OK, I haven't done real maths for four years! This may be total crap, but I'm thinking along these lines: [Trying to extend this to n^n^... = X ⇒ n = X ^ (1/X) ∀ X ∈ N]
p0 = X^(1/X), pi = p0 ^ pi-1
We already have:
∀ i, 1 < pi-1 < pi < X
The proof requires:
∀ 0 < y, ∃ k : X - y < pk < X
The case where y ≥ X - 1 is elementary:
∀ y ≥ X - 1, X - y ≤ 1 < pi < X
So we are left with showing:
∀ 0 < y < X - 1, ∃ k : X - y < pk < X
Thr section (0,1) seems to be easy. Since:
∀ x,y > 0, b > 1 : x < y ⇔ Logb(x) < Logb(y)
I can take the Log base p0 [call this LP] of the inequality ( :confused: ) to get:
LP(X) = ln(X) / ln(X ^ (1/X)) = ln(X) / ((1/X) ln (X)) = X
LP pk = pk-1
LP(X - y) = X × [ln(X - y) / ln(X)]
Let α = ln(X) - ln(X - y)
= ln(X) - ln(X) × ln(-y)
= ln(X) × [1 - 1/ln(y)]
∴ LP(X - y) = X × [1 - α / ln(X)]
= X × [1 - ln(X) × (1 - 1/ln(y)) / ln(X))
= X × [1 - (1 - 1/ln(y))]
= X / ln(y)
So the inequality is:
∀ 0 < y < 1, ∃ k : X / ln(y) < pk-1 < X
∀ 0 < y < 1, ln(y) < 0
∴ X / ln(y) < 0 < pk < X
So any value of k satisfies the inequality for (0,1). This leaves the section [1,X-1). Since the lower bound is included, we simply need to show:
∃ k : X - 1 < pk < X
In the case of X=2, any value of k satisfies this. Now, I just need proof for X > 2! I hope this make at least some sense, and doesn't read like the insane ramblings of a deranged lunatic! :) Cheers, Richard
