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Hyperoperators [n] basics for large n
#1
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 = 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.Smile

[edit]Some minor corrections made in this post[/edit]
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#2
(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.
Smile

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
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#3
(03/06/2011, 08:52 AM)bo198214 Wrote: PS: at a[n]1 = a you should add n>1
Indeed, thanks for checkingSmile

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.
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#4
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.

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#5
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


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#6
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.
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#7
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.
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#8
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.
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#9
Thanks, Dyitto.
Glancing through the proves, they all seem to be correct.
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#10
8.
Let a, b integer, 1 < a < b
Let c = 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.Blush
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