instance if x is transcendental then

−x and 2x are transcendental, while x − x = 0 is algebraic.

The product of two transcendental numbers may be algebraic or transcendental. For instance if x is

transcendental then 1/x and x

²

are transcendental, while x/x = 1 is algebraic.

The sum of an algebraic number and a transcendental number is transcendental. This is a consequence of the fact that the sum of two algebraic numbers is algebraic.

The product of a non–zero algebraic number and a transcendental number is transcendental. This is a

consequence of the fact that the product of two algebraic

numbers is algebraic.

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Pythagoreas school Hippasus of Metapontum (around 500 BC).

Sulba Sutras, Vedic civilization in India, ∼800-500 BC.

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Irrationality of 2

Pythagoreas school Hippasus of Metapontum (around 500 BC).

Sulba Sutras, Vedic civilization in India, ∼800-500 BC.

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Pythagoreas school Hippasus of Metapontum (around 500 BC).

Sulba Sutras, Vedic civilization in India, ∼800-500 BC.

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Irrationality of 2

Classical proof of the irrationality of √ 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

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Classical proof of the irrationality of 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

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Irrationality of 2

Classical proof of the irrationality of √ 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

(41)

Classical proof of the irrationality of 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

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Irrationality of 2

Classical proof of the irrationality of √ 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

(43)

Classical proof of the irrationality of 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

(44)

Irrationality of 2

Classical proof of the irrationality of √ 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

(45)

Classical proof of the irrationality of 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

(46)

Irrationality of 2

Classical proof of the irrationality of √ 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

(47)

Classical proof of the irrationality of 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

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Irrationality of 2

Classical proof of the irrationality of √ 2 Assume √

2 = p/q with p and q without common factor.

Hence one at least of p, q is odd.

By definition of √

2 we have p

²

/q

²

= 2, which means p

²

= 2q

²

.

Hence p is even : write p = 2a.

Now p

²

= 4a

²

and 4a

²

= 2q

²

. This yields

2a

²

= q

²

and q is even !

Contradiction.

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•

Start with a rectangle have side length 1 and 1 + √ 2.

•

Decompose it into two squares with sides 1 and a smaller rectangle of sides 1 + √

2 − 2 = √

2 − 1 and 1.

•

This second small rectangle has side lenghts in the proportion

√ 1

2 − 1 = 1 + √ 2, which is the same as for the large one.

•

Hence the second small rectangle can be split into two squares and a third smaller rectangle, the sides of which are again in the same proportion.

•

This process does not end.

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Irrationality of 2 : geometric proof

•

Start with a rectangle have side length 1 and 1 + √ 2.

•

Decompose it into two squares with sides 1 and a smaller rectangle of sides 1 + √

2 − 2 = √

2 − 1 and 1.

•

This second small rectangle has side lenghts in the proportion

√ 1

2 − 1 = 1 + √ 2, which is the same as for the large one.

•

Hence the second small rectangle can be split into two squares and a third smaller rectangle, the sides of which are again in the same proportion.

•

This process does not end.

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•

Start with a rectangle have side length 1 and 1 + √ 2.

•

Decompose it into two squares with sides 1 and a smaller rectangle of sides 1 + √

2 − 2 = √

2 − 1 and 1.

•

This second small rectangle has side lenghts in the proportion

√ 1

2 − 1 = 1 + √ 2, which is the same as for the large one.

•

Hence the second small rectangle can be split into two squares and a third smaller rectangle, the sides of which are again in the same proportion.

•

This process does not end.

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Irrationality of 2 : geometric proof

•

Start with a rectangle have side length 1 and 1 + √ 2.

•

Decompose it into two squares with sides 1 and a smaller rectangle of sides 1 + √

2 − 2 = √

2 − 1 and 1.

•

This second small rectangle has side lenghts in the proportion

√ 1

2 − 1 = 1 + √ 2, which is the same as for the large one.

•

Hence the second small rectangle can be split into two squares and a third smaller rectangle, the sides of which are again in the same proportion.

•

This process does not end.

(53)

•

Start with a rectangle have side length 1 and 1 + √ 2.

•

Decompose it into two squares with sides 1 and a smaller rectangle of sides 1 + √

2 − 2 = √

2 − 1 and 1.

•

This second small rectangle has side lenghts in the proportion

√ 1

2 − 1 = 1 + √ 2, which is the same as for the large one.

•

Hence the second small rectangle can be split into two squares and a third smaller rectangle, the sides of which are again in the same proportion.

•

This process does not end.

(54)

Rectangles with proportion 1 + 2

(55)

If we start with a rectangle having integer side lengths, then this process stops after finitely may steps (the side lengths are positive decreasing integers).

Also for a rectangle with side lengths in a rational proportion, this process stops after finitely may steps (reduce to a common denominator and scale).

Hence 1 + √

2 is an irrational number, and √

2 also.

(56)

Irrationality of 2 : geometric proof

If we start with a rectangle having integer side lengths, then this process stops after finitely may steps (the side lengths are positive decreasing integers).

Also for a rectangle with side lengths in a rational proportion, this process stops after finitely may steps (reduce to a common denominator and scale).

Hence 1 + √

2 is an irrational number, and √

2 also.

(57)

If we start with a rectangle having integer side lengths, then this process stops after finitely may steps (the side lengths are positive decreasing integers).

Also for a rectangle with side lengths in a rational proportion, this process stops after finitely may steps (reduce to a common denominator and scale).

Hence 1 + √

2 is an irrational number, and √

2 also.

(58)

The fabulous destiny of 2

• Benoˆıt Rittaud, ´ Editions Le Pommier (2006).

http://www.math.univ-paris13.fr/∼rittaud/RacineDeDeux

(59)

√

2 = 1, 414 213 562 373 095 048 801 688 724 209 . . .

satisfies √

2 = 1 + 1

√ 2 + 1 · Hence

√

2 = 1 + 1 2 + 1

√ 2 + 1

= 1 + 1

2 + 1 2 + 1

. ..

We write the continued fraction expansion of √

2 using the shorter notation

√

2 = [1; 2, 2, 2, 2, 2, . . . ] = [1; 2].

(60)

The number

√

2 = 1, 414 213 562 373 095 048 801 688 724 209 . . .

satisfies √

2 = 1 + 1

√ 2 + 1 · Hence

√

2 = 1 + 1 2 + 1

√ 2 + 1

= 1 + 1

2 + 1 2 + 1

. ..

We write the continued fraction expansion of √

2 using the shorter notation

√

2 = [1; 2, 2, 2, 2, 2, . . . ] = [1; 2].

(61)

√

2 = 1, 414 213 562 373 095 048 801 688 724 209 . . .

satisfies √

2 = 1 + 1

√ 2 + 1 · Hence

√

2 = 1 + 1 2 + 1

√ 2 + 1

= 1 + 1

2 + 1 2 + 1

. ..

We write the continued fraction expansion of √

2 using the shorter notation

√

2 = [1; 2, 2, 2, 2, 2, . . . ] = [1; 2].

(62)

The number

√

2 = 1, 414 213 562 373 095 048 801 688 724 209 . . .

satisfies √

2 = 1 + 1

√ 2 + 1 · Hence

√

2 = 1 + 1 2 + 1

√ 2 + 1

= 1 + 1

2 + 1 2 + 1

. ..

We write the continued fraction expansion of √

2 using the shorter notation

√

2 = [1; 2, 2, 2, 2, 2, . . . ] = [1; 2].

(63)

√

2 = 1, 414 213 562 373 095 048 801 688 724 209 . . .

satisfies √

2 = 1 + 1

√ 2 + 1 · Hence

√

2 = 1 + 1 2 + 1

√ 2 + 1

= 1 + 1

2 + 1 2 + 1

. ..

We write the continued fraction expansion of √

2 using the shorter notation

√

2 = [1; 2, 2, 2, 2, 2, . . . ] = [1; 2].

(64)

• H.W. Lenstra Jr,

Solving the Pell Equation, Notices of the A.M.S.

49 (2) (2002) 182–192.

(65)

A real number is rational if and only if its continued fraction expansion is finite.

A real number is rational if and only if its binary (or decimal, or in any basis b ≥ 2) expansion is ultimately periodic.

Consequence : it should not be so difficult to decide whether a given number is rational or not.

To prove that certain numbers (occurring as constants in analysis) are irrational is most often an impossible

challenge. However to construct irrational (even

transcendental) numbers is easy.

(66)

A real number is rational if and only if its continued fraction expansion is finite.

A real number is rational if and only if its binary (or decimal, or in any basis b ≥ 2) expansion is ultimately periodic.

Consequence : it should not be so difficult to decide whether a given number is rational or not.

To prove that certain numbers (occurring as constants in analysis) are irrational is most often an impossible

challenge. However to construct irrational (even

transcendental) numbers is easy.

(67)

A real number is rational if and only if its continued fraction expansion is finite.

A real number is rational if and only if its binary (or decimal, or in any basis b ≥ 2) expansion is ultimately periodic.

Consequence : it should not be so difficult to decide whether a given number is rational or not.

To prove that certain numbers (occurring as constants in analysis) are irrational is most often an impossible

challenge. However to construct irrational (even

transcendental) numbers is easy.

(68)

A real number is rational if and only if its continued fraction expansion is finite.

A real number is rational if and only if its binary (or decimal, or in any basis b ≥ 2) expansion is ultimately periodic.

Consequence : it should not be so difficult to decide whether a given number is rational or not.

To prove that certain numbers (occurring as constants in analysis) are irrational is most often an impossible

challenge. However to construct irrational (even

transcendental) numbers is easy.

(69)

A real number is rational if and only if its continued fraction expansion is finite.

A real number is rational if and only if its binary (or decimal, or in any basis b ≥ 2) expansion is ultimately periodic.

Consequence : it should not be so difficult to decide whether a given number is rational or not.

To prove that certain numbers (occurring as constants in analysis) are irrational is most often an impossible

challenge. However to construct irrational (even

transcendental) numbers is easy.

(70)

Euler’s Constant is γ= lim

n→∞

1 + 1

2 + 1

3 + · · · + 1

n − log n

= 0, 577 215 664 901 532 860 606 512 090 082 . . . Is–it a rational number ?

γ=

∞

X

k=1

1 k − log

1 + 1

k

= Z

∞

1

1 [x] − 1

x

dx

= − Z

1

0

Z

1

0

(1 − x)dxdy (1 − xy) log(xy) ·

Recent work by J. Sondow inspired by the work of

F. Beukers on Ap´ ery’s proof.

(71)

Euler’s Constant is γ= lim

n→∞

1 + 1

2 + 1

3 + · · · + 1

n − log n

= 0, 577 215 664 901 532 860 606 512 090 082 . . . Is–it a rational number ?

γ=

∞

X

k=1

1 k − log

1 + 1

k

= Z

∞

1

1 [x] − 1

x

dx

= − Z

1

0

Z

1

0

(1 − x)dxdy (1 − xy) log(xy) ·

Recent work by J. Sondow inspired by the work of

F. Beukers on Ap´ ery’s proof.

(72)

Euler’s Constant is γ= lim

n→∞

1 + 1

2 + 1

3 + · · · + 1

n − log n

= 0, 577 215 664 901 532 860 606 512 090 082 . . . Is–it a rational number ?

γ=

∞

X

k=1

1 k − log

1 + 1

k

= Z

∞

1

1 [x] − 1

x

dx

= − Z

1

0

Z

1

0

(1 − x)dxdy (1 − xy) log(xy) ·

Recent work by J. Sondow inspired by the work of

F. Beukers on Ap´ ery’s proof.

(73)

ζ(s) = X

n≥1

1 n

^s

was studied by Euler (1707– 1783) for integer values of s

and by Riemann (1859) for complex values of s.

Euler : for any even integer value of s ≥ 2, the number ζ(s) is a rational multiple of π

^s

.

Examples : ζ(2) = π

²

/6, ζ(4) = π

⁴

/90, ζ(6) = π

⁶

/945, ζ(8) = π

⁸

/9450 · · ·

Denominators : Bernoulli numbers.

(74)

The function ζ(s) = X

n≥1

1 n

^s

was studied by Euler (1707– 1783) for integer values of s

and by Riemann (1859) for complex values of s.

Euler : for any even integer value of s ≥ 2, the number ζ(s) is a rational multiple of π

^s

.

Examples : ζ(2) = π

²

/6, ζ(4) = π

⁴

/90, ζ(6) = π

⁶

/945, ζ(8) = π

⁸

/9450 · · ·

Denominators : Bernoulli numbers.

(75)

ζ(s) = X

n≥1

1 n

^s

was studied by Euler (1707– 1783) for integer values of s

and by Riemann (1859) for complex values of s.

Euler : for any even integer value of s ≥ 2, the number ζ(s) is a rational multiple of π

^s

.

Examples : ζ(2) = π

²

/6, ζ(4) = π

⁴

/90, ζ(6) = π

⁶

/945, ζ(8) = π

⁸

/9450 · · ·

Denominators : Bernoulli numbers.

(76)

The function ζ(s) = X

n≥1

1 n

^s

was studied by Euler (1707– 1783) for integer values of s

and by Riemann (1859) for complex values of s.

Euler : for any even integer value of s ≥ 2, the number ζ(s) is a rational multiple of π

^s

.

Examples : ζ(2) = π

²

/6, ζ(4) = π

⁴

/90, ζ(6) = π

⁶

/945, ζ(8) = π

⁸

/9450 · · ·

Denominators : Bernoulli numbers.

(77)

Leonhard Euler

(15 Avril 1707 – 1783)

Introductio in analysin infinitorum

(78)

Euler :

1 − 1 + 1 − 1 + 1 − 1 + · · · = 1 2

1 + 1 + 1 + 1 + 1 + · · · = − 1 2

1 + 2 + 3 + 4 + 5 + · · · = − 1 12

1

²

+ 2

²

+ 3

²

+ 4

²

+ 5

²

+ · · · = 0.

(79)

Euler :

1 − 1 + 1 − 1 + 1 − 1 + · · · = 1 2

1 + 1 + 1 + 1 + 1 + · · · = − 1 2

1 + 2 + 3 + 4 + 5 + · · · = − 1 12

1

²

+ 2

²

+ 3

²

+ 4

²

+ 5

²

+ · · · = 0.

(80)

Euler :

1 − 1 + 1 − 1 + 1 − 1 + · · · = 1 2

1 + 1 + 1 + 1 + 1 + · · · = − 1 2

1 + 2 + 3 + 4 + 5 + · · · = − 1 12

1

²

+ 2

²

+ 3

²

+ 4

²

+ 5

²

+ · · · = 0.

(81)

Euler :

1 − 1 + 1 − 1 + 1 − 1 + · · · = 1 2

1 + 1 + 1 + 1 + 1 + · · · = − 1 2

1 + 2 + 3 + 4 + 5 + · · · = − 1 12

1

²

+ 2

²

+ 3

²

+ 4

²

+ 5

²

+ · · · = 0.

(82)

Euler :

1 − 1 + 1 − 1 + 1 − 1 + · · · = 1 2

1 + 1 + 1 + 1 + 1 + · · · = − 1 2

1 + 2 + 3 + 4 + 5 + · · · = − 1 12

1

²

+ 2

²

+ 3

²

+ 4

²

+ 5

²

+ · · · = 0.

(83)

Euler :

1 − 1 + 1 − 1 + 1 − 1 + · · · = 1 2

1 + 1 + 1 + 1 + 1 + · · · = − 1 2

1 + 2 + 3 + 4 + 5 + · · · = − 1 12

1

²

+ 2

²

+ 3

²

+ 4

²

+ 5

²

+ · · · = 0.

(84)

Euler :

1 − 1 + 1 − 1 + 1 − 1 + · · · = 1 2

1 + 1 + 1 + 1 + 1 + · · · = − 1 2

1 + 2 + 3 + 4 + 5 + · · · = − 1 12

1

²

+ 2

²

+ 3

²

+ 4

²

+ 5

²

+ · · · = 0.

(85)

Euler :

1 − 1 + 1 − 1 + 1 − 1 + · · · = 1 2

1 + 1 + 1 + 1 + 1 + · · · = − 1 2

1 + 2 + 3 + 4 + 5 + · · · = − 1 12

1

²

+ 2

²

+ 3

²

+ 4

²

+ 5

²

+ · · · = 0.

(86)

•

Let x be a real number and n an integer. Consider the sum of n terms

S

_n

(x) = x + x

²

+ x

³

+ · · · + x

ⁿ

.

•

For instance S

_n

(1) = n for all n.

•

Formula : pour x 6= 1,

S

_n

(x) = x − x

ⁿ⁺¹

1 − x ·

(87)

•

Let x be a real number and n an integer. Consider the sum of n terms

S

_n

(x) = x + x

²

+ x

³

+ · · · + x

ⁿ

.

•

For instance S

_n

(1) = n for all n.

•

Formula : pour x 6= 1,

S

_n

(x) = x − x

ⁿ⁺¹

1 − x ·

(88)

•

Let x be a real number and n an integer. Consider the sum of n terms

S

_n

(x) = x + x

²

+ x

³

+ · · · + x

ⁿ

.

•

For instance S

_n

(1) = n for all n.

•

Formula : pour x 6= 1,

S

_n

(x) = x − x

ⁿ⁺¹

1 − x ·

(89)

Proof

S

_n

(x) = x + x

²

+ x

³

+ · · · + x

ⁿ

Multiply by x :

xS

n

(x) = x

²

+ x

³

+ x

⁴

+ · · · + x

ⁿ⁺¹

Substract

(1 − x)S

_n

(x) = x − x

ⁿ⁺¹

Since x 6= 1 we may divide by 1 − x :

S

n

(x) = x − x

ⁿ⁺¹

1 − x ·

(90)

S _n (x) = x + x + x + · · · + x =

(x − x ⁿ⁺¹ )/(1 − x) pour x 6= 1

Proof

S

_n

(x) = x + x

²

+ x

³

+ · · · + x

ⁿ

Multiply by x :

xS

n

(x) = x

²

+ x

³

+ x

⁴

+ · · · + x

ⁿ⁺¹

Substract

(1 − x)S

_n

(x) = x − x

ⁿ⁺¹

Since x 6= 1 we may divide by 1 − x :

S

n

(x) = x − x

ⁿ⁺¹

1 − x ·

(91)

Proof

S

_n

(x) = x + x

²

+ x

³

+ · · · + x

ⁿ

Multiply by x :

xS

n

(x) = x

²

+ x

³

+ x

⁴

+ · · · + x

ⁿ⁺¹

Substract

(1 − x)S

_n

(x) = x − x

ⁿ⁺¹

Since x 6= 1 we may divide by 1 − x :

S

n

(x) = x − x

ⁿ⁺¹

1 − x ·

(92)

S _n (x) = x + x + x + · · · + x =

(x − x ⁿ⁺¹ )/(1 − x) pour x 6= 1

Proof

S

_n

(x) = x + x

²

+ x

³

+ · · · + x

ⁿ

Multiply by x :

xS

n

(x) = x

²

+ x

³

+ x

⁴

+ · · · + x

ⁿ⁺¹

Substract

(1 − x)S

_n

(x) = x − x

ⁿ⁺¹

Since x 6= 1 we may divide by 1 − x :

S

n

(x) = x − x

ⁿ⁺¹

1 − x ·

(93)

Proof

S

_n

(x) = x + x

²

+ x

³

+ · · · + x

ⁿ

Multiply by x :

xS

n

(x) = x

²

+ x

³

+ x

⁴

+ · · · + x

ⁿ⁺¹

Substract

(1 − x)S

_n

(x) = x − x

ⁿ⁺¹

Since x 6= 1 we may divide by 1 − x :

S

n

(x) = x − x

ⁿ⁺¹

1 − x ·

(94)

S _n (1/2) = 1/2 + 1/4 + 1/8 + · · · + 1/2 ⁿ

•

Take x = 1/2 : S

_n

(1/2) = 1

2 + 1 4 + 1

8 + 1

16 + · · · + 1 2

ⁿ

•

Replace x by 1/2 in

x + x

²

+ x

³

+ · · · + x

ⁿ

= x − x

ⁿ⁺¹

1 − x · Since 1 − x = 1/2 we obtain

S

n

(1/2) = 1 − 1

2

ⁿ

·

(95)

•

Take x = 1/2 : S

_n

(1/2) = 1

2 + 1 4 + 1

8 + 1

16 + · · · + 1 2

ⁿ

•

Replace x by 1/2 in

x + x

²

+ x

³

+ · · · + x

ⁿ

= x − x

ⁿ⁺¹

1 − x · Since 1 − x = 1/2 we obtain

S

n

(1/2) = 1 − 1

2

ⁿ

·

(96)

For −1 < x < 1 we have

1 + x + x

²

+ x

³

+ · · · = 1 + x

1 − x = 1 1 − x ·

The right hand side at x = −1 is 1/2. Hence the value 1/2 given by Euler to the infinite sum

S = 1 − 1 + 1 − 1 + · · ·

We obtain the same value S = 1/2 by writing S − 1 = −S

.

(97)

1 + x + x

²

+ x

³

+ · · · = 1 + x

1 − x = 1 1 − x ·

The right hand side at x = −1 is 1/2. Hence the value 1/2 given by Euler to the infinite sum

S = 1 − 1 + 1 − 1 + · · ·

We obtain the same value S = 1/2 by writing S − 1 = −S

.

(98)

For −1 < x < 1 we have

1 + x + x

²

+ x

³

+ · · · = 1 + x

1 − x = 1 1 − x ·

The right hand side at x = −1 is 1/2. Hence the value 1/2 given by Euler to the infinite sum

S = 1 − 1 + 1 − 1 + · · ·

We obtain the same value S = 1/2 by writing S − 1 = −S

.

(99)

1 + x + x

²

+ x

³

+ · · · = 1 + x

1 − x = 1 1 − x ·

The right hand side at x = −1 is 1/2. Hence the value 1/2 given by Euler to the infinite sum

S = 1 − 1 + 1 − 1 + · · ·

We obtain the same value S = 1/2 by writing S − 1 = −S

.

(100)

For −1 < x < 1 we have

1 + x + x

²

+ x

³