Edgard Neuman

Ok ok i suggest ℕω because you can actually use a number as ω the + if fairly simple (a +b) ∈ ℕω if and only if a+b<ω if not, then it is not in.. ω is not in so it's not different in peano : S(S(.... S(0))) ∈ ℕω UNTIL you reach S(ω-1) which is not in ℕω (and so all its successor) so a = S(S... (S(0)))) so b = S(S... (S(0)))) a+b=S(S... (S(S(S... (S(0)))))))) is or is not in ℕω according the number of S<ω It's simpler to just consider that numbers works the same in this axiomatic excepts they are limited by ω Then you create new numbers in ℚω by adding " / " defined by a/b = c <=> a= b *c so 1/3 for instance is in ℚω but not in ℕω And because infinities are comparable (but not in ℕω of course) , when you write "...." you have to specify the number of terms in use ....[ ] for instance if S= 1+1+1+1 + ....[ω] P= 1+1+1+1 + ....[ω+1] P-S = 1 if Q= 1+2+3+4 + ....[ω] R= 1+2+3+4 + ....[ω+1] R-Q = ω +1 Q= (ω ² + ω) / 2 R= (ω ² + ω) / 2 + ω +1 1 + 3 + 5 + 7 + 9 + … [ω] = ω² or 1– 2 + 3 – 4 + 5 ….[ω] = 1 + 3 + 5 + 7 …[ω/2] – 2 – 4 – 6 – 8 …[ω/2] = 1 + 1 + 1 + 1 …[ω/2] + 0 + 2 + 4 + 6 …[ω/2] – 2 – 4 – 6 – 8 …[ω/2] = ω / 2 + 0 - 2 * (ω / 2) = - ω / 2 you can also group them by 2 1– 2 + 3 – 4 + 5 ….[ω] = (1-2)+ (3 -4) + (5-6) ….[ω/2] = -1 -1 -1 -1 ….[ω/2] = - ω / 2 Or another interesting one : c = 1 + r^1 + r ^ 2 + … [ω] =(1 – r^ω ) / (1 – r)

yes, but what makes your numbers better than mines ? (are they "more natural" ?) everywhere I look, in reality, in computers, infinities are strictly COMPARABLES, and epsilon quantities are not disappearing.. There is always more numbers in a set, that there is even numbers in it. ALWAYS. You want to make it somewhat equal at infinity, I don't. In my univers the "infinite" quantity is really a measurable quantity. And in your philosophy there is the problem that the quantity of number you use at the right side of n->n*2 is always different that the quantity you use at the left side.. that's a problem for me.

Things are only "CORRECT" or "INCORRECT" relatively to an axiomatic. I a propose an axiomatic, I don't care what your consider ""CORRECT" the axiomatic DEFINES what is correct. Yes that's true. That's because your "Card(ℕ) = Card(ℕ) + 1561651 " seems so absurd to me, I'll go further .. if your Card(ℕ) is not a quantity, if operations with it doesn't work the same way, it is simply not related to numbers.. You just created a different set of numbers with their weird rules. You can invent numbers with the same weird properties of Card(ℕ), that are not defined by Card(ℕ)> any n.. for me Card([0;n[) = n I don't even think it can be something else. I don't use your "strange" infinite things. In your view, everything become "different" at infinity. In my view Infinity is only the limit of finite, in every proposition.

I understand why we don't understand each others : When I write Second, if you defineCard(ℕ) =ω I my mind, it's self explanatory than there is ω elements in ℕ So no, it's not "your usual" ℕ

Ok I'll explain : ℕ= [0;ω[ ℕ= {0 ;1 ;2 ;3 ; ..... ; ω-2 ;ω-1} All your question about all "+" "*" and number construction are answered by that. and you have ℚ and ℝ in the first post. It is not a very complicated idea in fact. By the way when I write Card(ℕ)=ω It means exactly that there is ω elements in ℕ, not one more, not one less element. It really is the "upper boundary" of the set. ok here is the weird part .. To know if a number is in N Q or R isn't always trivial For instance ω+1 is not in ℕ but (ω+1) -2 is in ℕ in ℚ it's easy because if a and b are in ℕ, + - a/b is in ℚ In ℝ it's tricky again.. for instance x= 0,0000.....000025 = 25/ (10^a) It is in ℚ if and only if 10^a<ω, so if a<log10(ω) To be in ℝ, means there is a way to write it in the form of a sum of ω number from ℚ, that is not trivial But if found that every number in ℝ (when you compute it outside of the axiomatic of course) , are multiples of 1/A003418(ω-1)

Well, there is something that I don't understand, each time I speak with people Why don't you THINK BY YOURSELF ? Everything is written is THE FIRST POST. Can't you figure out that when I write "when a+b>=ω it's not in N " It means exactly what is written ? I get really annoyed I have to explain how I "build numbers". So annoyed that I get confused myself in my own idea. At some point I said I used Peano or {{{}}} or even any system that generate new elements different than the precedents Because that don't matter here. It's NOT in the FIRST POST. What next ? You want me to explain the sign " +" the sign ">", the usage of the alphabet ? You should be able - to REALLY read the hypothesis, the axioms, understand them and what they mean. (why does it take 20 post for you to figure out what the axiom means ? ) - to think about everything they imply (EVERYTHING) BY THEMSELF, and certainly not by OTHER AXIOM YOU KNOW (why did you ever suppose that ω-1 is not in N ? Did I once said ω is a ordinal number ?? WHERE ? It's in the hypothesis ! Do I really have to explain each and every axiom I DON'T USE because somebody can't figure out that a letter is just a letter ? ) So no, "I don't have to demonstrate" it is "consistent". (What does that even mean ? I think a axiom while contradict some other axiom ?) You do that, according to the hypothesis, not you own math. And if you find something, you tell me. That's why I'm here. Stop asking question about thing that you should be able to figure out from the hypothesis.

that's a axiom I add. It's different that what you know, and It solve the problem. I define ω SO THAT Peano stops at ω.. so yes, In my axiomatic, (a+b) ISN'T always in N, that's exactly how I built it. What I use is a limited version of Peano.. that's how I calculate the Card(Q) and Card(R).. It is in my FIRST POST, right there : You want to leave a infinite space between ω and the integer, because you want to respect Peano infinite. I try to solve all the bijection problem, by stating that the only proper way to work with infinite set that is always coherent, is to treat them as the limit case of finite sets.. because if all proposition are true together, you can have a first set of Peano elements there and a bigger (when it have some new elements built in, it's bigger) one somewhere else in the same universe (Especially at the two ends of the same bijection). That's the whole philosophy of it. You asked me how I build integer so I said the usual way : with Peano.. I never said that my set of axiom is the whole set of axiom from Peano, uncontradicted somewhere. ω is just the variable I play with to access this limit. So in my axiomatic N = [0;ω[ and yes ω-1 ∈ ℕ (I remember now that I had the same kind of arguments years ago, so that's why I wanted to come up with a way of comparing infinite set closer to what "intuition about quantities" tells us) To come up with A018805 and A003418, I assumed ω is an integer and I computed the Card(Q) and Card(R) with this limited set [0;ω[..

That's not how maths works.. So then ω is obviously not Card(N) which contradict (1) To be > any n isn't enough to be the cardinal of a set. The cardinal of {1;2;3;4} isn't 12. (-1) is not an addition.. so I can simply assert (by axiom) that ω-1 is in N I can't wait to see what your French looks like So ω + 1 or ω-1 doesn't make sense.. or is it =ω ? let's assume ω + 1 = ω So if I take S = 1+1+1+1+... and I take the first 1 and I leave the rest.. the R So we have S= 1 +R because R= 1+1+1 + .... and ω = ω +1 R= 1+1+1+ ( 1 + .... ) so R = S = R+0 So 0 = 1 We come to a contradiction, right ? So if you really do "serious" math, it prove that ω + 1 != ω How sorry that's right you can't subtract infinite value from an equation, I get that. So R =R +1 = R +1 +1 = R + 1+1+1 .... = 2*R So R= 2 * R but you can't divide right.. So R = 2* R= R *2 *2 = R * 2 *2 *2 *2 * .... = R * (2 ^R ).. right ? So Cantor was wrong with his diagonal argument.. right ? You just have to use the logic that "everything infinite" = "everything else infinite" to quickly come to something nobody never came up with, and you genius mathematician didn't add an axiom like "you can't do that, because it's forbidden by the bijection" .. and over the years you add a lot of those rules .. like an infinity of them Why do you think I am here ? I have a nice video from numberphile that says 1+2+3+4+5+ .... = -1/12 .. but because ... = ... +1 ? Isn't it 1+2+3+4+5+ .... = -1/12 +1 ? I'm confused

Do you even read what I write ??? I constructed ω so ω-1 is in N. I don't care about you preconcieve notion of what you call ω. It's a ω I DEFINED SO, as someone asked me to explain. If you don't use my definition, why do you ask for it ?? I can copy and paste : ω is defined by :(1) for any n ∈ ℕ, we have n<ω, (2) there is no n with n<ω and n not in ℕ- so ω-1 ∈ ℕ, and ω don't ∈ ℕ ω-1 ∈ ℕ because of (2). You just can't suppose that Card(ℕ)=ω and ω-1 is not in ℕ because then It would contradict (2) SO YES ω-1 ∈ ℕ because of (2). why the hell would you create Card(ℕ)=ω if ω-1 is not in ℕ ??? That would make no sense.. Card([0;4[) = 4; Card([0;4[) -1 = 3 ; 3 ∈ [0;4[ Card([0;5[) = 5; Card([0;5[) -1 = 4 ; 4 ∈ [0;5[ Card([0;6[) = 6; Card([0;6[) -1 = 5 ; 5 ∈ [0;6[ .... Card([0;n[) = n; Card([0;n[) -1 = n-1 ;n-1 ∈ [0;n[ ... and IF [0;ω[ = ℕ Card([0;ω[) = ω; Card([0;ω[) -1 = ω-1 ;ω-1 ∈ [0;ω[ Card(ℕ) = ω; Card(ℕ) -1 = ω-1 ;ω-1 ∈ ℕ

The second case is tricky so let me explain again.. so n is an integer and n*2 is an integer.. so you have n->n*2.. but if n*2 is an integer, to prove the bijection, n*2 must also have a counterpart.. right ? let's try n*4... but does n *4 have a counterpart ? lets calculate .. and so on CAN YOU PROVE ME that n*2*2*2*2*2*.... is an integer ? I don't think so. so the bijection is it proven for ALL integer ?

it doesn't prove the bijection.. how does it ? I've repeated all along that you can ETHER assume Card(ℕ)=ω or NOT ASSUME. In the first case, (ω-1) * 2 is not in ℕ => NO BIJECTION In the second case, you can't prove the bijection because it's a AND operation that is never finished and the last FALSE can always disprove it. => NO BIJECTION

"since omega is not finite" You accuse me of being imprecise, and you use the adjective "finite" to describe a element. (but Ok that maybe a english way of speach) assume you meant positioned at infinite. . And as I said before, you do your math based on construction you made to accommodate the bijection. I don't. You wanted a precise definition of ω I already gave you one : ω is defined by :(1) for any n ∈ ℕ, we have n<ω, (2) there is no n with n<ω and n not in ℕ- so ω-1 ∈ ℕ, and ω don't ℕ ω-1 ∈ ℕ because of (2). You just can't suppose that Card(ℕ)=ω and ω-1 is not in ℕ because then It would contradict (2) For some reason, current mathematician decided to leave a "space" between ℕ and ω for numbers ω-1, I don't. because then Card(ℕ) wouldn't be ω

What did I ignore ?

But it is enough to prove the bijection ? I don't think so. You said the set is unlimited.. so you must prove your "etc" part for the bijection to be complete.. go ahead.. (take the infinite time in the univers)

ok that's not what I meant.. I meant there is no m (in ℕ of course) so m <n that is not in [0;n[

Card([0;n[) = n Card([0;n-1[) = n -1 = Card([0;n[) -1 and n -1 ∈ [0;n[ so Card([0;n[) -1 ∈ [0;n[ and because every n is in ℕ Card(ℕ) -1 ∈ ℕ

hum... Card([0;n[) = n and Card(N)-1 just can't be = Card(N) I know you created a class of specific "not number" with there all set of properties and operation, only because you accept the bijection. But I don't, and so I don't assume Card(N) behave diffently that any number and I don't see why I should

sorry it's not very easy to systematicaly copy an paste the "ℕ" character .. I thought it was obvious enough since we are talking about integer.. (the set ℕ)

If you have Card(N), you have Card(N) -1 that is an integer but (Card(N)-1) * 2 is not. If you don't have Card(N), then there's no point in bijections anyway, because you can't compare or equal a quantity that doesn't exist. And again, "adding a room" contradict the very idea that the hotel contains ALL the rooms. People can't really ADD numbers to sets. Sets are equal to themselves only. A set union a new element, is a different set. Numbers are in or are not. You can create a set, by deciding a number has the property to be in the set.. (and then it's a DIFFERENT set after you added your new number) The Hilbert hotel can't be a "two state" hotel, with a "before" and a "after". You yourself state that bijection is not a stepped process. the Hotel is neither. The hotel is infinite and full. Any room you can imagine build, or create, IS ALREADY full.

And again, if A contains every integer, B= {a*2 } contains elements that are NOT in A, therefor if A = N, those added numbers are NOT integers. You can't "create them" at this imaginary point in your mind. And so your "couple" for the bijection doesn't exist. At most, if you say your a*2 is in A, you then have to check that a*2 has is own element in B. and because this never end, you really DON'T know. So no, as strange as it sound, "For every number, it is NOT possible to double that number". You can't double infinite and except it to fit in itself. For the number you can reach, yes, but, not for ALL.. "all" means ALL, to the END of the set. You don't see it, because you mistake reaching ALL for a recursive process.. But in reality, you recursive thinking NEVER really reach ALL. You do for one, for the other etc.. and you think it's ALL, but it's not. By definition, you can't reach ALL, because it's a infinite set. The only thing you can do, is put a unit of measurement, (ω) and see how your set behave according to ω. (that's what I did).. . In one sentence : if you assume N has a last number, this number doesn't have any counterpart in the bijection, and if you assume that N doesn't have a last element, you bijection is never proven true (because bijection requires all couples to exists). In fact, there is a very simple to prove it, but you will reject it because you trained to see it as something to "avoid" because of infinity instead of simply accepting it.. For every set if A = B union C and B and C are not empty and does not intersect. Card(A) = Card(B) + Card(C) and Card(A) > Card(B) and Card(A) > Card(C) So if N = {even numbers} union {odd numbers} Card(N) > Card(even numbers). NOT EQUAL. If this is not a proof, what is it ? There you have a reasoning, that consider all numbers together, absolutely out of time and steps, that doesn't involve knowing or building something from the elements, and simply consider the properties of numbers (being even or odd). And for some reason, modern math decided to go the other way that is just creating a lot of problems for no reason. It's like the Hilbert hotel.. the infinite hotel is full.. the paradox state that you can slide everybody to the next room. But really can you ? The hotel is full, so there is NO empty room at the end of the hotel.

I don't see "speculations" here.. but ok to obtain Card(Q) and Card(R), I simply counted the set according to values of ω, and it matches the sequences you add a lot of axiom everywhere to avoid the annoying paradoxes created solely by those bijections like Card(N) = Card(Even numbers) and Card(N) > Card(Even numbers) .. so for instance if S = 1 +1 +1 +1 ... and you separate the 1 in two equal sets S1 and S2, if you use your bijection, you get S=S1+S2 and S=S1=S2 so S=0.. yeah good job with your bijections ! In theory if your math come to a contradiction, you don't admire it as a magical mystery, you don't add a specific axiom to remove the one proposition you can't handle. you simply look for the wrong hypothesis.. (the bijection, obviously) The modern math is full of these absurdities to the point where a guy explain to you seriously that 1+2 + 3 +.... = - 1/12 And you call my idea "speculations".. I'm way better with my ideas, thanks

you can use Peano or the {{{{}}}} to create integers. It could also be simplified by "elements" that are different from each other, you don't even need to order them to define a bijection. In any case, if one of hypothesis is you have "all" the elements, you can't then use the creation process.. creating a new element different than all you already have means you didn't had all in the first place. my definition of bijection is the standard one.. I suppose You create a bijection between to sets A and B when each element of A is paired with a unique element of B and vice-versa (you can call it a "function" if you want, it's a set of couples from A * B) in other words : exist(bijection) = exist(a[1];b[1]) AND exist(a[2];b[2]) AND exist(a[3];b[3]) .. with obviously all the a[] and b[] different elements and no remaining elements from a or from b. If you give me an other one, without adding extra axiom to suit you demonstration, i will work with it. [Of course if you arbitrarily add "except for infinite sets", or you add "when ordered a certain way" you just added an axiom to avoid the contradiction i'm talking about.. If you add special axiom to the definition for infinite sets, why not add anything else you want ? ] "Why did you not take the opportunity to answer when you had more than 20 minutes logged on here after your last post?" (because I'm in France, in lunchtime, and in fact I don't think questioning my definition of the bijection is relevant) I start with assuming that Card(ℕ)=ω ω here is like a imaginary boundary - for any n ∈ ℕ, we have n<ω, - there is no n with n<ω and n not in ℕ - so ω-1 ∈ ℕ, and ω don't ℕ ℕ are integers using Peano or anything else that define the usual a+b nothing new here so : ℕ=[0;ω[ (so of course) ℚ={any q= a/b} with -ω<a<ω and 0<b<ω (a and b from ℕ) with the ordinary identities.. we can add that PGCD(a;b)=1 to remove equal couples a/b=c means a=b*c, so it implies a/b= a*x/b*x... (a,b,c,x integers and without the problematic zeroes) ℝ={any x=sum(q[n] with n=1..ω} (q[n] from ℚ) with -ω<x<ω I think I sometimes speak and come as an annoying person, I hope I didn't offend you in any way. that should'nt have any impact on the "math" though.

You would admit that, in your example, B can't be a part of A (because some elements in B are always bigger than the biggest in A) So it works for finit sets, but not with ℕ Because if A is ℕ, and B can't be a part of A some elements in B are not in ℕ (so it's not a bijection) you have something like : 1->2 2->4 n->n*2 ... it's Ok.. but : card(ℕ)-1 -> (card(ℕ)-1)*2 card(ℕ)-1 is in ℕ but (card(ℕ)-1)*2 >= card(ℕ) so it's NOT in ℕ You have an element in A that have no counterpart in B if B is a set of elements from ℕ you have A = {0 ;1;2;3;4;...} B = {0 ;2;4;6;8;...} in that case, you can't have A include B and Card(A) = Card(B) if Card(A) = Card(B) some elements of B are not in A you can't add new numbers to ℕ when needed And, in the other hand, if you don't define Card(ℕ)=ω, you can't prove that you have all the couples of elements from A and B (that is needed to say the bijection exists). Building a bijection is not a mathematical induction.. the fact that can always get new elements from A and B doesn't imply that the bijection exists or doesn't, because to do so, you have to reach the end.. (it's in fact, the opposite) You say that n->n*2 always exists for any n, but how can you prove that ? the fact that n -> n*2 doesn't imply that (n+1)->(n+1)*2 exists at all..

I understand that, but if you start with all the elements from ℕ as an hypothesis, you can't then use numbers out of that set. You can use n-> n (fortunately, I wouldn't contradict that!), but you can't have n -> n*2 or n-> n+1 because if you assume that you have ALL the elements from ℕ in the left set, some elements from the right set are always bigger, so they are not in the ℕ you started with (unless you didn't have ALL, which is a contradiction with the hypothesis, and in that case, the bijection wouldn't be decided) I came up with the construction of the bijection because I wanted to avoid using ω (sorry for the multiple editing)

I think I have proven that what's usually accepted as bijection and cardinality is all wrong, and nobody seems to believe me or care.. what should I do ? I know you don't easily reject old accepted ideas, but can you at least really think about it ? First : bijection between two infinite set are not true, they are UNDECIDABLE. When you build a bijection, it is true only when you don't get a new element from one of the sets or both, not just when you can still get elements from the sets. So with 2 infinite set, the bijection is NEVER proven true, because you can always get new elements from both. It just keep infinitely UNDECIDABLE. So you can NEVER prove a bijection between for instance ℕ and ℚ (precisely because you can always get new elements from both).. There's an infinite number of ways of ordering elements in ℚ (even in ℕ), that doesn't define the bijection as true or anything. (that seems important to the people I spoke to for some reason) Second, if you define Card(ℕ) =ω You can in fact easily find Card(ℚ) and Card(ℝ) : Card(ℕ) =ω implies that ( a ∈ ℕ ) if and only if ( a<ω ) so ( a+b ∈ ℕ ) if and only if ( a+b<ω ) and so ( (a/b) ∈ ℚ ) if and only if (-ω<a<ω ) and (b<ω) So you can actually count the number of element in ℚ according to ω : Card(ℚ+)=A018805 (ω-1) +1 Card(ℚ)=Card(ℚ+)*2-1 = A018805 (ω-1) *2 +1 Card(ℚ)=~ ω² * (3/Pi²) Numbers in ℚ are spreaded like a fractal and are symmetrical by the transformation x=> 1/x Then I build each x in ℝ as a sum of ω numbers from ℚ : ℝ={any x=Sum(a[n]/b[n]; n=1..ω} with -ω<a[n]<ω et 0<b[n]<ω and -ω<x<ω (That sum can actually be > ω) I find that : Card(ℝ)=(A003418(ω-1)*ω*2)-1 and guess what : Number in ℝ are spreaded perfectly evenly in ]-ω;ω[ with a step of 1/A003418(ω-1) (that is not obvious at all given the spreading of numbers in ℚ) I know it's extraordinary claims.. very different from what is currently the mainstream math.. but can you prove me wrong ?