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Gelfond–Schneider constant

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The Gelfond–Schneider constant, named after Aleksandr Gelfond and Theodor Schneider, is

2 
2√  2
.

The Gelfond–Schneider constant was proved to be a transcendental number by Rodion Kuzmin in 1930.

Decimal expansion of the Gelfond–Schneider constant

The decimal expansion of the Gelfond–Schneider constant is

2 
2√  2
  ≈  2.6651441426902251886502972498731398482742113137146594928359795933649204461787059548676091800051964169419893638542353...

giving the sequence of decimal digits (see A007507)

{2, 6, 6, 5, 1, 4, 4, 1, 4, 2, 6, 9, 0, 2, 2, 5, 1, 8, 8, 6, 5, 0, 2, 9, 7, 2, 4, 9, 8, 7, 3, 1, 3, 9, 8, 4, 8, 2, 7, 4, 2, 1, 1, 3, 1, 3, 7, 1, 4, 6, 5, 9, 4, 9, 2, 8, 3, 5, 9, 7, 9, 5, 9, 3, 3, 6, 4, 9, 2, 0, 4, 4, 6, 1, 7, 8, 7, 0, 5, 9, 5, 4, ...}

Binary expansion of the Gelfond–Schneider constant

The binary expansion of the Gelfond–Schneider constant is

2 
2√  2
  ≈  10.1010101001000110111000101111001111111011000000000110001011100011000101101100011000101110110111100100000111001000001...2

giving the sequence of binary digits (see A??????)

{1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, ...}

Continued fraction expansion of the Gelfond–Schneider constant

The continued fraction expansion of the Gelfond–Schneider constant is

     
2 
2√  2
  =  2 + 
1
1 + 
1
1 + 
1
1 + 
1
72 + 
1
3 + 
1
4 + 
1
1 + 
1
⋱

giving the sequence of integer part and partial denominators (see A062979)

{2, 1, 1, 1, 72, 3, 4, 1, 3, 2, 1, 1, 1, 14, 1, 2, 1, 1, 3, 1, 3, 1, 2, 1, 1, 2, 1, 1, 1, 2, 9, 1, 2, 1, 4, 1, 1, 6, 4, 8, 1, 6, 2, 1, 1, 1, 1, 1, 5, 1, 6, 1, 1, 2, 2, 6, 68, 1, 3, 3, 4, 10, 8, 4, 1, 16, 10, 1, 1, 3, 1, 25, 2, 3, 2, 1, 3, 6, 2, ...}

Square root of the Gelfond–Schneider constant

The square root of the Gelfond–Schneider constant gives

2√  2 
2√  2
 
 =  (2 
2√  2
 )  
1
2
  =  (2 
1
2
) 
2√  2
  = 
2√  2
 
2√  2
,
2√  2 
2√  2
 
 =  2 
2√  2
2
  =  2 
1
2√  2
  = 
2√  2
√  2
.

The square root of the Gelfond–Schneider constant is a transcendental number.

This shows that an irrational number (transcendental number) to the power of an irrational number (irrational algebraic number) can sometimes produce a rational number, since this number raised to the power of
2√  2
 is equal to 2:
(
2√  2
 
2√  2
 ) 
2√  2
  =  (
2√  2
 ) 2  =  2.

Decimal expansion of the square root of the Gelfond–Schneider constant

The decimal expansion of the square root of the Gelfond–Schneider constant is

2√  2 
2√  2
 
 = 
2√  2
 
2√  2
  = 
2√  2
√  2
 ≈  1.6325269194381528447734953810247196020791088570531141172477806843830352059986166422478555075066260414...

giving the sequence of decimal digits (see A078333)

{1, 6, 3, 2, 5, 2, 6, 9, 1, 9, 4, 3, 8, 1, 5, 2, 8, 4, 4, 7, 7, 3, 4, 9, 5, 3, 8, 1, 0, 2, 4, 7, 1, 9, 6, 0, 2, 0, 7, 9, 1, 0, 8, 8, 5, 7, 0, 5, 3, 1, 1, 4, 1, 1, 7, 2, 4, 7, 7, 8, 0, 6, 8, 4, 3, 8, 3, 0, 3, 5, 2, 0, 5, 9, 9, 8, 6, 1, 6, 6, 4, 2, ...}

Binary expansion of the square root of the Gelfond–Schneider constant

The binary expansion of the square root of the Gelfond–Schneider constant is

2√  2 
2√  2
 
 = 
2√  2
 
2√  2
  = 
2√  2
√  2
 ≈  1.1010000111101101010010001100000011010011100101010000111001010001001111101001000110000111101010010101...2

giving the sequence of binary digits (see A??????)

{1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, ...}

n 
n √  n

The Gelfond–Schneider constant may be viewed as the following special case:

n 
n √  n
, n  =  2.

n-th root of n 
n √  n

n√  n 
n√  n
 
 =  (n 
n√  n
 )  1 / n  =  (n 1 / n ) 
n√  n
  =  ( 
n√  n
 ) 
n√  n
,
where
n, n   ≥   2,
 is a positive integer.
This shows that an irrational number (transcendental number) to the power of an irrational number (irrational algebraic number) can sometimes produce a rational number, since this number raised to the power of
(
n √  n
  )  n  − 1
 is equal to the positive integer
n
:
( 
n√  n
 )  
n√  n
 ( 
n√  n
 )  n − 1  =  ( 
n√  n
 ) ( 
n√  n
 )  n  =  ( 
n√  n
 )  n  =  n.

Also,

( 
n√  n
 )  ( 
n√  n
 ) k ( 
n√  n
 )  n − k  =  ( 
n√  n
 ) ( 
n√  n
 )  n  =  ( 
n√  n
 )  n  =  n, 1 ≤ k ≤ n − 1.

bn, k  ^  en, k = n

Transcendental numbers (red background excepted*)
bn, k

n = 2
2√  2
  ^  
2√  2
3
3√  3
  ^  
3√  3
3√  3
  ^   (
3√  3
) 2
4
4√  4
  ^  
4√  4
4√  4
  ^   (
4√  4
) 2
4√  4
  ^   (
4√  4
) 3
5
5√  5
  ^  
5√  5
5√  5
  ^   (
5√  5
) 2
5√  5
  ^   (
5√  5
) 3
5√  5
  ^   (
5√  5
) 4
6
6√  6
  ^  
6√  6
6√  6
  ^   (
6√  6
) 2
6√  6
  ^   (
6√  6
) 3
6√  6
  ^   (
6√  6
) 4
6√  6
  ^   (
6√  6
) 5
7
7√  7
  ^  
7√  7
7√  7
  ^   (
7√  7
) 2
7√  7
  ^   (
7√  7
) 3
7√  7
  ^   (
7√  7
) 4
7√  7
  ^   (
7√  7
) 5
7√  7
  ^   (
7√  7
) 6
8
8√  8
  ^  
8√  8
8√  8
  ^   (
8√  8
) 2
8√  8
  ^   (
8√  8
) 3
8√  8
  ^   (
8√  8
) 4
8√  8
  ^   (
8√  8
) 5
8√  8
  ^   (
8√  8
) 6
8√  8
  ^   (
8√  8
) 7
k = 1 2 3 4 5 6 7
^

Irrational algebraic exponents (red background excepted**)
en, k

n = 2
2√  2
3 (
3√  3
) 2
3√  3
4 (
4√  4
) 3 		(
4√  4
) 2
4√  4
5 (
5√  5
) 4 		(
5√  5
) 3 		(
5√  5
) 2
5√  5
6 (
6√  6
) 5 		(
6√  6
) 4 		(
6√  6
) 3 		(
6√  6
) 2
6√  6
7 (
7√  7
) 6 		(
7√  7
) 5 		(
7√  7
) 4 		(
7√  7
) 3 		(
7√  7
) 2
7√  7
8 (
8√  8
) 7 		(
8√  8
) 6 		(
8√  8
) 5 		(
8√  8
) 4 		(
8√  8
) 3 		(
8√  8
) 2
8√  8
k = 1 2 3 4 5 6 7
_______________
* When
n
 is a perfect power,
bn, k
 is not transcendental when its exponent is not irrational, thus among the set listed below.
** When
n
 is a perfect power, the exponent is not irrational for
en, k ∈ {(
4√  4
) 2, (
16√  16
) 8, (
64√  64
) 32, ...} ∪ {(
27√  27
) 9, (
729√  729
) 243, (
19683√  19683
) 6561, ...} ∪ {(
256√  256
) 64, ...} ∪ {(
3125√  3125
) 625, ...} ∪ ⋯

For example,

3√  3
  
3√  3
 ( 
3√  3
 ) 2  =  ( 
3√  3
 ) ( 
3√  3
 ) 3  =  ( 
3√  3
 ) 3  =  3.

Also,

3√  3
  ( 
3√  3
 ) 2 
3√  3
  =  ( 
3√  3
 ) ( 
3√  3
 ) 3  =  ( 
3√  3
 ) 3  =  3.

See also

External links

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