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 2 [n] b and 3 [n] b for (large) integer n, b dyitto Junior Fellow  Posts: 13 Threads: 3 Joined: Mar 2011 03/08/2011, 10:56 PM There's one thing I can easily proof about this: 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. dyitto Junior Fellow  Posts: 13 Threads: 3 Joined: Mar 2011 03/09/2011, 10:31 PM 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) dyitto Junior Fellow  Posts: 13 Threads: 3 Joined: Mar 2011 03/12/2011, 10:52 PM 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) « Next Oldest | Next Newest »

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