Difficulties With Dedekind Cuts MP3/MP4 Free Download

  • Insights into Mathematics
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Download Difficulties with Dedekind cuts | Real numbers and limits Math Foundations 116 | N J Wildberger MP3 music or Mp4 Video file at 320kbps audio quality and Full HD. Difficulties with Dedekind cuts | Real numbers and limits Math Foundations 116 | N J Wildberger music file uploaded on 01 December 2014 by Insights into Mathematics.

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Difficulties With Dedekind Cuts

Difficulties with Dedekind cuts | Real numbers and limits Math Foundations 116 | N J Wildberger

  • Insights into Mathematics
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Public Response On Difficulties With Dedekind Cuts

bvoq7 Months ago
It's really a bit more complicated. I tried to make the distinction between algorithm and choice of the cuts and their equivalences: 1. An algorithm that selects all (finite) possible selections and terminates <=> there is no choice analogy 2. An algorithm that selects countably many rational numbers and selects them in an order (for example Stern-Brocott tree order) <=> A choice has to be decidable in a finite time of computation [For example the choice of x in sqrt(2) can be checked in finite amount of time: x < 0 or x^2 < 2]. 3. An algorithm that selects countably many rational numbers and selects them in any order <=> A choice (if it is in the set) has to be computable in finite time but not if it isn't. [also called recursively enumerable, the pi^2/6 choice] 4. Statements involving any choice over rational numbers. There are 2^Q choices, which is not countable! This leads to paradoxes of the axiom of choice. Look up the infinite prisoners hat dilemma for a great illustration. The reason it leads to a problem/paradox is that we assume it is possible to create a random (martin-löb random) sequence of hats. I'm not saying it's an issue to create countably many prisoners but selecting a random hat for each one is where the problem lies. You can't write down the order of those hats in any way using an algorithm. It is 4 that really is the issue. I can't possibly select an arbitrary cut (that's the equivalent of selecting a real number uniformly at random, it's not possible).
Max Percer7 Months ago
There is another thorny question that comes to my mind. Can we construct every irrational number using set builder notation and rational numbers, e.g. A = {a \in Q | a^2 < 2 or a <0 }. But what about irrational numbers like e. Even if we use limits such as e = lim { n → oo } { (1 + 1/n) ^n} , how do we know that such limit expressions adequately cover all irrational numbers. It seems that dedekind cuts just moves the goal post of how to deal with the continuum and relegates it to set theory and set builder notation.
Neilcourtwalker7 Months ago
In my simple understanding I would say: It is not surprising, that you can't do arithmetic with this definition, because, if you want to define irrational numbers with rational numbers and these cuts, you define them by expressing what they arent, but not what they are. Would you agree with this statement?
Santiago Erro Alvarez7 Months ago
30:40 π²/6 = { a∈ℚ | ∃N∈ℕ: a < ∑(n=1,N) (1/n²)} That's not an infinite amount of conditions.
Paket Baand7 Months ago
I'm sorry but not knowing or not trying to find out, how subsets (and ultimately dedekind cuts) are justified in set theory, doesnt mean that they arent. If your source material doesnt define set/subsets rigorously then you should look if there is one that does. Also, just because you cant use Dedekind Cuts to do arithmetic in the real numbers doenst mean they dont exist. They are just a proof that a complete ordered field exists. (which the rational numbers are not)
Vincenzo7 Months ago
"Little one, it's a simple calculus. This universe is finite, its resources finite." - Thanos
brenda williams7 Months ago
I mean the reflections are there.
brenda williams7 Months ago
It seems polytopes and geometry makes a pretty strong case.
Verschlungen7 Months ago
Beautiful presentation! Thank you for being the voice of reason! Here is another way, quite different from yours, that leads to the same view of the Dedekind cut: The term 'irrational number' (as applied to root 2, pi, etc.) I regard as a kind of carnival huckster's trick in that it focuses one's attention on the adjective 'irrational' when our attention should be on 'number': I.e., IS root 2 a number? IS pi a number? No. Rather, there exists an algorithm to produce digits of root 2 or digits of pi For Ever, but the output from an algorithm that runs For Ever is not a number, it is (again) the output from an algorithm that runs For Ever. In other words, since root 2 and pi, etc. are not numbers, they have no place on the number line, so Dedekind need not have wasted his time trying to find a clever way to put them there. Being non-numbers, they live in a separate, non-number space of their own, a space where computer algorithms run. (And those algorithms run not 'to infinity', which is an infantile babble-phrase, but FOR EVER, which is a grown-up concept that actually works.)
MGSnyder7 Months ago
Why did I watch this before doing my homework on field theory? 😭

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    Public Response On Difficulties With Dedekind Cuts

    bvoq7 Months ago
    It's really a bit more complicated. I tried to make the distinction between algorithm and choice of the cuts and their equivalences: 1. An algorithm that selects all (finite) possible selections and terminates <=> there is no choice analogy 2. An algorithm that selects countably many rational numbers and selects them in an order (for example Stern-Brocott tree order) <=> A choice has to be decidable in a finite time of computation [For example the choice of x in sqrt(2) can be checked in finite amount of time: x < 0 or x^2 < 2]. 3. An algorithm that selects countably many rational numbers and selects them in any order <=> A choice (if it is in the set) has to be computable in finite time but not if it isn't. [also called recursively enumerable, the pi^2/6 choice] 4. Statements involving any choice over rational numbers. There are 2^Q choices, which is not countable! This leads to paradoxes of the axiom of choice. Look up the infinite prisoners hat dilemma for a great illustration. The reason it leads to a problem/paradox is that we assume it is possible to create a random (martin-löb random) sequence of hats. I'm not saying it's an issue to create countably many prisoners but selecting a random hat for each one is where the problem lies. You can't write down the order of those hats in any way using an algorithm. It is 4 that really is the issue. I can't possibly select an arbitrary cut (that's the equivalent of selecting a real number uniformly at random, it's not possible).
    Max Percer7 Months ago
    There is another thorny question that comes to my mind. Can we construct every irrational number using set builder notation and rational numbers, e.g. A = {a \in Q | a^2 < 2 or a <0 }. But what about irrational numbers like e. Even if we use limits such as e = lim { n → oo } { (1 + 1/n) ^n} , how do we know that such limit expressions adequately cover all irrational numbers. It seems that dedekind cuts just moves the goal post of how to deal with the continuum and relegates it to set theory and set builder notation.
    Neilcourtwalker7 Months ago
    In my simple understanding I would say: It is not surprising, that you can't do arithmetic with this definition, because, if you want to define irrational numbers with rational numbers and these cuts, you define them by expressing what they arent, but not what they are. Would you agree with this statement?
    Santiago Erro Alvarez7 Months ago
    30:40 π²/6 = { a∈ℚ | ∃N∈ℕ: a < ∑(n=1,N) (1/n²)} That's not an infinite amount of conditions.
    Paket Baand7 Months ago
    I'm sorry but not knowing or not trying to find out, how subsets (and ultimately dedekind cuts) are justified in set theory, doesnt mean that they arent. If your source material doesnt define set/subsets rigorously then you should look if there is one that does. Also, just because you cant use Dedekind Cuts to do arithmetic in the real numbers doenst mean they dont exist. They are just a proof that a complete ordered field exists. (which the rational numbers are not)
    Vincenzo7 Months ago
    "Little one, it's a simple calculus. This universe is finite, its resources finite." - Thanos
    brenda williams7 Months ago
    I mean the reflections are there.
    brenda williams7 Months ago
    It seems polytopes and geometry makes a pretty strong case.
    Verschlungen7 Months ago
    Beautiful presentation! Thank you for being the voice of reason! Here is another way, quite different from yours, that leads to the same view of the Dedekind cut: The term 'irrational number' (as applied to root 2, pi, etc.) I regard as a kind of carnival huckster's trick in that it focuses one's attention on the adjective 'irrational' when our attention should be on 'number': I.e., IS root 2 a number? IS pi a number? No. Rather, there exists an algorithm to produce digits of root 2 or digits of pi For Ever, but the output from an algorithm that runs For Ever is not a number, it is (again) the output from an algorithm that runs For Ever. In other words, since root 2 and pi, etc. are not numbers, they have no place on the number line, so Dedekind need not have wasted his time trying to find a clever way to put them there. Being non-numbers, they live in a separate, non-number space of their own, a space where computer algorithms run. (And those algorithms run not 'to infinity', which is an infantile babble-phrase, but FOR EVER, which is a grown-up concept that actually works.)
    MGSnyder7 Months ago
    Why did I watch this before doing my homework on field theory? 😭
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