Comments by "MC116" (@angelmendez-rivera351) on "How to Add" video.

  1. All hyperoperations can be defined in this fashion. Given the μth hyperoperation, denoted %, the S(m)th hyperoperation, denoted #, can be defined by having m#0 = 1, m#S(n) = m%(m#n). Even more direcrly, we can define a function H : N^3 —> N such that H(m, n, 0) := S(m) & H(m, S(n), S(μ)) := H(m, H(m, n, S(μ)), μ). This uniquely defines every hyperoperation all at once. This is related the Ackermann function.
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  2. This is incoherent nonsense.
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  3.  @simongunkel7457  I don't think that is what OP was referring to.
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  4. This is wrong. Numbers do care if they are well-defined or not. There is no sense in which something "just is." That is not how it works.
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  5. Relax. We have not even discussed integers yet.
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  6.  @rmsgrey  The problem is that when you are working with real numbers or the algebraic numbers, it becomes impractical and conceptually useless to work with explicit constructions and encodings. To properly give a formal introduction to the algebraic numbers, you need ring theory, and to properly give a formal introduction to the real numbers, you need lattice theory on top of ring theory. Set-theoretic constructions are not appropriate when dealing with these higher-level mathematical objects. For instance, axiomatically, it is very easy to write a list of simple axioms that uniquely define what the real numbers are. Talking about the algebraic numbers is even easier: the field of algebraic numbers is the algebraic closure of the field of rational numbers. However, while it is very easy to understand the axioms, actually constructing these objects using nothing but sets is complicated, and to be honest, a waste of time. That is not to say that it cannot be done, but rather, that it should not be done.
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  7. Subtraction is just addition of integers. So we only need to construct the integers.
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  8. The sets in the hierarchy 0, {0}, {0, 1}, {0, 1, 2}, etc., are special, and so they get their own special names. This is not so for every set.
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  9.  @AGENTX506  Well, that is not entirely true. A defining feature of the successor function is that it is injective. So, if the predecessor does exist, then it is unique. The predecessor of n can easily be defined as the unique quantity m such that S(m) = n, if it exists.
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  10.  @AGENTX506  Agree with you that using the successor to start with is easier, but you are severely exaggerating the difficulties of dealing with the predecessor function. The only natural number with no predecessor is 0, and this is not difficult to prove, and there is no need to even deal with that, because of the n + 1 and n·1 cases made explicit in the definitions.
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  11.  @AGENTX506  Over the whole numbers? You said later in your comment it is defined over the integers, which is the same thing. This video deals with the natural numbers. As I already pointed out to you, there is no need to null-check every instance of the predecessor, and null-checking is also relatively trivial, it does not make it particularly difficult. At worst, it is just slightly annoying.
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  12.  @AGENTX506  What do you mean by working with "the whole numbers"? We are working with the natural numbers here.
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  13. The better way to phrase that is that for any natural number n, the succesor S(n) is defined as the set {0, 1, 2, ..., n}, where 0 = {}.
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  14. This video already covered multiplication.
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  15. The video has captions in English.
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  16. A bracket is just a grouping symbol. It is not an operation, so there is no need to define how it works.
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  17.  @ready1fire1aim1  I am not uneducated, I have graduate level education in these topics. I know you do not.
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  18.  @ready1fire1aim1  For example, I know that quantum field interactions do not carry a specific dimensionality to them, and those interactions are not forced, they are energy state transitions of fields.
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  19. He is just doing it so that the recursive definition is intuitively seen. Defining n·1 = n is completely redundanr, as you point out.
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  20.  @vojtechsejkora1554  No. I meant what I said, and what I said is true. Nothing you have said disproves what I said. Also, yes, I am aware that we have only constructed the natural numbers, not the integers, but this is irrelevant, since we know there will be a future video on this, and that video will cover subtraction as addition as well.
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