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.

[edit]Some minor corrections made in this post[/edit]

(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

(03/06/2011, 08:52 AM)bo198214 Wrote: [ -> ]PS: at a[n]1 = a you should add n>1

Indeed, 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.

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.

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

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.

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.

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.

Thanks, Dyitto.

Glancing through the proves, they all seem to be correct.

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.