# Tetration Forum

Full Version: 2 [n] b and 3 [n] b for (large) integer n, b
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n > 2 -> 3 [n] 2 > 2 [n] 3

Proof:

n = 3:
3  2 = 3 ^ 2 = 9 > 8 = 2 ^ 3 = 2  3

Suppose for some n > 2 : 3 [n] 2 > 2 [n] 3
Then we wish to prove that 3 [n+1] 2 > 2 [n+1] 3

3 [n+1] 2 = 3 [n] 3
= 3 [n-1] (3 [n] 2)
> 3 [n-1] (2 [n] 3)
> 2 [n-1] (2 [n] 3)
= 2 [n] 4
= 2 [n] (2 [n+1] 2)
= 2 [n+1] 3

But now I also suspect that for each n:

2 [n+1] b > 3 [n] b

will be true for sufficiently large b

n = 1:
b > 3 -> 2 * b > 3 + b

n = 2:
b > 3 -> 2 ^ b > 3 * b

But haven't yet proved it for any n > 2.
n = 3:
b > 3 -> 2  b > 3  b

Proof:

For b = 4:
2  4 = 2 ^ (2 ^ (2 ^ 2)) = 2 ^ (2 ^ 16) = 2 ^ 65536 > 81 = 3 ^ 4 = 3  4

Let's assume that 2  b > 3  b for some b > 3.
Then we wish to prove that 2  (b + 1) > 3  (b + 1)

2  (b + 1) = 2 ^ (2  b)
> 2 ^ (3  b)
> 3 * (3  b)
= 3  (b + 1)
n = 4:
b > 3 -> 2  b > 3  b

Proof:

For b = 4:

Apply lemma 8 having a = 2, b = 3, c = 2, m = 4, k = 2:

2  4 > 2 * (3 + 2) is certainly true
-> 2  7 >= 3  4
2  4 = 2  65536 > 2  7 >= 3  4

Let's assume that 2  b > 3  b for some b > 3.
Then we wish to prove that 2  (b + 1) > 3  (b + 1)

2  (b + 1) = 2  (2  b)
> 2  (3  b)
> 3  (3  b)
= 3  (b + 1)