aarjaneiro/planetmath huggingface dataset:
unimodular lattice

unimodular lattice

A

lattice

L

in

ℝ

n

is

unimodular

if a

fundamental domain

(

http://planetmath.org/MinkowskisTheorem

) of

L

has volume 1. This is to

L

being unimodular as a

ℤ

-module with the

bilinear form

from

ℝ

n

.

Title

unimodular lattice

Canonical name

UnimodularLattice

Date of creation

2013-03-22 18:43:26

Last modified on

2013-03-22 18:43:26

Owner

monster (22721)

Last modified by

monster (22721)

Numerical id

4

Author

monster (22721)

Entry type

Definition

Classification

msc 11H06

Generated on Fri Feb 9 15:34:59 2018 by

LaTeXML
using Minkowski’s constant to find a class number

using Minkowski’s constant to find a class number

We will use the

theorem

of Minkowski (see the parent entry (

http://planetmath.org/MinkowskisConstant

)).

Theorem

(Minkowski’s Theorem).

Let

K

be a

number field

and let

D

K

be its

discriminant

. Let

n

=

r

1

+

2

⁢

r

2

be the degree of

K

over

Q

, where

r

1

and

r

2

are the number of

real and complex embeddings

, respectively. The

class group

of

K

is denoted by

Cl

⁡

(

K

)

. In any ideal class

C

∈

Cl

⁡

(

K

)

, there exists an ideal

A

∈

C

such that:

|

𝐍

⁢

(

𝔄

)

|

≤

M

K

⁢

|

D

K

|

where

N

⁢

(

A

)

denotes the

absolute norm

of

A

and

M

K

=

n

!

n

n

⁢

(

4

π

)

r

2

.

Example 1

.

The discriminants of the

quadratic fields

K

2

=

ℚ

⁢

(

2

)

,

K

3

=

ℚ

⁢

(

3

)

and

K

13

=

ℚ

⁢

(

13

)

are

D

K

2

=

8

,

D

K

3

=

12

and

D

K

13

=

13

respectively. For all three

n

=

2

=

r

1

and

r

2

=

0

. Therefore, the Minkowski’s

constants

are:

M

K

i

=

1

2

⁢

|

D

K

i

|

,

i

=

2

,

3

,

13

so in the three cases:

M

K

i

≤

1

2

⁢

13

=

1.802

⁢

…

Now, suppose that

C

is an arbitrary class in

Cl

⁡

(

K

i

)

. By the theorem, there exists an ideal

𝔄

, representative of

C

, such that:

|

𝐍

⁢

(

𝔄

)

|

<

1.802

⁢

…

<

2

and therefore

𝐍

⁢

(

𝔄

)

=

1

. Since the only ideal of norm one is the

trivial

ideal

𝒪

K

i

, which is principal, the class

C

is also the trivial class in

Cl

⁡

(

K

i

)

. Hence there is only one class in the class group, and the

class number

is one for the three fields

K

2

,

K

3

and

K

13

.

Example 2

.

Let

K

=

ℚ

⁢

(

17

)

. The discriminant is

D

K

=

17

and the Minkowski’s bound reads:

M

K

=

1

2

⁢

17

=

2.06

⁢

…

Suppose that

C

is an arbitrary class in

Cl

⁡

(

K

)

. By the theorem, there exists an ideal

𝔄

, representative of

C

, such that:

|

𝐍

⁢

(

𝔄

)

|

<

2.06

⁢

…

and therefore

𝐍

⁢

(

𝔄

)

=

1

or

2

. However,

2

=

-

3

+

17

2

⋅

3

+

17

2

so the ideal

2

⁢

𝒪

K

is split in

K

and the

prime ideals

(

-

3

+

17

2

)

,

(

3

+

17

2

)

are the only ones of norm

2

. Since they are principal, the class

C

is the trivial class, and the class group

Cl

⁡

(

K

)

is trivial. Hence, the class number of

ℚ

⁢

(

17

)

is one.

Title

using Minkowski’s constant to find a class number

Canonical name

UsingMinkowskisConstantToFindAClassNumber

Date of creation

2013-03-22 15:05:38

Last modified on

2013-03-22 15:05:38

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Example

Classification

msc 11H06

Classification

msc 11R29

Related topic

ClassNumbersAndDiscriminantsTopicsOnClassGroups

Generated on Fri Feb 9 15:35:01 2018 by

LaTeXML
Dirichlet’s approximation theorem

Dirichlet’s approximation theorem

Theorem

(Dirichlet, c. 1840): For any

real number

θ

and any

integer

n

≥

1

, there exist integers

a

and

b

such that

1

≤

a

≤

n

and

|

a

⁢

θ

-

b

|

≤

1

n

+

1

.

Proof: We may assume

n

≥

2

.
For each integer

a

in the

interval

[

1

,

n

]

, write

r

a

=

a

⁢

θ

-

[

a

⁢

θ

]

∈

[

0

,

1

)

, where

[

x

]

denotes
the greatest integer less than

x

. Since the

n

+

2

numbers

0

,

r

a

,

1

all lie in the same

unit interval

, some two
of them differ (in

absolute value

) by at most

1

n

+

1

.
If

0

or

1

is in any such pair, then the other element of the
pair is one of the

r

a

, and we are done.
If not, then

0

≤

r

k

-

r

l

≤

1

n

+

1

for some distinct

k

and

l

. If

k

>

l

we have

r

k

-

r

l

=

r

k

-

l

, since each side is in

[

0

,

1

)

and the

difference

between them is an integer. Similarly,
if

k

<

l

, we have

1

-

(

r

k

-

r

l

)

=

r

l

-

k

. So, with

a

=

k

-

l

or

a

=

l

-

k

respectively, we get

|

r

a

-

c

|

≤

1

n

+

1

where

c

is

0

or

1

, and the result follows.

It is clear that we can add the condition

gcd

⁡

(

a

,

b

)

=

1

to the

conclusion

.

The same statement, but with the weaker conclusion

|

a

⁢

θ

-

b

|

<

1

n

,
admits a slightly shorter proof, and is sometimes also referred to
as the

Dirichlet approximation theorem

. (It was that shorter proof
which made the “

pigeonhole principle

” famous.) Also, the theorem
is sometimes restricted to

irrational

values of

θ

, with the
(nominally stronger) conclusion

|

a

⁢

θ

-

b

|

<

1

n

+

1

.

Title

Dirichlet’s

approximation theorem

Canonical name

DirichletsApproximationTheorem

Date of creation

2013-03-22 13:15:37

Last modified on

2013-03-22 13:15:37

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

7

Author

Koro (127)

Entry type

Theorem

Classification

msc 11J04

Synonym

Dirichlet approximation theorem

Related topic

IrrationalityMeasure

Generated on Fri Feb 9 15:35:03 2018 by

LaTeXML
Hurwitz equation

Hurwitz equation

Given a length

k

and a multiplier

m

, a

Hurwitz equation

for a set of integers

x

1

,

x

2

,

…

,

x

k

states that

∑

i

=

1

k

x

i

2

=

m

⁢

∏

i

=

1

k

x

i

.

There is no solution to the equation if

m

>

k

. Solutions can sometimes be found for

m

≤

k

by starting from a

k

-long set of 1s. The

Markov equation

for the

Markov numbers

has

k

=

m

=

3

.

References

1

R. K. Guy,

Unsolved Problems in

Number Theory

New York: Springer-Verlag 2004: D12

Title

Hurwitz equation

Canonical

name

HurwitzEquation

Date of creation

2013-03-22 17:26:13

Last modified on

2013-03-22 17:26:13

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

4

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 11J06

Classification

msc 11D72

Generated on Fri Feb 9 15:35:05 2018 by

LaTeXML
uniqueness conjecture for Markov numbers

uniqueness conjecture for Markov numbers

Conjecture

. Given a

Markov number

z

>

1

, there are several other Markov numbers

x

and

y

such that

x

2

+

y

2

+

z

2

=

3

⁢

x

⁢

y

⁢

z

, but there is only set of values of

x

and

y

satisfying the inequality

z

>

y

≥

x

.

The conjecture is easy enough to check for small values. For example, for

z

=

5

, we could even test

x

and

y

we know not to be Markov numbers with minimum increase in computational overhead. Trying the triples in order: (1, 1, 5), (1, 2, 5), (1, 3, 5), (1, 4, 5), (2, 1, 5), … (4, 4, 5) against

15

⁢

x

⁢

y

-

(

x

2

+

y

2

+

25

)

we obtain the

sequence

-

12

, 0, 10, 18, 0, 27, 52, 75, 10, 52, 92, 130, 18, 75, 130, 183. It doesn’t take significantly larger Markov numbers to show the need for a general proof of uniqueness. Many attempted proofs have been submitted, but Richard Guy dismisses them all as seemingly faulty.

A divide-and-conquer approach to the problem has yielded encouraging results, however. Baragar proved the uniqueness of prime Markov numbers

p

as well as

semiprimes

2

⁢

p

. Schmutz then proved the uniqueness of Markov numbers of the forms

p

n

and

2

⁢

p

n

. Ying Zhang used these results to extend this to

4

⁢

p

n

and

8

⁢

p

n

.

References

1

R. K. Guy,

Unsolved Problems in

Number Theory

New York: Springer-Verlag 2004: D12

2

Ying Zhang, “

Congruence

and Uniqueness of Certain Markov Numbers”

Acta Arithmetica

128

3 (2007): 297

Title

uniqueness conjecture for Markov numbers

Canonical

name

UniquenessConjectureForMarkovNumbers

Date of creation

2013-03-22 17:26:16

Last modified on

2013-03-22 17:26:16

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

5

Author

PrimeFan (13766)

Entry type

Conjecture

Classification

msc 11J06

Synonym

unicity conjecture for Markov numbers

Generated on Fri Feb 9 15:35:07 2018 by

LaTeXML
Davenport-Schmidt theorem

Davenport-Schmidt theorem

For any real

ξ

which is not

rational

or quadratic

irrational

, there are infinitely many rational or real quadratic irrational

α

which satisfy

∣

ξ

-

α

∣

<

C

⋅

H

⁢

(

α

)

-

3

,

where

C

=

{

C

0

,

if

⁢

∣

ξ

∣

<

1

,

C

0

⋅

ξ

2

,

if

⁢

∣

ξ

∣

>

1

.

C

0

is any fixed number

greater

than

160

9

and

H

⁢

(

α

)

is the of

α

.

[

DS

]

The

of the rational or quadratic irrational number

α

is

H

⁢

(

α

)

=

max

⁡

(

|

x

|

,

|

y

|

,

|

z

|

)

where

x

,

y

,

z

are from the unique equation

x

⁢

α

2

+

y

⁢

α

+

z

=

0

such that

x

,

y

,

z

are not all zero relatively prime

integral

coefficients

.

[

DS

]

References

DS

Davenport, H. Schmidt, M. Wolfgang: Approximation to

real numbers

by quadratic irrationals. Acta Arithmetica XIII, 1967.

Title

Davenport-Schmidt theorem

Canonical name

DavenportSchmidtTheorem

Date of creation

2013-03-22 13:32:58

Last modified on

2013-03-22 13:32:58

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

9

Author

Daume (40)

Entry type

Theorem

Classification

msc 11J68

Generated on Fri Feb 9 15:35:08 2018 by

LaTeXML
Liouville approximation theorem

Liouville approximation theorem

Given

α

, a real

algebraic number

of degree

n

≠

1

, there is a

constant

c

=

c

⁢

(

α

)

>

0

such that for all

rational numbers

p

/

q

,

(

p

,

q

)

=

1

, the inequality

|

α

-

p

q

|

>

c

⁢

(

α

)

q

n

holds.

Many mathematicians have worked at strengthening this

theorem

:

•

Thue: If

α

is an algebraic number of degree

n

≥

3

, then there is a constant

c

0

=

c

0

⁢

(

α

,

ϵ

)

>

0

such that for all rational numbers

p

/

q

, the inequality

|

α

-

p

q

|

>

c

0

⁢

q

-

1

-

ϵ

-

n

/

2

holds.

•

Siegel: If

α

is an algebraic number of degree

n

≥

2

, then there is a constant

c

1

=

c

1

⁢

(

α

,

ϵ

)

>

0

such that for all rational numbers

p

/

q

, the inequality

|

α

-

p

q

|

>

c

1

⁢

q

-

λ

,

λ

=

min

t

=

1

,

…

,

n

⁡

(

n

t

+

1

+

t

)

+

ϵ

holds.

•

Dyson: If

α

is an algebraic number of degree

n

>

3

, then there is a constant

c

2

=

c

2

⁢

(

α

,

ϵ

)

>

0

such that for all rational numbers

p

/

q

with

q

>

c

2

, the inequality

|

α

-

p

q

|

>

q

-

2

⁢

n

-

ϵ

holds.

•

Roth: If

α

is an

irrational

algebraic number and

ϵ

>

0

, then there is a constant

c

3

=

c

3

⁢

(

α

,

ϵ

)

>

0

such that for all rational numbers

p

/

q

, the inequality

|

α

-

p

q

|

>

c

3

⁢

q

-

2

-

ϵ

holds.

Title

Liouville approximation theorem

Canonical name

LiouvilleApproximationTheorem

Date of creation

2013-03-22 11:45:45

Last modified on

2013-03-22 11:45:45

Owner

KimJ (5)

Last modified by

KimJ (5)

Numerical id

13

Author

KimJ (5)

Entry type

Theorem

Classification

msc 11J68

Classification

msc 46-01

Classification

msc 46N40

Related topic

ExampleOfTranscendentalNumber

Generated on Fri Feb 9 15:35:10 2018 by

LaTeXML
proof of Liouville approximation theorem

proof of Liouville approximation theorem

Let

α

satisfy the equation

f

⁢

(

α

)

=

a

n

⁢

α

n

+

a

n

-

1

⁢

α

n

-

1

+

…

+

a

0

=

0

where the

a

i

are

integers

. Choose

M

such that

M

>

max

α

-

1

≤

x

≤

α

+

1

⁡

|

f

′

⁢

(

x

)

|

.

Suppose

p

q

lies in

(

α

-

1

,

α

+

1

)

and

f

⁢

(

p

q

)

≠

0

.

|

f

⁢

(

p

q

)

|

=

|

a

n

⁢

p

n

+

a

n

-

1

⁢

p

n

-

1

⁢

q

+

…

+

a

0

⁢

q

n

|

q

n

≥

1

q

n

since the

numerator

is a non-zero integer.

By the

mean-value theorem

1

q

n

≤

|

f

⁢

(

p

q

)

-

f

⁢

(

α

)

|

=

|

(

p

q

-

α

)

⁢

f

′

⁢

(

x

)

|

<

M

⁢

|

p

q

-

α

|

.

Title

proof of Liouville approximation theorem

Canonical name

ProofOfLiouvilleApproximationTheorem

Date of creation

2013-03-22 13:19:22

Last modified on

2013-03-22 13:19:22

Owner

lieven (1075)

Last modified by

lieven (1075)

Numerical id

6

Author

lieven (1075)

Entry type

Proof

Classification

msc 11J68

Generated on Fri Feb 9 15:35:12 2018 by

LaTeXML
alternative proof that sqrt(2) is irrational

alternative proof that

2

is irrational

Following is a proof that

2

is

irrational

.

The

polynomial

x

2

-

2

is

irreducible

over

ℤ

by Eisenstein’s criterion with

p

=

2

. Thus,

x

2

-

2

is irreducible over

ℚ

by Gauss’s lemma (

http://planetmath.org/GausssLemmaII

). Therefore,

x

2

-

2

does not have any roots in

ℚ

. Since

2

is a root of

x

2

-

2

, it must be irrational.

This method generalizes to show that any number of the form

n

r

is not

rational

, where

r

∈

ℤ

with

r

>

1

and

n

∈

ℤ

such that there exists a prime

p

dividing

n

with

p

2

not dividing

n

.

Title

alternative proof that

2

is irrational

Canonical name

AlternativeProofThatsqrt2IsIrrational

Date of creation

2013-03-22 16:55:15

Last modified on

2013-03-22 16:55:15

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

8

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 11J72

Classification

msc 12E05

Classification

msc 11J82

Classification

msc 13A05

Related topic

Irrational

Related topic

EisensteinCriterion

Related topic

GausssLemmaII

Generated on Fri Feb 9 15:35:15 2018 by

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e is not a quadratic irrational

e is not a quadratic irrational

We wish to show that

e

is not a quadratic

irrational

, i.e.

ℚ

⁢

(

e

)

is not a

quadratic extension

of

ℚ

. To do this, we show that it can not be the root of any quadratic polynomial with

integer

coefficients

.

We begin by looking at the Taylor series for

e

x

:

e

x

=

∑

k

=

0

∞

x

k

k

!

.

This

converges

for every

x

∈

ℝ

, so

e

=

∑

k

=

0

∞

1

k

!

and

e

-

1

=

∑

k

=

0

∞

(

-

1

)

k

⁢

1

k

!

. Arguing
by

contradiction

, assume

a

⁢

e

2

+

b

⁢

e

+

c

=

0

for integers

a

,

b

and

c

. That is the same as

a

⁢

e

+

b

+

c

⁢

e

-

1

=

0

.

Fix

n

>

|

a

|

+

|

c

|

, then

a

,

c

∣

n

!

and

∀

k

≤

n

,

k

!

∣

n

!

.
Consider

0

=

n

!

⁢

(

a

⁢

e

+

b

+

c

⁢

e

-

1

)

=

a

⁢

n

!

⁢

∑

k

=

0

∞

1

k

!

+

b

+

c

⁢

n

!

⁢

∑

k

=

0

∞

(

-

1

)

k

⁢

1

k

!

=

b

+

∑

k

=

0

n

(

a

+

c

⁢

(

-

1

)

k

)

⁢

n

!

k

!

+

∑

k

=

n

+

1

∞

(

a

+

c

⁢

(

-

1

)

k

)

⁢

n

!

k

!

Since

k

!

∣

n

!

for

k

≤

n

, the first two terms are integers. So the third term should be an integer. However,

|

∑

k

=

n

+

1

∞

(

a

+

c

⁢

(

-

1

)

k

)

⁢

n

!

k

!

|

≤

(

|

a

|

+

|

c

|

)

⁢

∑

k

=

n

+

1

∞

n

!

k

!

=

(

|

a

|

+

|

c

|

)

⁢

∑

k

=

n

+

1

∞

1

(

n

+

1

)

⁢

(

n

+

2

)

⁢

⋯

⁢

k

≤

(

|

a

|

+

|

c

|

)

⁢

∑

k

=

n

+

1

∞

(

n

+

1

)

n

-

k

=

(

|

a

|

+

|

c

|

)

⁢

∑

t

=

1

∞

(

n

+

1

)

-

t

=

(

|

a

|

+

|

c

|

)

⁢

1

n

is less than

1

by our

assumption

that

n

>

|

a

|

+

|

c

|

. Since there is only one integer which is less than

1

in

absolute value

, this means that

∑

k

=

n

+

1

∞

(

a

+

c

⁢

(

-

1

)

k

)

⁢

1

k

!

=

0

for every sufficiently large

n

which is not the case because

∑

k

=

n

+

1

∞

(

a

+

c

⁢

(

-

1

)

k

)

⁢

1

k

!

-

∑

k

=

n

+

2

∞

(

a

+

c

⁢

(

-

1

)

k

)

⁢

1

k

!

=

(

a

+

c

⁢

(

-

1

)

n

+

1

)

⁢

1

(

n

+

1

)

!

is not identically zero. The contradiction

completes

the proof.

Title

e is not a quadratic irrational

Canonical name

EIsNotAQuadraticIrrational

Date of creation

2013-03-22 14:04:06

Last modified on

2013-03-22 14:04:06

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

11

Author

mathcam (2727)

Entry type

Proof

Classification

msc 11J72

Classification

msc 26E99

Related topic

EIsIrrationalProof

Related topic

ErIsIrrationalForRinmathbbQsetminus0

Related topic

EIsTranscendental

Generated on Fri Feb 9 15:35:18 2018 by

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proof that sqrt(2) is irrational

proof that

2

is irrational

Assume that the

square root

of

2

(

http://planetmath.org/SquareRootOf2

) is

rational

. Then we can write

2

=

a

b

,

where

a

,

b

∈

ℕ

and

a

and

b

are relatively prime. Then

2

=

(

2

)

2

=

(

a

b

)

2

=

a

2

b

2

. Thus,

a

2

=

2

⁢

b

2

. Therefore,

2

∣

a

2

. Since

2

is prime, it must divide

a

. Then

a

=

2

⁢

c

for some

c

∈

ℕ

. Thus,

2

⁢

b

2

=

a

2

=

(

2

⁢

c

)

2

=

4

⁢

c

2

, yielding that

b

2

=

2

⁢

c

2

. Therefore,

2

∣

b

2

. Since

2

is prime, it must divide

b

.

Since

2

∣

a

and

2

∣

b

, we have that

a

and

b

are not relatively prime, which contradicts the

hypothesis

. Hence, the initial

assumption

is false. It follows

2

is

irrational

.

With a little bit of work, this

argument

can be generalized to any

positive integer

that is not a square. Let

n

be such an

integer

. Then there must exist a prime

p

and

k

,

m

∈

ℕ

such that

n

=

p

k

⁢

m

, where

p

∤

m

and

k

is odd. Assume that

n

=

a

/

b

, where

a

,

b

∈

ℕ

and are relatively prime. Then

p

k

⁢

m

=

n

=

(

n

)

2

=

(

a

b

)

2

=

a

2

b

2

. Thus,

a

2

=

p

k

⁢

m

⁢

b

2

. From the

fundamental theorem of arithmetic

, it is clear that the maximum powers of

p

that divides

a

2

and

b

2

are even. Since

k

is odd and

p

does not divide

m

, the maximum power of

p

that divides

p

k

⁢

m

⁢

b

2

is also odd. Thus, the same should be true for

a

2

. Hence, we have reached a

contradiction

and

n

must be irrational.

The same argument can be generalized even more, for example to the case of nonsquare irreducible fractions and to higher order roots.

Title

proof that

2

is irrational

Canonical name

ProofThatsqrt2IsIrrational

Date of creation

2013-03-22 12:39:13

Last modified on

2013-03-22 12:39:13

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

11

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 11J72

Related topic

Irrational

Related topic

Surd

Generated on Fri Feb 9 15:35:23 2018 by

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rational and irrational

rational and irrational

The sum,

difference

,

product

(

http://

planetmath

.org/Ring

) and

quotient

of two non-zero

real numbers

, from which one is

rational

and the other

irrational

, is irrational.

Proof.

Let

a

be a rational and

α

irrational number.  Here we prove only that

a

α

is irrational — the other cases are

similar

.  If

a

α

were a

rational number

r

≠

0

,  then also

α

=

a

⁢

r

-

1

would be rational as a product of two rationals.  This

contradiction

shows that

a

α

is irrational.

Note.

In the result, the words real,

rational and irrational

may be replaced resp. by the words

complex

,

algebraic

and transcendental or resp. by the words complex, real and (the last here meaning, as commonly in Continental Europe, a

complex number

having non-zero

imaginary part

).

Title

rational and irrational

Canonical

name

RationalAndIrrational

Date of creation

2013-03-22 14:58:33

Last modified on

2013-03-22 14:58:33

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Result

Classification

msc 11J72

Classification

msc 11J82

Related topic

ExamplesOfPeriodicFunctions

Related topic

CommensurableNumbers

Related topic

SolutionsOfXyYx

Generated on Fri Feb 9 15:35:24 2018 by

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root(2,n) is irrational for n≥3 (proof using Fermat’s last theorem)

2

n

is irrational for

n

≥

3

(proof using Fermat’s last theorem)

Theorem 1

.

If

n

≥

3

, then

2

n

is

irrational

.

The below proof can be seen as an example of a

pathological

proof.
It gives no information to “why” the result holds, or
how non-trivial the result is.
Yet, assuming Wiles’ proof does not use the above

theorem

anywhere,
it proves the statement. Otherwise, the below proof would be an
example of a

circular argument

.

Proof.

Suppose

2

n

=

a

/

b

for some

positive integers

a

,

b

. It follows that

2

=

a

n

/

b

n

, or

b

n

+

b

n

=

a

n

.

(1)

We can now apply a recent result of Andrew Wiles

[

1

]

,
which states that there are no non-zero

integers

a

,

b

satisfying equation (

1

).
Thus

2

n

is irrational.
∎

The above proof is given in

[

2

]

, where it is attributed to W.H. Schultz.

References

1

A. Wiles,

Modular elliptic curves and Fermat’s last theorem

,
Annals of Mathematics, Volume 141, No. 3 May, 1995, 443-551.

2

W.H. Schultz,

An observation

,
American Mathematical Monthly, Vol. 110, Nr. 5, May 2003.
(submitted by R. Ehrenborg).

Title

2

n

is irrational for

n

≥

3

(proof using Fermat’s last theorem)

Canonical

name

sqrtn2IsIrrationalForNge3proofUsingFermatsLastTheorem

Date of creation

2013-03-22 13:38:32

Last modified on

2013-03-22 13:38:32

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

13

Author

matte (1858)

Entry type

Proof

Classification

msc 11J72

Generated on Fri Feb 9 15:35:26 2018 by

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example of four exponentials conjecture

example of four exponentials conjecture

Taking

x

1

=

i

⁢

π

,

x

2

=

i

⁢

π

⁢

2

,

y

1

=

1

,

y

2

=

2

, we see that this conjecture implies that one of

e

i

⁢

π

,

e

i

⁢

π

⁢

2

, or

e

i

⁢

2

⁢

π

is

transcendental

. Since the first is

-

1

and the last is

1

, the conjecture states that second must be transcendental, that is,

e

i

⁢

π

⁢

2

is (conjecturally) transcendental.

In this particular case, the result is known already, so the conjecture is verified. Using Gelfond’s theorem, take

α

=

e

i

⁢

π

and

β

=

2

and it follows that

α

β

is transcendental.

Title

example of

four exponentials conjecture

Canonical name

ExampleOfFourExponentialsConjecture

Date of creation

2013-03-22 14:09:09

Last modified on

2013-03-22 14:09:09

Owner

archibal (4430)

Last modified by

archibal (4430)

Numerical id

6

Author

archibal (4430)

Entry type

Example

Classification

msc 11J81

Generated on Fri Feb 9 15:35:28 2018 by

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four exponentials conjecture

four exponentials conjecture

:

Given four

complex numbers

x

1

,

x

2

,

y

1

,

y

2

, either

x

1

/

x

2

or

y

1

/

y

2

is

rational

, or one of the four
numbers

exp

⁡

(

x

i

⁢

y

j

)

is

transcendental

.

This

conjecture

is stronger than the six

exponentials

theorem

.

References

1

Waldschmidt, Michel,

Diophantine approximation on

linear algebraic groups

.
Transcendence

properties

of the

exponential function

in several

variables

. Grundlehren der Mathematischen
Wissenschaften
[Fundamental Principles of Mathematical Sciences], 326. Springer-Verlag, Berlin, 2000. xxiv+633 pp.
ISBN 3-540-66785-7.

Title

four exponentials conjecture

Canonical

name

FourExponentialsConjecture

Date of creation

2013-03-22 13:40:51

Last modified on

2013-03-22 13:40:51

Owner

Kevin OBryant (1315)

Last modified by

Kevin OBryant (1315)

Numerical id

7

Author

Kevin OBryant (1315)

Entry type

Conjecture

Classification

msc 11J81

Synonym

4 exponentials

Related topic

sixexponentialstheorem

Related topic

exponentialfunction

Related topic

ExponentialFunction

Related topic

SixExponentialsTheorem

Generated on Fri Feb 9 15:35:30 2018 by

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