Re: [math-fun] Is e^e^e^79 an integer?

Start with owning the numbers 0 and 1. Each step, with two numbers you already own (perhaps identical), compute A+B, A-B, A*B, A/B, or 2^A. CONJECTURE: the greatest finite real number reachable in N>0 steps is 2^2^2^...^2 with N-1 uses of "^" and evaluating right to left.

OK, in the first step, I compute 1 + 1 = 2 . In the next step, compute 2^2 = 4 (which may as well be 2 + 2 or 2 * 2 ). But in the third step you want me to do 2^4 , whereas I'd rather do 4^4 , since I already "pwn" 4 . Did I misunderstand something?