Hᵤ(2, 2) = 4 for all u>0

A bit of recreational mathematics. I'd be (pleasantly) shocked if this is novel.

2+2 = 2×2 = 2² = 4 🤔

Hᵤ is the set of hyperoperations, e.g.

H₁(a, b) = a + b

H₂(a, b) = a × b

H_u(a, b) = H_u-1(a, H_u(a, b-1)) when u≥3 & b≠0

∴ H_u(2, 2) = H_u-1(2, H_u(2, 1)) when u≥3

also H_u(2, 1) = H_u-1(2, H_u(2, 0)) when u≥3

also H_u(2, 0) = 1 when u≥3

∴ H_u(2, 1) = H_u-1(2, 1) when u≥3

∴ H₃(2, 1) = H₂(2, 1) = 2

∴ H₃(2, 2) = H₂(2, 2) ⇒ 2² = 2 × 2 = 4

also H_u(2, 1) = 2 when u≥3

∴ H_u(2, 2) = H_u-1(2, 2) when u≥3

∴ ∀ u ≥ 3 H_u(2, 2) = 4

as H₁(2, 2) = 4 and H₂(2, 2) = 4, ∀ u ≥ 1 H_u(2, 2) = 4

i.e., for any hyperoperation ∘ other than the zeroth (successor) operator, 2 ∘ 2 = 4

Original post timestamp: Mon, 27 Jul 2020 07:55:11 +0000