Transcendental Numbers: e, pi

The French Mathematician and Physicist Joseph Fourier proved
e is irrational,

Another French mathematician Charles Hermite went further: e belongs to another mathematical world:
e is transcendental.

Hermite’s German student Lindermann followed the same method, proved:
pi is also transcendental.

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Transcendental Number

Transcendental numbers: e, Π, L…

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

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

Gelfond-Schneider Theorem

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

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

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

Proof:
10^{log 2} = 2

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

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

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