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 Hyperoperators [n] basics for large n dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/06/2011, 09:41 PM 6. (a + 1)[n]b > a[n]b Proof: For n = 1 & 2 lemma 6 is evident. Let's assume lemma 6 to be true for a given n > 1. b = 1 ===== (a + 1)[n+1] 1 = a + 1 > a = a [n+1] 1 Assume (a + 1) [n+1] b > a [n+1] b for some b > 0 Then we wish to prove that (a + 1) [n+1] (b + 1) > a [n+1] (b + 1) This is proved by: (a + 1) [n+1] (b + 1) = (a + 1) [n] ((a + 1) [n+1] b) > (a + 1) [n] (a [n+1] b) > a [n] (a [n+1] b) = a [n+1] (b + 1) This proves lemma 6 for n + 1 by induction applied to b. So lemma 6 also has been proved for all n by induction applied to n. « Next Oldest | Next Newest »

 Messages In This Thread Hyperoperators [n] basics for large n - by dyitto - 03/06/2011, 01:20 AM RE: Hyperoperators [n] basics for large n - by bo198214 - 03/06/2011, 08:52 AM RE: Hyperoperators [n] basics for large n - by dyitto - 03/06/2011, 03:19 PM RE: Hyperoperators [n] basics for large n - by dyitto - 03/06/2011, 03:24 PM RE: Hyperoperators [n] basics for large n - by dyitto - 03/06/2011, 03:32 PM RE: Hyperoperators [n] basics for large n - by dyitto - 03/06/2011, 08:56 PM RE: Hyperoperators [n] basics for large n - by dyitto - 03/06/2011, 09:41 PM RE: Hyperoperators [n] basics for large n - by dyitto - 03/07/2011, 11:26 AM RE: Hyperoperators [n] basics for large n - by bo198214 - 03/07/2011, 01:35 PM RE: Hyperoperators [n] basics for large n - by dyitto - 03/12/2011, 10:19 PM

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