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What are the additive and multiplicative qualities?

That doesn’t prove that the continuum hypothesis is true because it doesn’t say anything about the actual value of the continuum. What’s important to understand when dealing with CH is that we don’t know which ℵ number 2^ℵ0 actually evaluates to. It’s a bit like if somebody asked you to calculate 2×3 but you didn’t know what 3 meant. The reason we know CH is unprovable is that we can describe mathematical universes in which the continuum is exactly ℵ1 and other universes in which the continuum is provably larger than that. The former is true in Gödel’s constructible universe L and the latter is true in Cohen’s forcing extension. Both are somewhat difficult to understand.

the same like in every field - you can't -since divide by zero is multiply by the reciprocal of 0, but the reciprocal of zero is undefined (since 0 * any = 1 means 0 = 1 means FALSE)

Me when I can’t find 2-3 because it’s impossible to subtract 3 from 2 and trying to do so would be meaningless speculation

“[…] by definition it has no size”. Yeah, no, this depends on the definition of “size”, or rather “cardinality”. If we define it in such a way that it makes sense for infinite sets, then by definition they will have a size.

Most mathematics is "fanciful speculation" and also "meaningless jiberish". For instance, there isn't any infinities in the physical universe. Mathematics IS NOT about physical universe, it is about relationships between abstract entities - it doesn't need to have connection to anything real.

There are more things in Heaven and Earth (and Mathematics), @nickosc88, than are dreamt of in your philosophy.

It's a definition by recursion; every addition in the definition involves an earlier created number. Eventually, you'll get to 0, which has no options, and the process stops. An explicit example: 1+ 1 = {0|} + {0|} = {1+0,1+0|}; 1 + 0 = {0|} + {|} = {0+0|}; 0 + 0 = {|} + {|} = {|} = 0. Thus 0 + 0 = 0; 1 + 0 = {0+0|} = {0|} = 1; 1 + 1 = {1+0|} = {1|} = 2. A bit lengthy, but way shorter than Principia Mathematica. The surreals are said to be well-founded, in that it is impossible to find an infinite sequence a0, a1, a2,..., where a1 is an option of a0, a2 is an option of a1, etc. (There are plenty of surreal numbers where you can construct such a sequence of any given finite length, but never one of infinite length.) This allows us to make definitions such as this one for addition, provided that we make sure to always involve earlier created numbers. For more information, look up transfinite recursion and transfinite induction, both of which are used extensively with the surreals. Edit: It's not transfinite recursion and induction, but e-recursion and e-induction, where 'e' should be the 'is an element of' (epsilon) symbol, which my phone keyboard lacks.

I don't know enough about the hyoerreals to answer your question concerning them. As for the surreals, omega is still the smallest infinite ordinal, where ordinals are defined to be surreal numbers with no right options.

Well, in hyperreal setting. It's really a relabeling of lemniscata symbol ∞ for ω, given that a concrete definition, that is its the ordinal representing the cardinality of ℕ. Said that, they are just different relative to others transfinite numbers of the same order, that is, ω + 1 < ω + 2 < ω², or their reciprocal ε, such that ε = 1/ω. But compared to the finite numbers, is just as ∞ in the standard analysis. Example: (ω²+2ω)/ω = ω+1 -ω/√ω = -√ω (3ε²+2ε)/ε = 3ε+2 5ε/ε = 5 When taken the "standard part" that will be simplified to a finite number or a regular old infinity. The examples above have std part equal to: ∞, -∞, 2, and 5, respectively.

As an easier example, let m=3 and n=2. Then an element of m^λ is a ternary sequence f of length λ and an element of n^λ is a binary sequence g of length λ. Take any ternary f and reinterpret it in binary to get a binary g. Go from binary to ternary to get the other map direction. If you can show these are both injective or both surjective, then the Cantor-Bernstein theorem implies they have the same cardinality. Alternatively if n<m, the easiest map is just the inclusion map from n^λ into m^λ that does nothing to f. Then all you have to do is show the map in the other direction is an injection. Another way, pick any fast growing function s from λ to itself. Then for a given f∈m^λ, for every ordinal α<λ, 1. if f(α)<2, make h(s(α))=f(α) and h(β)=0 for everything else. 2. if f(α)≥2, then set h(s(α))=0 and h(s(α)+1)=f(α)+1 3. h(β)=0 for everything else. Show this is an injection. It is basically just carrying like in basic arithmetic but with enough gaps to avoid any infinite carrying.

Obviously, there are uncountably many irrational numbers. There are countably many rational numbers, but uncountably many real numbers; and union of two countably infinite sets can't be uncountable. It's also fairly easy to construct an explicit one-to-one mapping between real numbers and irrational numbers.

It's possible to construct an uncontable infinite set from a cartesian product between two countable infinite sets. Though, it's not in any way mappable to a countable set. My argument in favor of the cardinality: |ℝ| > |ℤ| is that you can imagine the real numbers as the Cartesian product between two natural sets, but with a key twist. While the "integer part" of the reals grows in absolute value: {X ∈ ℕ | 0 ≤ |X| ≤ ...99999} (where we can pair numbers from the interval ]-∞, +∞[ to the interval [0,+∞[ ), the "decimal portion" decreases: {Y ∈ ℕ | 1 > 0.9999.... ≥ Y ≥ 0} (also mappable to the naturals, but read backwards). If we plotted this Cartesian product X × Y, we would have an infinite set of decimals for each of the infinite real integer values, i.e., ∞ ^ ∞ (which is mappable to 2^∞, the power sets of the naturals). However, it is still impossible to list all real numbers, as we can always add a new digit to the left in the integer part or to the right in the decimal part. Moreover, even if you try to "move the decimal point" all the way to the right, attempting to transform reals into integers, it is always possible to add a decimal interval after this integer, turning it back into a real number. This is trying to illustrate, in an intuitive way, that you already "exhaust" the set of naturals/integers just to count between 0 and 1, thus the cardinality of the reals is greater than that of the integers and naturals.

@@linuxp00 "It's possible to construct an uncountable infinite set from a cartesian product between two countable infinite sets." No, that's not possible. Cartesian product of two countably infinite sets is countably infinite; and this result doesn't require axiom of choice. (As opposed to the more general statement that union of a countably infinite collection of countably infinite sets is countably infinite, which *does* require axiom of choice, or a weak variant thereof - axiom of countable choice. For example, it's consistent in absence of axiom choice that the set of real numbers is a union of countably many countable sets; and that doesn't contradict Cantor's result that there are uncountably many real numbers.) There's no such natural number as ...99999. Any natural number is finite in magnitude, and therefore its decimal (or base-n) expansion has finitely many digits. And that's by definition; a set is finite, if its number of elements is equal to some natural number. (It is an empty set, or it has exactly one element, or it has exactly two elements, or it has exactly three elements, or... and so on. Of course, the "and so on" part is a bit involved; you can't have a formula of an infinite length.) The mathematical structure towards which you are heading are the n-adic numbers; in this structure, any base-n expansion (infinite sequence of base-n digits) is an element of this structure, and then you define mathematical operations on these base-n expansions. For example, the number ...999 is equal to -1 - you can verify that indeed ...001+...999=0. (Usually, n is taken to be a prime number, because otherwise you have divisors of zero - a pair of non-zero numbers a and b such that a*b=0.) But the problem is that there are uncountably many n-adic numbers - any infinite sequence of digits corresponds to a unique n-adic number. (From Cantor's diagonal proof it follows that n-adic numbers can't be mapped one-to-one with natural numbers; however, they can be mapped one-to-one with real numbers.)

@@MikeRosoftJH I'm inspired by adics, but I'm not using an actual adic arithmetics. I'm just using naturals as infinite strings of digits in reverse. Naturals are infinitely countable, so are their digits. So, as I see it the whole set of Naturals are contained in the interval [0,1[. Though, it's the same for all the gaps between two consecutive integers, so it is impossible to count the reals. No need for Diagonalization, just Zenon's Paradox in it's essence.

@@linuxp00Again, every natural number is finite in magnitude, and therefore has finitely many non-zero digits. If you reverse the decimal expansion of a natural number, you get the set of all real numbers in an interval whose decimal expansion has finitely many digits after the decimal point. And that there are countably many such numbers is not a new result; all such numbers are rational, and that there are countably many rational numbers was proven by Cantor himself. And that one specific function from natural numbers to real numbers doesn't cover all reals doesn't by itself prove that real numbers are uncountable. It's a well-known result that an infinite set can be mapped one-to-one with its strict superset or subset. (Precisely, this is true for every set which has a countably infinite subset.) For example, consider the function n->2*n on natural numbers; or, conversely, the same function on an interval from 0 to 1. So in order to prove that set A can't be mapped onto set B, it doesn't suffice that one specific function between the two sets doesn't cover all elements of B; you have to prove that no function does. (Such as by using the diagonal proof, or the earlier Cantor's proof by nested intervals.) In set theory, an infinite sequence is not a process. It's a function whose domain is the set of natural numbers. And a function is itself a set: set of ordered pairs. For example, the function f: x->2*x on the set {1,2,3} is the set {[1,2], [2,4], [3,6]}; if the pair [a,b] is an element of the set f, we by convention say that f(a)=b. Zeno's paradox isn't a theorem of mathematics. In fact, Zeno was heading towards perfectly well-defined concepts of calculus, such as limit of a sequence, infinite sum, or derivative of a function. (Except for that he wasn't able to understand the infinite case, and so he instead concluded that the infinite case is impossible.)

I didn’t know this was disputed, but I was mind blown when my teacher showed me a proof for 0.999 repeating equaling 1.0 but honestly I just accepted it because I figured he knew more about math than I did. Now you’ve got me questioning the proof

You're assuming that surreal numbers are necessarily ordered; that, for any surreal a and b, exactly one of a>b, a=b, a<b holds. This is true, but needs to first be proven.

First time i encountered this set (learned maths in french). So i know : N, Z, D, Q, R, C

Algebraic numbers

@@Xnoob545 Thanks!

You can't prove (without additional axioms) that there is such a set whose cardinality is strictly between Aleph-0 and P(Aleph-0); and you can't prove that there isn't such a set. In other words: there's a model of set theory where the set in question doesn't exist (such as the constructible universe); and there is a model in which the set does exist. (The proof uses the technique of forcing: start with some model of set theory, such as the constructible universe, and then add elements to it so that the new model still satisfies the axioms of set theory, and also satisfies the proposition whose consistency with the rest of set theory axioms we are trying to prove; for example, continuum hypothesis is not true, or real numbers can't be well-ordered, or the like.)

Well, according to my current understanding, any real number can be represented as a triplet: [a sign (-1 or 1), a natural number for the integer part (digits to the left of the decimal point), and a natural number for the decimal part (digits to the right of the decimal point)]. So, between any two consecutive natural or integer (naturals + sign) numbers, there are infinite natural numbers as decimal digits, which hint at a discretization, but that are uncontable because just as integers grow unbound, decimals shrink in the same manner, making their count unfeasible. Because of that I think that if continuum hypothesis is set to false, that also allow us to define infinitesimals conceptually, with their fractal nature (i.e. it's possibly to do magnifications on the real number line and it will self similiar, either by zoom in or zoom outs). P.S: Writing π as 3 + 0.14596... = (+1, 3, ...69541). It is possible to have the infinite digits of π, or any irrational.

@@linuxp00 It's indeed possible to define infinitesimals, by creating an alternate model of first-order theory of real numbers (i.e. the hyperreals). But continuum hypothesis has nothing to do with it; on real numbers there are no non-zero infinitesimals, regardless of whether or not continuum hypothesis is true. Likewise, fractals have nothing to do with this.

@@MikeRosoftJH yeah, I mistook my words on it. But, surreals have this fractal design with their higher-orders of infinities and infinitesimals. May it help consider them in the CH problem.

This is just pure math, no real world implications at all. Basically, mathematicians said: what if we pack all numbers in one black box and name this box "omega"? One can object: but numbers are infinite, so how can one fit them all in a box? But we should not forget that this is just math, it is all only about being consistent with your first assumptions no matter whether they are from reality (like biology and physics) or just from mere mind. So, by using this box "omega" as a new unit, we can now count the boxes: first omega box, second omega box, ... millionth omega box and go on. Here again you can see that the series of boxes is endless, so why do not pack all these boxes in a new bigger box? this new box has "omega" number of smaller boxes, which by turn has "omega" numbers each. The new box is called then "Omega times Omega". And we continue counting this way, and getting more bigger embedding boxes. This recursive process is what we call "ordinal generation".

@@boubakersoltani4566 Thank you so much for helping me to better understand this! It truly means a lot! I am currently pursuing a major in mathematics. Is there any chance I could email you to ask about this further please?

@@masonmireles9295Hi Mason! I wanted to really thank you for your trust, but indeed I am not that expert in set theory. Maybe you won't gain much from me.

@@boubakersoltani4566 Okay. Thank you anyways for your help! I truly appreciate it! Not a lot of people go out of their way to help others out anymore. You helping me truly made my day.

@@masonmireles9295 glad to here that ^^ I wish you all success.

I just came across your comment and was curious. How'd the thesis go?

e0 is omega to the power omega to the power omega...[repeat omega times]. e0 is much much smaller than w1

@@hypercosmologyexplanations5178 You mean greater as it is an infinite tetration/Hyper"w" ?

@@bakert11 wonderful explanation.

Hi, the idea is you assume that the set is listable/countable. Set of natural numbers, N is listable as you can write them down : 1, 2, 3, ..... Given any natural number, I can find it in the above list. I can do the same with Z and Q. But not possible with R. Simple reason for me is, N, Z, Q are all discrete sets whereas R is continuous, so if you suppose it is listable, put them in an order no matter how, I can always construct a new real number which was not in the above list. The same way interval (0, 1) is not listable. Here is a try : Suppose the open interval (0, 1) is countable. So lets list them. Here are the decimal numbers. a_11, a_12 etc ... are the digits. A1 : 0.a_11 a_12 a_13 .... a_1n ... A2 : 0.a_21 a_22 a_23 .... a_2n ... A3 : 0.a_31 a_32 a_33 .... a_3n .... ..... An : 0.a_n1 a_n2 a_n3 .... a_nn .... Now if we can find a decimal number which is outside the above list, then our assumption is wrong. Can we find a new number ? Yes we can. Consider this element A = 0.a_1 a_2 a_3 ... a_n ... with the condition that a_1 ≠ a_11 , a_2 ≠ a_22, a_3 ≠ a_33 .... a_n ≠ a_nn. The 1st inequality above proves A ≠ A_1, 2nd inequality proves A ≠ A_2 etc ... nth inequality proves A ≠ A_n etc ... Hence A is not equal to any of the numbers above. Therefore A was not listed and hence the list was not complete. If (0, 1) is not countable then R can't be either as (0, 1) ⊂ R. Here is video on Numberphile useful. ruclips.net/video/elvOZm0d4H0/видео.html

The reason is: every natural number is finite, and so it has finitely many digits. (And that's by definition; a set is finite, if its number of elements is equal to some natural number - it's an empty set, or it has exactly one element, or it has exactly two elements, or ... and so on.) If you try to apply the diagonal procedure on the sequence of all natural numbers, you get a sequence which has infinitely many non-zero digits, which doesn't represent any natural number. Likewise, when you apply the diagonal procedure to a sequence of all rational numbers - and such a sequence exists - then the resulting number must be irrational.

My argument in favor of the cardinality: |ℝ| > |ℤ| is that you can imagine the real numbers as the Cartesian product between two natural sets, but with a key twist. While the "integer part" of the reals grows in absolute value: {X ∈ ℕ | 0 ≤ |X| ≤ ...99999} (where we can pair numbers from the interval ]-∞, +∞[ to the interval [0,+∞[ ), the "decimal portion" decreases: {Y ∈ ℕ | 1 > 0.9999.... ≥ Y ≥ 0} (also mappable to the naturals, but read backwards). If we plotted this Cartesian product X × Y, we would have an infinite set of decimals for each of the infinite real integer values, i.e., ∞ ^ ∞ (which is mappable to 2^∞, the power sets of the naturals). However, it is still impossible to list all real numbers, as we can always add a new digit to the left in the integer part or to the right in the decimal part. Moreover, even if you try to "move the decimal point" all the way to the right, attempting to transform reals into integers, it is always possible to add a decimal interval after this integer, turning it back into a real number. This is trying to illustrate, in an intuitive way, that you already "exhaust" the set of naturals/integers just to count between 0 and 1, thus the cardinality of the reals is greater than that of the integers and naturals.

There ARE actually surcomplex numbers.

@@ranjitsarkar3126 damn

Yes dude👏👏👏👏

if you mean i as in the imaginary unit then no, however according to wikipedia there is an extension of the surreal numbers called the surcomplex numbers, although the surcomplex numbers aren't a proper class

naively, I would think you could just invent a new system where you "rotate" about the left and right "handed-ness" of this system.

I don't know for sure, but I suspect not. I think the surreals can be ordered, as positive or negative, obeying the usual arithmetical rules (negative times negative equals positive, etc), but the complex numbers cannot be. For instance, the usual laws of positive and negative arithmetic would suggest that if i.i=-1 then -i.i=+1, but it doesn't. It just equals -1 again. However, you can obviously extend the surreal numbers to surreal complex numbers by adding i in.

@@donaldb1 Uhm... -i.i equals +1...

@@joelproko Oh yes, so it does. Haha. Um.

What I _meant to say_ is, if you can have positive or negative then, 1) For any number x either x or -x (but not both) should be greater than zero 2) the product of any two numbers greater than 0 should also be greater than zero So, by 1) either i or -i should be greater than zero, in which case their square should be greater than zero, by 2). But they both square to -1. So they can't be positive or negative. Is that better?