Transcendental Number

Transcendental numbers: e, Π, L...

What about $Latex e^{e}, \pi^{\pi} ,\pi^{e} $ ?

Aleksander (Alexis) Osipovich Gelfond (1906-68):

Gelfond-Schneider Theorem

$Latex a^ b $ transcendental if
a is algebraic, not 0 or 1
b irrational algebraic number

Examples:
$latex \sqrt{6}^{\sqrt{5}}, 3^{\sqrt{7}}$
Hilbert Number: $latex 2^{\sqrt {2}}$ (Hilbert Problem proven by Gelfond}

Is log 2 transcendental ?
[log = logarithm Base 10]

Proof:
$latex 10^{log 2} = 2$

1) Sufficient to prove log 2 irrational
Assume log 2 rational
log 2= p/q, p and q integers
$latex 10^ {log 2} = 2 = 10^ {p/q}$
raise power q
$latex 2^{q} = 10^{p} = (2.5)^{p}$
$latex 2^{q} = 2^{p}.5^{p}$

Case 1: p>q
$latex 1 = 2^{p-q}.5^{p}$
=> False

Case 2: q>p
$latex 2^{q-p}= 5^{p}$
Left is even : $latex 2^{m} \text { = even} $
Right is odd: $latex 5^{n} \text {= ....5} $
=> False

Therefore p,q do not exist,
=> log 2 irrational

Reference: Top 15 Transcendental Numbers: http://sprott.physics.wisc.edu/pickover/trans.html

Irrational 'e'

In secondary school, we know how to prove √2 is irrational, how about e ?

e= 1 + 1/1!  +1/2!  + 1/3!  + 1/4!  +...

Prove by Reductio ad absurdum (contradiction):

Assume e= p/q as rational
multiply both sides by q!
LHS: e. q!= (p/q) .q! = p.(q-1)!  => integer

RHS:  q!+q! + (3.4...q)+ (4.5...q) +...1 + 1/(q+1) +.... => fraction

Contradiction !

Therefore e is irrational.

Google Math

Google = Googol means '1 followed by 100 zeroes'

Google raised Capital at IPO = $2.718281828 Billion = e Billion
(e= 2.718281828...)

Google’s 3 building names: e, pi, ø (Golden ratio)