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Hyperoperators [n] basics for large n - dyitto - 03/06/2011
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.
[edit]Some minor corrections made in this post[/edit]
RE: Hyperoperators [n] basics for large n - bo198214 - 03/06/2011
(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
RE: Hyperoperators [n] basics for large n - dyitto - 03/06/2011
(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.
RE: Hyperoperators [n] basics for large n - dyitto - 03/06/2011
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.
RE: Hyperoperators [n] basics for large n - dyitto - 03/06/2011
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
RE: Hyperoperators [n] basics for large n - dyitto - 03/06/2011
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.
RE: Hyperoperators [n] basics for large n - dyitto - 03/06/2011
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.
RE: Hyperoperators [n] basics for large n - dyitto - 03/07/2011
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.
RE: Hyperoperators [n] basics for large n - bo198214 - 03/07/2011
Thanks, Dyitto.
Glancing through the proves, they all seem to be correct.
RE: Hyperoperators [n] basics for large n - dyitto - 03/12/2011
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/hyperoperators_basics_lemma8_proof.pdf
I hope I'll find some time to put this content in Tex.
