The Star Wars Room :--O--:

Holo

12:00 AM

I get, you are trying to find a model where omega is singular but is still similar to ZF

Forget C for now

But omega is defined using ZF :/

Secret

yup

Holo

Can you define omega without axiom of infinity/union/pairing?

Moreover, if A is a theory and Mod(A)⊇Mod(PA) then cof(omega) is ill-defined or omega itself

So your model will have to be either very weak, or very different

Secret

so far our consensus with 21820 and Leaky and Alessandro is highly unlikely but we don't know

Holo

Because it will have to be a weird theory

Secret

That's the point. The things I am interested in often breaks so much of the nice rules that it is a whole different world altogether

Holo

12:08 AM

This won't break the world, this is defining new world

Secret

previously I played with systems that allowed division by zero before and did some sorts of reverse mathematical study on what makes 0x=0 for example

my preliminary conclusion is that asdociativity and many of its weaker variations have to break else you get left null semigroups which are relatively boring on their own

Holo

Let me rephrase it

Secret

yup. In particular, it breaks induction (unless you modify the distributive laws like in wheels) because 1+1=1 in many of these systems since 0+0=0 must hold by additive identity

Holo

Our axioms are talking about ×, we define / by being the inverse of ×, if × is not bijective we need choice function define the "inverse", but it is possible

(And not unique)

Secret

But even with an inverse defined by choice, once we have 0x=0, then dividing by zero must give x=1

fir all x

so it seems choice alone won't get around that

Holo

12:16 AM

This is why I said "inverse"

It is easy to prove that inverse of F is defined if and only if F is bijective

Secret

yup

Holo

So I don't see the point to defining / as inverse function+dividing by 0

You will have to completely redefine /

Secret

That's the path taken by wheels and meadows in the literature

whereas the path I took is instead make F surjective or injective by somehow prevent 0x=0 or x0=0 or both to be proved from a new set of axioms

Holo

So you can't have a identity element for addition

Or you have to lose some other axioms of the field

What axioms did you thought about?

Secret

There are many possiblities based on the investigation donevwith other users a year ago: first we need to know the minimal definition of what 0 is. 0 is an element that has some or all of the following properties:

1. It is an (one sided or two sided) additive identity

2. It is an (one sided or two sided) multiplicative absorber

3. It is the minimum or bottom element in a lattice

Thus the key to define division by zero is to modify or throw away the usual axiom of rings without end up nuking two many of the above properties that makes the element 0 to cease be 0

The first thing to do is to run through the proofs that gives 0x=0 or x0=0 in the first place

here: we can have x+0x=x(1+0)=x1=x

and then by cancellation, we have 0x=0

so the first axiom that has to go is either the left or right additive inverses

Next, because we can have x+x0=x(1+0)=x1=x

So If we for some reason we found x+x0=x undesirable, then the left/right or right/left additive identity or multiplicative identity has to go

Finally, because we still have a pair of additive or multiplicative identity that can lead to x to absorb 0x or x0, the other distributive law has to go

Holo

12:37 AM

So we left with very little

Secret

This thus give one way to define division by zero, where the most important property is 1+1=1 and x+y=y if y > x where > is an ordering generated from 1 > 0

Holo

How do you define x+y if y≤x≠1?

Secret

To be more precise x+y=y comes from this:

0+1=1 or 1+0=1 has to be true since zero is an identity (and in this route it's pretty much the only thing that inform us 0 is indeed zero). Now multiply both sides by any element y, you get:

y0+y1=y1 or y1+y0=y1

So any element generated from 1 will absorb any element that has a zero in its expression

Holo

From one side

Secret

Yup, and that is a common property for division by zero via this route, which heavily restricts the relations between the elements

Holo

12:45 AM

It is so restricted so we lose our normal arithmetic

Secret

The simplest nontrivial finite example is having $\Bbb{Z}/p\Bbb{Z}$ as the multiplication structure and either a left or right null semigroup as the addition structure, which is relatively boring

Another route one can take is to blow up associativity, alternatively and (forgot). This allows you to retain more axioms, but I currently still studying the properties of it. Regardless, 1+1=1 is a theorem unless you modify the distributive law

Currntly the evidence from experimentation and other proofs strongly suggests most interesting division by zero systems often can only retain power associativity

Holo

Hmm

Secret

what can potentially gained from defining division by zero, is that the zero terms 0x, x0 0x0, x0x etc. often have properties very different from the algebraic systems we work with, thus may open doors to new worlds

Holo

0,0x,0xx,0xxx,0xxx,0xxxx,0xxxxx...

Secret

in particular, I currently have a conjecture is that if the restrictiveness of algebraic systems can be arranged in a hierarchy, division.by zero is some of the most constrained

Holo

12:53 AM

Still, you now want finite sum of finite elements to be infinite...(Or all countable sum of finite elements be finite)

This is a bit harder to achieve without throwing away our definitions

Secret

yeah. I often like to define highly pathological objects and wedge them into the usual systems to see what happens

Holo

Lets say we successfully define that object, how would you wedge it together?

Secret

Defining the "Interfinite sets" (which the definition is basically generalising what you said above) is very tricky because finite numbers are closed under addition. And while I really want to wedge them in, ultimately it isogic that decides whether I can do so. Thus most of my investigation have two possibilities:

1. Either the object is successfully defined and thus further studies can be carried out on them

2. Or that the most general proof of their impossibility is generated, thus writing off almost all the models that can produces it, thus demonstrating its non existence in the most extreme sense

Holo

Yes, but what is you definition of addition? Generally models need not to have addition definied

Secret

My investigation on Division by zero is initially motivated by trying to show that division by zero is impossible in all conceivable systems, but logic showed otherwise

i am not sure yet. It would be nice to retain as much of the usual addition. I usually don't change definition of things unless I hit a no go theorem

In order to define interfinite sets, one must understood fully what "finite" means which is why I am reading that paper as the author does not define them using the naturals

Holo

1:03 AM

I can define: ${\cal L}=\{f,+,0\}$ where $f$ is 1-ary relation, + 2-ary function, 0 0-ary function with the following axioms: $f(0)=\Bbb F$, $x+0=0\leftrightarrow x=0$, $x\ne 0\rightarrow f(x)=\Bbb T$(or if you want: $\lnot f(0)$, $x+0=0\leftrightarrow x=0$, $x\ne 0\rightarrow f(x)$)

Then define being infinite as $f(x)$

You have to be more specific on how close do you want it to be to a field

Secret

For me, I usually don't expect a field, as often the objects I want to create is too pathological for fields. Often I just settle with rings or semirings

Holo

Still, how close do you want to lend?

My theory has a model with interfinite sets

But it is trivial

And not what you search for

Secret

1:22 AM

Yeah I do not expect to be easy. It is often either they are impossible in the most extreme sense (a no go theorem) or it is just very nontrivial

. Currently, I am thinking about that $0\in S \land n \in S \implies n+1 \in S$ holds until some large finite number $M$. Otherwise, I am thinking about assuming they exists, then try to apply set operations on them to see what is produced to figure out what properties are to be expected for these sets in a Boolean algebra

and in any moment a contradiction is hit, trace back to the theorems that lead to the contradiction to be possible, blow up some axioms there and retry

> All countable sums of finite numbers being finite

i think the set of all converging series will be a good lead to a model for these kinds of interfinite

Holo

Idk, because this set is not closed under multiplication

Maybe the set absolute converging series?

Secret

1:38 AM

Yeah sounds reasonable

anyway need to go offline for a few hours, will be back later

Holo

Bye

Secret

3:05 AM

Back

Secret

5:17 AM

→ 1 message moved to Trash

in Logic, Mar 24 at 15:11, by user21820

Let Q(n) denote "Forall k in N ( k<n implies P(k) )".
Then Q is a property on N.
If Forall n in N ( Forall k in N ( k<n implies P(k) ) implies P(n) ):
	Forall n in N ( Q(n) implies P(n) ).
	// Now we shall prove "Forall n in N ( Q(n) )" by ordinary induction //
	...
	Thus Q(0).
	Given n in N:
		If Q(n):
			Forall k in N ( k<n implies P(k) ).
			Given k in N:
				If k<n+1:
					k<n or k=n.	// Are you convinced this follows from "k<n+1"?
					If k<n:
						...
					If k=n:
						...
					P(k).
			Forall k in N ( k<n+1 implies P(k) ).

Let Q(n) denote "Forall k in N (k<n implies P(k))"

Then Q is a property on N.

If Forall n in N (Q(n) implies P(n)):

----Forall n in N (Q(n) implies P(n))

---- There is no k such that k < 0, hence "k<n" and P(k) are empty

----Thus Q(0) by vacuous truth

----Given n in N:

--------If Q(n):

-------------Forall k in N (Q(n))

------------Given k in N:

-----------------If k<n+1:

-------------------------If k>n:

----------------------------k<n+1

----------------------------n<k<n+1

----------------------------There is no such k

-------------------------k<=n

-------------------------If k<n:

→ 1 message moved to Trash

Secret

6:58 AM

----------------------------If Q(k+1):

--------------------------------Forall k in N (k+1< n implies P(k+1))

Secret

7:18 AM

--------------------------------Given k in N:

-------------------------------------If k+1>n:

----------------------------------------k+1>n>k

----------------------------------------No such k

-------------------------------------k+1<=n

-------------------------------------Q(k+1)

-------------------------------------P(k+1)

--------------------------------Forall k in N(P(k+1))

Secret

7:57 AM

RESTART

Let Q(n) denote "Forall k in N(k<n implies P(k))"

Then Q(n) is a property on N

If Forall n in N(Q(n) implies P(n)):

=Forall n in N(Q(n) implies P(n))

=There is no k such that k<0, hence "k<0" and P(0) are both empty

=Thus Q(0) by vacuous truth

=Given n in N:

==If Q(n):

===Q(n)

===Given k in N:

====If k<n+1:

=====If k>n:

======k<n+1

======No such k

======n<k<n+1

=====k<=n

=====If k<n:

======If Q(n+1):

=======Forall k in N(k<n+1 implies P(k))

=======Given k in N:

========If k<n+1:

=========If k>n:

==========n<k<n+1

==========No such k

=========k<=n

=========If k=n:

==========n<n+1

Secret

8:47 AM

→ 1 message moved to Trash

==========k<n+1 implies P(k)

==========P(k)

=========If k<n:

==========k<n+1 implies P(k)

==========k<n<n+1

==========P(k)

=========P(k)

========k<n+1 implies P(k)

→ 1 message moved to Trash

========k<n<n+1

========P(k)

=======Forall k in N(P(k))

Secret

9:57 AM

RESTART

Let Q(n) denote "Forall k in N(k<n implies P(k))". Then Q(n) is a property on N

If Forall n in N(Q(n) implies P(n)):

=Forall n in N(Q(n) implies P(n))

=There is no k in N such that k<0, hence k<0 and P(0) are both empty

=Thus Q(0) by vacuous truth

=Given n in N:

==If Q(n):

===Q(n)

Secret

10:41 AM

===Given k in N:

====If k<n+1:

=====If k>n:

======n<k<n+1

======No such k

=====k<=n

=====If k<n:

======If Q(n+1):

=======Forall k in N(k<n+1 implies P(k))

Secret

11:56 AM

=======If k>n+1:

========n+1<k<n

========No such k

=======k<=n+1

=======If k=n+1:

========n+1<n

========FALSE

=======k<n+1

=======k<n+1 implies P(k)

=======P(k)

Secret

4:54 PM

======Q(n+1) implies P(k)

=====If k=n:

======If Q(n+1):

=======Forall k in N(k<n+1 implies P(k))

=======If k>n+1:

========n>n+1

========FALSE

=======k<=n+1

=======If k=n+1:
========n+1<n
========FALSE
=======k<n+1
=======k<n+1 implies P(k)
=======P(k)

======Q(n+1) implies P(k)
=====Q(n+1) implies P(k)

====k<n+1 implies (Q(n+1) implies P(k))

===Forall k in N(k<n+1 implies (Q(n+1) implies P(k)))

===m<n+1 implies (Q(n+1) implies P(m))

==Q(n) implies (m<n+1 implies (Q(n+1) implies P(m)))

=Forall n in N(Q(n) implies (m<n+1 implies (Q(n+1) implies P(m))))

Forall n in N(Q(n) implies P(n)) implies (Forall n in N(Q(n) implies (m<n+1 implies (Q(n+1) implies P(m)))))

Holo

Oh that is a lot of msgs

Secret

5:18 PM

in Logic, 22 secs ago, by Secret

I think I am stuck at how to get P(k) out from Q(n+1)

Will discuss tomorrow. It is late here at 2:00 am

Holo

Pretty late

Bye

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$Mathematics$ The Star Wars Room :--O--: