Irrational^Irrational

In an earlier post I proved that $\sqrt{2}$ is an irrational number. What about $\sqrt{2}^{\sqrt{2}}$? We can actually use this number to prove that it's possible to have an irrational$^{\text{irrational}}$ = rational, without actually constructing an example.

• Either $\sqrt{2}^{\sqrt{2}}$ is rational $\checkmark$
• Or it is irrational and $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = 2$ is rational $\checkmark$

The Gelfond Schneider Theorem tells us that $\sqrt{2}^{\sqrt{2}}$ is transcendental, and since it is real, it is irrational.

• If $\alpha$ and $\beta$ are algebraic numbers with $\alpha \neq 0,1$ and if $\beta$ is not a rational number, then any value of $\alpha^{\beta} = e^{\beta \text{log} \alpha}$ is a transcendental number.

In fact, it's not too difficult to find examples for all of the 8 possible rational/irrational triples $\alpha^{\beta} = \gamma$. Let's explore a bit further and look at infinite tetrations. For your amusement I wrote some hackish code that prompts you for a number $n>1$, then outputs the infinite tetration that converges to it. It is a surprising result that $$ \sqrt{2}^{{\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^{...}}}} = 2 $$

 from math import *
 LIMIT,em = 2,2**-52
 
 def tetration(b, LIMIT):
  a = 1
  for i in xrange(1000):
   try:
    a = b**a
    if a > LIMIT: return False
   except OverflowError:
    return False
  return a
 
 while LIMIT >= 1:
  LIMIT = input('\nTetration Limit: ')
  em = 2**-52
  lo, hi = 1.0,2.0
  
  while hi-lo > em:
   mid = (hi+lo)/2
   tetmid = tetration(mid, LIMIT)
   if tetmid: lo = mid
   else: hi = mid
   
  print lo, tetration(lo, LIMIT)

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