Comments by "MC116" (@angelmendez-rivera351) on "Infinitesimal Monad (extra footage)" video.

  1. Ian moseley That is not atomic physics per se, and the uncertainty principle has nothing to do with infinitesimal numbers. In fact, the statistics of the uncertainty principle come from real-valued mathematics with no infinitesimal numbers in it.
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  2. Generic Internetter Not at all. There are no infinitesimals in sub-atomic physics specifically.
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  3. maciej53 The hyperreal numbers are OBJECTIVELY SIMPLER than the ε-δ definition of limits. The set of axioms and definitions contains less quantifiers and the conditionals contain less connectives, and most proofs are deductively shorter. It literally is simpler by definition. The hyperreal numbers are an objectively better and more useful way of doing mathematics with exactly the same level of formal axiomatization and rigor.
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  4. PiaNoONE They are not going back to a messy definition. If anything, limits are messier than infinitesimals, AND at best, only equally as rigorous as the hyperreal numbers. Abraham Robinson proved this. The proofs are simpler, the axioms are simpler as they require less connectives and quantifiers, and they are objectively more useful as there is more that can be done with infinitesimals.
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  5. Tim Williams Because mathematicians are the worst at nomenclature.
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  6. FBI Wrong. 1/k does have a a monad, and the monad is given by the set of all numbers of the form 1/k + r/k^2, where r is a real standard number.
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  7. Fabrice Nonez No, that is wrong. The hierarchy of infinitesimals in the hyperreal numbers ALREADY is infinite by definition. The monad of 1/K is 1/K + r/K^2 where r is real, and this is different from the monad of 0 in that it is strictly a subset.
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  8. Thomas Bridge The power set of the countable infinite cardinal K is, to be precise, 2^K, and this is strictly bigger than K. However, it is not known and it is furthermore not provable that 2^K is equal to the cardinality of the real numbers.
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  9. Jacob Marburger R* is the set of hyperreal numbers, and N* the set of hypernatural numbers. The construction can be found very easily if you search “hyperreal numbers” on Google. The theory is super rigorous and expansive.
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  10. Jaap Versteegh Yes.
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  11. aman makhija Incorrect description. The fact that two numbers share the same monad does not disprove the idea of both numbers still having a monad.
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  12. particle No. Incorrect. The monad of 1/e is simply the set of numbers of the form 1/e + 1/e^2*r, where r is a real number.
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  13. Grugknuckle It is not useless for everyday life. Maybe get more educated on the subject?
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  14. Matti Kauppinen No, 0.(9) and 1 are THE SAME NUMBER in every Euclidean number system, even in the surreal numbers, which contain the hyperreal numbers. This is true by definition. There is no describable or effectively constructible and formally axiomatizable number system which preserves the definition of rational numbers and addition such that 0.(9) is infinitely close to 1 yet not equal to 1. There exists a number which is infinitely close to 1 of the form 1 - ε, but whatever this number is, it cannot be 0.(9), because infinitesimal numbers, by construction, have no finite or infinite decimal expansion.
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  15. Martin Savc The real numbers are not limited to fractional notation. This is why 1/K cannot be real.
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  16. Fabrice Nonez 0.(9) = 1 in the real numbers, in the hyperreal numbers, and in the surreal numbers, because 0.(9) by construction is merely a real number whose infinitesimal part is 0. There are multiple ways of dealing with irrational numbers and infinite decimal expansions for real numbers without ever needing to invoke the limit. In fact, it would be somewhat of a mistake to invoke it. You can appeal to the Dedekind construction of the real numbers and prove that the number is 1, not because the infinitesimal part of it cannot be described, but because THERE IS NO INFINITESIMAL PART.
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  17. Tau being the better circle constant is completely debatable.
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  18. notoriouswhitemoth Between any two hyperreal numbers, there are infinitely many surreal numbers. Between any two surreal numbers, there are infinitely many surreal numbers, because the set of surreal numbers has the biggest cardinality of any ordered field and the longest measure. The cardinality of the surreal numbers is the same as that of the universe of set theory.
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  19. Giannhs Polychronopoulos You are asking if it is possible? Possible it is, but very difficult. I think the key to the fact is that the surreal numbers technically do not form a set. They’re too much to be able to form one. Instead, they form what is called a class. Though you cannot have a set of all sets, you can have a class of all ZFC sets, for instance. These classes form a more general type of object than a set. It is in the study of clases that you can prove a statement such as the cardinality of the class of surreal numbers is that of the set theoretic universe. Edit: I know I did use the term set in my first comment, which I realize is misleading. I did not want to delve into the technicalities initially, and using the word “collection” instead does not feel better, though.
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  20. Giannhs Polychronopoulos ah, I don’t the know the specifics, to be honest. I watched John Conway’s lecture about games and how his theory of games led to his discovery of the surreal numbers. However, the class of games is a super-class of the surreal numbers and is said to be “bigger” in some sense. In this lecture, he did say specifically the cardinal number describing the class of surreal numbers is that of the set theoretic universe. He did not specify an explanation of proof of this, which I imagine he intended for us to take for granted in the lecture, since the lecture is not about classes and cardinalities in general. I have read online that the surreal numbers form a class, not a set, but I don’t know of any specific books to read. Actually, I’m looking for books of my own myself to read about it too. I can link you his lecture, though, if it helps
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  21. The hyperreal numbers are useful in physics, economics, and game theory.
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  22. Chuck Zhang No, there is no such a thing as the last digit of an infinitesimal. Infinitesimals, by construction, do not have decimal expansions. Only the real numbers do that. 0.000... 1 does not exist.
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