03/09/2011, 10:31 PM
n = 3:
b > 3 -> 2 [4] b > 3 [3] b
Proof:
For b = 4:
2 [4] 4 = 2 ^ (2 ^ (2 ^ 2)) = 2 ^ (2 ^ 16) = 2 ^ 65536 > 81 = 3 ^ 4 = 3 [3] 4
Let's assume that 2 [4] b > 3 [3] b for some b > 3.
Then we wish to prove that 2 [4] (b + 1) > 3 [3] (b + 1)
2 [4] (b + 1) = 2 ^ (2 [4] b)
> 2 ^ (3 [3] b)
> 3 * (3 [3] b)
= 3 [3] (b + 1)
b > 3 -> 2 [4] b > 3 [3] b
Proof:
For b = 4:
2 [4] 4 = 2 ^ (2 ^ (2 ^ 2)) = 2 ^ (2 ^ 16) = 2 ^ 65536 > 81 = 3 ^ 4 = 3 [3] 4
Let's assume that 2 [4] b > 3 [3] b for some b > 3.
Then we wish to prove that 2 [4] (b + 1) > 3 [3] (b + 1)
2 [4] (b + 1) = 2 ^ (2 [4] b)
> 2 ^ (3 [3] b)
> 3 * (3 [3] b)
= 3 [3] (b + 1)
