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Zeration
#41
GFR Wrote:
bo198214 Wrote:
GFR Wrote:I hope that everybody would be patient enough to read it, without ... fainting,
At least I do it! Smile
Do you mean ... fainting?

No, no, no! That meant: I (am patient enough to) read it without fainting!!! Smile

Quote: By the way, did you mean "at least" or "at last"?

And probably I was even the first one that read it!

Quote:In fact, in my opinion, this would mean that zeration does not exist at all, as a binary operation.
First of all a binary operation is just an operation that takes two inputs and delivers one output. Whether the output depends on both operands is another question. So in this sense the mother-law-zeration is a binary operation.

Quote:And this would be very sad, at least for me.

But why? The hyper operations become simpler and simpler with decreasing degree and an operation that only depends on the right operand shows an extreme simplicity as one would expect from zeration.

But mathematics is mathematics, I also find it sad that I can not divide by zero, but its just the way it is. Thats a higher joy.

Quote:Only to that, for any "a"?
We have always to be a bit cautious of the base. As long as we work with inverse operations we have to ascertain that they indeed exist (which is the base requirement in quasigroups). Thatswhy I assume in all my derivations that the base is in the bijectivity domain. That means arbitrary x for x[1]y, that means x>0 for x[2]y, and that means x>1 for x[n]y, n>2. Let R(n) denote the corresponding base range, then I think the functions f(x)=x[n]y : R(n) -> R(n) are always bijective. And the functions g(x)=b[n]x : R(0) -> R(n) , b in R(n), are also bijective. The exception is however n=0, where f is not bijective.

Quote:
Quote:And now under the assumption that y[s]0=1:
y [s] 2 = y [s] (1+1+0) = y[s-1](y[s-1](y[s]0)) = x
y = x [s]/ 2
Which, of course, should read y = x/ [s]2. You see? In the best families!

Haha, yes, though I dont know even whether my family belongs to the best Smile

Quote:But, ... Wink ... Henryk! This is the Slash Algebra.
Its quasigroups!

Quote:
bo198214 Wrote:However the assumption is wrong for s=2, y[2]0=0.
Yes, it is wrong, but I don't see the point.
The point is that the limit of the sequence y <= (y[0](y[1]\ x)) /[1]2 = y[0](x-y) -2 is not x /[2] 2 (if it exist at all).

Quote:
bo198214 Wrote:
Quote:Pillar 4 - The Hyper-means.
The hyper means (a[n]a)/[n+1] 2 = a
follow directly from the assumption that a[n]a=a[n+1]2 which is equivalent to a[n+1]1=a.
Stop a moment, please! I don't believe in a generalisation of: a[n+1]1 = a. In fact, for n=0, a[1]1 = a+1 = a is wrong!

Yes exactly, and thatswhy (a[0]a)-2=a is wrong.

Quote:
bo198214 Wrote:Asserting that a[0]a=a+2 is a bit like asserting that a[1]1=a.
Do you see the similarity?
Not really! At level a[s]1, we have different behaviours. In fact, as we have seen:
a[2]1 = a*1 = a
a[1]1 = a+1 (and tht's it!)
while:
a[2]a = a*a = a^2
a[1]a = a+a = a*2
As I already wrote, we have a[n]1=a for all n>1 and are not allowed to generealize to a[1]1=a, in the same way we have a[n]a=a[n+1]2 for all n>0 and are not allowed to generalize to a[0]a=a[1]2.

Quote:But, Henryk, the problem is that the formula mentioned in Pillar 4 may work, and the choice of -oo as neutral element of zeration (as we defined it) is shared by other researchers in this field.
-oo as well as +oo is a neutral element for your zeration as well as for the mother-law-zeration. This comes from -oo and +oo not being a number (not being in the bijectivity domain). You know you can prove everything (e.g. 1=0) by treating infinities like numbers. In so far my pillar 5 was a bit misleading as for s=0 the limit does not exist (is not a number).
However I dont know about other publishing researchers considering zeration. Why dont they post on this forum in this thread?
Reply
#42
I wonder why. Let's try to find them. I'll do my best. The first one is ... KAR. Sooner or ... later.
Reply
#43
Wink 
I try to make some definition of [0] along back, but I stopped at nonassociativity of [0] and I expect that formula,,, a [0] a [0] ...[0] a {b+1 times} = a [0] (a [0] a [0] ... [0] a {b times}) is not correct.

How I can see your definition of [0] make correct results. It is very interesting that the [0] is not associative and ,,, a [0] a [0] ...[0] a {b+1 times} = a [0] (a [0] a [0] ... [0] a {b times}) is correct

I realize that the formula can be corect, bat other bracket construction(with identical number of operands) needn't to be equal betwen each other ,,, and than ,,, if some bracket construction isn't equal to another,the operation cannot be associative.

Good question is ,whether are here some other formulas with bracket construction ,which are equal betwen each other?

it could by interesting to find out behaviour of this special nonasociativity (I think ,this special behaviour is cauced by combination {comutativity vs. nonassociativity of [0]})
Reply
#44
Hey Igor,

let me welcome you to the forum (if however somewhat late).
So are there some questions left for you,
or did any problems about zeration already resolve?
Reply
#45
(02/14/2008, 10:41 PM)GFR Wrote:
Ivars Wrote:Have You considered "+" zeration and "-" zeration?

Not really. But there is some work carried on for "mapping" the numbers obtained by its inverse operation. They ... risk ... to be multivalued objects, similar, in some cases, to the logarithms of negative real numbers. There might perhaps also exist hyperoperations with negative ranks, giving, for instance:

a@a = ao2
aoa = a+2
a+a = a*2
......

But, this requires a ... more active intervention of KAR in this discussion.

GFR

------
That "@" operation or the s=-1 operation, I find interesting. I did my own research and realized the operations below zero must not be functions when you use the operation that's not the opposite.
I discovered that
2@4= 4 and 5
3@5= 5 and 6
6@7= 7 and 8
6@8= 8 and 9
This only works if you accept Kar's zeration.
below s=0, they are all +1 operation, but they are +1 for a longer range of numbers before its +2, ie 6@6@6 is 6 zeration 3 which is 7 until 6@6@6@6@6@6 which is 8.
let me know what you think.
Reply
#46
(04/05/2008, 10:14 AM)bo198214 Wrote:
GFR Wrote:
bo198214 Wrote:
GFR Wrote:I hope that everybody would be patient enough to read it, without ... fainting,
At least I do it! Smile
Do you mean ... fainting?

No, no, no! That meant: I (am patient enough to) read it without fainting!!! Smile

Quote: By the way, did you mean "at least" or "at last"?
----

And probably I was even the first one that read it!

Quote:In fact, in my opinion, this would mean that zeration does not exist at all, as a binary operation.
First of all a binary operation is just an operation that takes two inputs and delivers one output. Whether the output depends on both operands is another question. So in this sense the mother-law-zeration is a binary operation.

Quote:And this would be very sad, at least for me.

But why? The hyper operations become simpler and simpler with decreasing degree and an operation that only depends on the right operand shows an extreme simplicity as one would expect from zeration.

But mathematics is mathematics, I also find it sad that I can not divide by zero, but its just the way it is. Thats a higher joy.

Quote:Only to that, for any "a"?
We have always to be a bit cautious of the base. As long as we work with inverse operations we have to ascertain that they indeed exist (which is the base requirement in quasigroups). Thatswhy I assume in all my derivations that the base is in the bijectivity domain. That means arbitrary x for x[1]y, that means x>0 for x[2]y, and that means x>1 for x[n]y, n>2. Let R(n) denote the corresponding base range, then I think the functions f(x)=x[n]y : R(n) -> R(n) are always bijective. And the functions g(x)=b[n]x : R(0) -> R(n) , b in R(n), are also bijective. The exception is however n=0, where f is not bijective.

Quote:
Quote:And now under the assumption that y[s]0=1:
y [s] 2 = y [s] (1+1+0) = y[s-1](y[s-1](y[s]0)) = x
y = x [s]/ 2
Which, of course, should read y = x/ [s]2. You see? In the best families!

Haha, yes, though I dont know even whether my family belongs to the best Smile

Quote:But, ... Wink ... Henryk! This is the Slash Algebra.
Its quasigroups!

Quote:
bo198214 Wrote:However the assumption is wrong for s=2, y[2]0=0.
Yes, it is wrong, but I don't see the point.
The point is that the limit of the sequence y <= (y[0](y[1]\ x)) /[1]2 = y[0](x-y) -2 is not x /[2] 2 (if it exist at all).

Quote:
bo198214 Wrote:
Quote:Pillar 4 - The Hyper-means.
The hyper means (a[n]a)/[n+1] 2 = a
follow directly from the assumption that a[n]a=a[n+1]2 which is equivalent to a[n+1]1=a.
Stop a moment, please! I don't believe in a generalisation of: a[n+1]1 = a. In fact, for n=0, a[1]1 = a+1 = a is wrong!

Yes exactly, and thatswhy (a[0]a)-2=a is wrong.

Quote:
bo198214 Wrote:Asserting that a[0]a=a+2 is a bit like asserting that a[1]1=a.
Do you see the similarity?
Not really! At level a[s]1, we have different behaviours. In fact, as we have seen:
a[2]1 = a*1 = a
a[1]1 = a+1 (and tht's it!)
while:
a[2]a = a*a = a^2
a[1]a = a+a = a*2
As I already wrote, we have a[n]1=a for all n>1 and are not allowed to generealize to a[1]1=a, in the same way we have a[n]a=a[n+1]2 for all n>0 and are not allowed to generalize to a[0]a=a[1]2.

Quote:But, Henryk, the problem is that the formula mentioned in Pillar 4 may work, and the choice of -oo as neutral element of zeration (as we defined it) is shared by other researchers in this field.
-oo as well as +oo is a neutral element for your zeration as well as for the mother-law-zeration. This comes from -oo and +oo not being a number (not being in the bijectivity domain). You know you can prove everything (e.g. 1=0) by treating infinities like numbers. In so far my pillar 5 was a bit misleading as for s=0 the limit does not exist (is not a number).
However I dont know about other publishing researchers considering zeration. Why dont they post on this forum in this thread?
---
of course -oo and +oo is a number. Are you saying 0 is not a number? its 1/oo and -1/oo. you can make -oo a number since its 0-oo and 0-delta oo.
tell me your thoughts
Reply
#47
(07/30/2009, 06:21 PM)mathamateur Wrote: of course -oo and +oo is a number. Are you saying 0 is not a number? its 1/oo and -1/oo. you can make -oo a number since its 0-oo and 0-delta oo.
tell me your thoughts

the infinities are not (normally) considered numbers (that you can treat like ordinary numbers) because if you do, you lose some important algebraic properties. There are extensions of real and complex numbers to include infinity/ies. The "extended number line" has the positive and the negative infinity you want. The "real projective line" tapes the two infinities together to make it one infinity. But you still cant do infinity - infinity. Also you don't know if 1/0 is positive or negative unless there are both positive and negative zeros. Actually it's neither. infinity doesn't behave like numbers all the time, so mathematicians just like to be careful.

But for a simple operation like deltation, it may be useful to treat it like a "zerative" identlty. Some info on limiting values of zeration may be needed tho.
Reply
#48
(07/30/2009, 11:40 PM)Tetratophile Wrote:
(07/30/2009, 06:21 PM)mathamateur Wrote: of course -oo and +oo is a number. Are you saying 0 is not a number? its 1/oo and -1/oo. you can make -oo a number since its 0-oo and 0-delta oo.
tell me your thoughts

the infinities are not (normally) considered numbers (that you can treat like ordinary numbers) because if you do, you lose some important algebraic properties. There are extensions of real and complex numbers to include infinity/ies. The "extended number line" has the positive and the negative infinity you want. The "real projective line" tapes the two infinities together to make it one infinity. But you still cant do infinity - infinity. Also you don't know if 1/0 is positive or negative unless there are both positive and negative zeros. Actually it's neither. infinity doesn't behave like numbers all the time, so mathematicians just like to be careful.


But for a simple operation like deltation, it may be useful to treat it like a "zerative" identlty. Some info on limiting values of zeration may be needed tho.

----
Thanks, interesting thoughts to ponder. Now i've heard a suggestion that -oo for zeration ought to get a new symbol, a zero with a line through it. That makes me think that you can actually reach negative infinity just line one can reach zero even though fractions go on getting smaller forever.
Maybe +infinity is the only "number-type symbol" that can't be reached. We have delta numbers before -oo, but nothing after +oo or even +ooi. unless some s=? operation can create it.
(03/26/2008, 12:50 AM)James Knight Wrote: I think the Delta Numbers are elements of the hyperreal sets. In addition, I think that Deltation will revolutionize calculus once it has been well defined. I don't think anymore that deltation values are complex or undefined, but are either infinitely far or infinitesimally close to all real numbers. I think that it is rather interesting that the delta symbol was chosen to represent deltatation as it has to do with calculus and infinitesimal quantities. I am beginng to wonder whether or not hyper real positive vs negative quantites exist. (ie does positive and negative infinity mean the same thing?)

Deltation -> Hyper Real Infinite and Infinitesimal
Subtraction -> Integer Negative Numbers
Division -> Rational Fractions
Roots and Logarithms -> Irrational and Complex/Imaginary
etc.

Notice how you cant produce the "new" number type in a previous level without using that number type.
ie. you can't get a rational number by subtracting two integers
ie. you can't get an irrational number by diving two rationals

I'm not saying you can't have negative infity, but what I am asking is whether you can result negative infinity from deltation?
----
Sure you can result -oo from deltation. I think the order of things goes backwards in terms of hyperoperations. division creates fractions below one (approaches zero). subtraction creates negative numbers below zero (approaches -00). then delation creates numbers below -infinity (approaches delta +infinity). goes from s=2 to s=1 to s=0
-------
so can a hyper real really be infinitely negative when negativity doesn't exist in the natural number set
Possibly, because Knightation subtracts so it might produce negative numbers... (also a better name for Knightation might me Jeration... )

I am also pondering when in Knightation whether there a limit to how far back you can go like with logarithms. (ie you can't take the logarithm of zero. This would mean an asymptotic relationship for zeration and knightation. This might be something to look in to.
Well, I think I got my post quota for today!

James

(03/26/2008, 12:50 AM)James Knight Wrote: I think the Delta Numbers are elements of the hyperreal sets. In addition, I think that Deltation will revolutionize calculus once it has been well defined. I don't think anymore that deltation values are complex or undefined, but are either infinitely far or infinitesimally close to all real numbers. I think that it is rather interesting that the delta symbol was chosen to represent deltatation as it has to do with calculus and infinitesimal quantities. I am beginng to wonder whether or not hyper real positive vs negative quantites exist. (ie does positive and negative infinity mean the same thing?)

Deltation -> Hyper Real Infinite and Infinitesimal
Subtraction -> Integer Negative Numbers
Division -> Rational Fractions
Roots and Logarithms -> Irrational and Complex/Imaginary
etc.

Notice how you cant produce the "new" number type in a previous level without using that number type.
ie. you can't get a rational number by subtracting two integers
ie. you can't get an irrational number by diving two rationals

I'm not saying you can't have negative infity, but what I am asking is whether you can result negative infinity from deltation?
----
-------
so can a hyper real really be infinitely negative when negativity doesn't exist in the natural number set
Possibly, because Knightation subtracts so it might produce negative numbers... (also a better name for Knightation might me Jeration... )

I am also pondering when in Knightation whether there a limit to how far back you can go like with logarithms. (ie you can't take the logarithm of zero. This would mean an asymptotic relationship for zeration and knightation. This might be something to look in to.
Well, I think I got my post quota for today!

James

-----

Deltation -> Hyper Real Infinite and Infinitesimal
Subtraction -> Integer Negative Numbers
Division -> Rational Fractions
Roots and Logarithms -> Irrational and Complex/Imaginary
etc.

Notice how you cant produce the "new" number type in a previous level without using that number type.
ie. you can't get a rational number by subtracting two integers
ie. you can't get an irrational number by diving two rationals

I'm not saying you can't have negative infity, but what I am asking is whether you can result negative infinity from deltation?
----
Sure you can result -oo from deltation. I think the order of things goes backwards in terms of hyperoperations. division creates fractions below one (approaches zero). subtraction creates negative numbers below zero (approaches -00). then delation creates numbers below -infinity (approaches delta +infinity). goes from s=2 to s=1 to s=0
-mathamateur
Reply
#49
Two things I don't get:

a+1=a o (a + 0)=a o a=a+2

By the definition of addition as iterated zeration, and the definition of zeration. Plus,

a o -inf = a

Since -inf is the unit element. But by the definition of zeration, a o -inf should be a+1. Why is this?
Reply
#50
(07/05/2010, 12:00 AM)73939 Wrote: a o a=a+2

Though this is intuitive - because true for all operations [n] with n>0 on the left side - this is not true for zeration - if we require the following (mother) law to hold:

a [n+1] (x+1) = a [n] ( a [n+1] x )

which you already used on the left side for n=0, x=0; [0] is zeration, [1] is addition.

Actually not only *you* were confused about the failure of (*) a [n+1] m = a [n] ( a [n] ( ....( a[n]a ) ) ) - right side containing m times a - for n=0. But if one takes a close look one sees that one can only expect (from the mother law):
a [n+1] m = a [n] ( a [n] ( ... a [n] ( a [n+1] 1 ) ) ) ), where the right side contains m times a.

if now a [n+1] 1 = a (which is the case for all n>0) then (*) follows. But n=0 remains the exception, because a [0+1] 1 = a+1 != a.
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