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 Hyperoperators [n] basics for large n dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/06/2011, 01:20 AM (This post was last modified: 03/12/2011, 10:15 PM by dyitto.) Hyperoperators [n] basics for large n We could inductively define hyperoperators as follows: a, b, n being positive integers: a[1]b := a + b n > 1 -> a[n]1 := a a[n+1](b + 1) := a[n](a[n+1]b) From this some lemmas can be proven: 1. a[2]b = a * b 2. a[3]b = a ^ b 3. 2[n]2 = 4 4. n > 2 -> 1[n]b = 1 5. a > 1 -> a[n](b + 1) > a[n]b 6. (a + 1)[n]b > a[n]b 7. ((a > 2 or b > 2) and a > 1 and b > 1) -> a[n+1]b > a[n]b 8. 1 < a < b c = $^a\log(b)$ rounded up to integer m > 0, k >= 0 Then: a [4] m >= c * (b + k) -> a [4] (m + k + 1) >= b [4] (k + 2) Is this the common definition here? Have proofs been given somewhere for the lemmas? I wrote them down long time ago, and I was about to do it again before I discovered this forum. Some minor corrections made in this post[/edit] bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/06/2011, 08:52 AM (This post was last modified: 03/06/2011, 08:54 AM by bo198214.) (03/06/2011, 01:20 AM)dyitto Wrote: Is this the common definition here? yes, thats the common definition here. Quote:Have proofs been given somewhere for the lemmas? I wrote them down long time ago, and I was about to do it again before I discovered this forum. I dont think proofs were written down already. But it sounds a very good idea to have these elementary statements safe. And it should not be too difficult. So could you do that? I (and I guess the other forum members too) would very appreciate it. PS: at a[n]1 = a you should add n>1 dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/06/2011, 03:19 PM (This post was last modified: 03/06/2011, 03:36 PM by dyitto.) (03/06/2011, 08:52 AM)bo198214 Wrote: PS: at a[n]1 = a you should add n>1Indeed, thanks for checking 1. a[2]b = a * b Proof: For b = 1: By definition a[2]1 = a = a * 1 If a[2]b = a * b for a given b, then we wish to prove that a[2](b + 1) = a * (b + 1): a[2](b + 1) = a[1](a[2]b) = a + (a[2]b) = a + (a * b) = a * (b + 1) So it is proven by induction. 2. a[3]b = a ^ b Proof: For b = 1: By definition a[3]1 = a = a ^ 1 If a[3]b = a ^ b for a given b, then we wish to prove that a[3](b + 1) = a ^ (b + 1): a[3](b + 1) = a[2](a[3]b) = a * (a[3]b) = a * (a ^ b) = a ^ (b + 1) So it is proven by induction. dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/06/2011, 03:24 PM 3. 2 [n] 2 = 4 Proof: 2 [1] 2 = 2 + 2 = 4 Suppose 2 [n] 2 = 4 for a given n, then we wish to prove that 2 [n+1] 2 is also 4. 2 [n+1] 2 = 2 [n] (2 [n+1] 1) = 2 [n] 2 = 4 So it is proven by induction. dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/06/2011, 03:32 PM 4. n > 2 -> 1[n]b = 1 Proof: n = 3 gives 1 ^ b = 1 which is indeed true for any b Now suppose that for a given n > 2, it is proven that 1 [n] b = 1 (for any b) Then we wish to prove that 1 [n+1] b = 1 (for any b) b = 1 : 1 [n+1] 1 = 1 by definition b > 1: 1 [n+1] b = 1 [n] (1 [n+1] (b - 1)) = 1 because of the induction hypothesis dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/06/2011, 08:56 PM 5a. a > 1 -> a[n]b > b 5b. a > 1 -> a[n](b + 1) > a[n]b Proof: For n = 1 & 2 5a and 5b are evident. Let's assume 5a and 5b to be true for a given n > 1 ==> (i). Proof of 5a for n + 1: a [n+1] 1 = a and so a [n+1] 1 > 1 Now assume a [n+1] b > b for some b a [n+1] (b + 1) = a [n] (a [n+1] b) > a [n] b (Since (i) says: x > y -> a [n] x > a [n] y) Furthermore a [n] b >= b + 1, so: a [n+1] (b + 1) > b + 1 So 5a has been proven for n + 1 by induction applied to b Proof of 5b for n + 1: a [n+1] (b + 1) = a [n] (a [n+1] b) > a [n+1] b (Since (i) says: a [n] x > x) So 5a and 5b have been proven for any n by induction applied to 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. dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/07/2011, 11:26 AM (This post was last modified: 03/07/2011, 11:27 AM by dyitto.) 7. ((a > 2 or b > 2) and a > 1 and b > 1) -> a[n+1]b > a[n]b Proof: ====== The case n = 1: we wish to prove that a > 1 and b > 1 and (a > 2 or b > 2) -> a * b > a + b 2 * 3 > 2 + 3 a > 2 -> a * 2 = a + a > a + 2 Now suppose a * b > a + b for any a, b > 1 a * (b + 1) = a * b + a > a + b + a > a + (b + 1) Now let's consider lemma 7 true for some n > 0. 2 [n+2] 3 = 2 [n+1] (2 [n+2] 2) = 2 [n+1] 4 > 2 [n+1] 3 a > 2 -> a [n+2] 2 = a [n+1] (a [n+2] 1) = a [n+1] a > a [n+1] 2 Now suppose for some a, b: a > 1 and b > 1 and (a > 2 or b > 2) and a [n+2] b > a [n+1] b a [n+2] (b + 1) = a [n+1] (a [n+2] b) > a [n+1] (a [n+1] b) > a [n] (a [n+1] b) (mind that a [n+1] b > b >= 2) = a [n+1] (b + 1) This proves lemma 7 to be true for n + 1. So lemma 7 also has been proved for all n by induction applied to n. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/07/2011, 01:35 PM Thanks, Dyitto. Glancing through the proves, they all seem to be correct. dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/12/2011, 10:19 PM (This post was last modified: 08/13/2012, 10:19 PM by dyitto.) 8. Let a, b integer, 1 < a < b Let c = $^a\log(b)$ rounded up to the next integer Let k, m integer, m > 0, k >= 0 a [4] m >= c * (b + k) -> a [4] (m + k + 1) >= b [4] (k + 2) Proof: http://www.scrybqj.com/scrybqjdocuments/..._proof.pdf I hope I'll find some time to put this content in Tex. « Next Oldest | Next Newest »

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