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wtf last won the day on March 6

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  1. This forum allows this nonsense and bans a discussion of the ordinal numbers?
  2. No, I don't think u did. All u did is say how can I handle your example. I can kick myself for not seeing this. I like to think of myself as being extraordinarily imaginative. I should have anticipated this. I know u don't want to discuss this further, but its occurred to me before the 2 of u have  more problems with this example: 1) What then are u going to do with the "everything else"? 2) Could your example therefore be an actual WO for the set in question?

  3. It's refuted by the order "1/2 <* 1/4 <* everything else." We've been over this many times already. If you say, "Ok just apply the idea twice," I'll give you "1/2 <* 1/4 <* 1/8 <* everything else." If you say, "Apply it countably many times" I'll give you [math]\omega + 1[/math]. And I can keep on going like this right up to the first uncountable ordinal and beyond. You have NEVER responded satisfactorily to this refutation of your argument. It's clear that whatever context your prof used this P idea, it wasn't this one.
  4. Your earlier arguments that there's no well-order of the reals are wrong. This paragraph is "not even wrong." It's incoherent. It says nothing and means nothing. Is Property P like Preparation H?
  5. This business with P is half baked because you haven't yet incorporated it into a complete argument about anything. Secondly if you are retracting your admission that you're wrong, please respond to the question I asked earlier. How does your argument handle the case of the order 1/2 <* 1/4 * <* 1/8 <* ... <* everything else?
  6. The axiom of choice says that you can choose a legislature. If you have a country with a bunch of states, you can pick a resident from each state and form them into a lawmaking body. Suppose in the future they prove the universe is infinite and there is some country with infinitely many states. Why can't they form a legislature to organize their affairs? The axiom of choice is more natural than its negation. Now the next time you hear about the US Senate, just think of two applications of the axiom of choice. Two representatives from each state. That's what the axiom of choice says. If you have a collection of nonempty sets, then Mitch McConnell is involved.
  7. The usual generalization of the reals, complex numbers, quaternions, octonions, sedenions, etc., is the Cayley-Dickson construction. It's also related to the subject of Clifford algebras. https://en.wikipedia.org/wiki/Cayley–Dickson_construction https://en.wikipedia.org/wiki/Clifford_algebra I haven't perused the code but if the OP would express his/her idea in a couple of sentences perhaps we'd be interested to know more. Source code is not a good medium for communicating ideas to people. Its sole purpose is to allow trained programmers to communicate ideas to computers. I thought everyone knew that. Hey let's see the machine code! There's a nice visualization that [math]\sqrt{i} = \pm (\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} i)[/math]. In the plane, [math]i[/math] makes an angle of [math]\frac{\pi}{2}[/math] with the positive [math]x[/math]-axis. Now the geometric insight is that when you multiply complex numbers, you add the angles. To find a (not "the") square root of [math]i[/math], we ask ourselves: What angle, when added to itself, gives [math]\frac{\pi}{2}[/math]? Then it's obvious, it's just [math]\frac{\pi}{4}[/math]. A line making an angle of [math]\frac{\pi}{4}[/math] with the positive [math]x[/math]-axis crosses the unit circle at the point [math](\cos \frac{\pi}{4}, \ \sin \frac{\pi}{4})[/math] in trig form, giving one square root. The other one is its negative, which is the point directly opposite it on the unit circle. ps -- I just noticed the question was for [math]- i[/math]. Same visualization applies with suitable modifications, which I didn't feel like making. You take half of [math]\frac{3 \pi}{2}[/math] which is [math]\frac{3 \pi}{4}[/math], etc. Doing it that way gives the negative of the answer you got. My value's in the second quadrant and yours is in the fourth. I don't think there's a convention as to which is preferred. This was in response to @Passenger writing, "Real numbers and imaginary numbers, in turn, are real subsets of a complex number." If you substitute the word actual, or legitimate, or proper, for the second use of "real," then this makes sense. Perhaps that's what was meant. On the other hand the phrasing was the real subsets of '"a complex number" and NOT "the complex numbers." Making me think that perhaps they mean the real and imaginary parts of a complex number, which are always real numbers by definition. That would be another interpretation. FWIW zero is the only number that is both real and imaginary.
  8. > But this just can't be because for all x in S\{1/2}, 1/2 <* x so 1/2 is the greatest lower bound for S\{1/2}. If 1/4 is the min for S\{1/2}.than its also the greatest lower bound for S\{1/2}.. 1/2 <* 1/4 because 1/4 is in S\{1/2} and for all x in S\{1/2}, 1/2 <* x. So, this is a contradiction. You know I read through that and as far as I can see IT'S EXACTLY RIGHT. 1/2 is the min, then when you delete that 1/4 is the next min, and when you delete that 1/8 is the new min. But it's NOT a contradiction. It's exactly the way it might be ... which refutes your argument. Also you're confusing your intuitions of greatest lower bounds in the real numbers in the STANDARD order, with what's going on here. Again, think of the naturals 0, 1, 2, 3, 4, 5. What's the greatest lower bound of {x >= 4}? It's 4. What's the GLB of {x >= 5}? It's 5. It's a discrete order. Thinking about limit points in the real numbers is what's throwing you off.
  9. Suppose I give you the following linear order: 1/2 <* 1/4 <* 1/8 <* 1/16 <* ... <* everything else. What happens to your argument?
  10. That's like me saying I'm interested in a scoop of tuna salad between two slices of bread, but that I am NOT interested in a tuna salad sandwich! The two things are synonymous. If you have a well-ordered set, its order type is an ordinal. If you have an ordinal, it's a well-ordered set. It's not possible to be interested in well-orders and not be interested in ordinals. I can't fathom where you're coming from with a statement like this, after I've been explaining it to you for a year. The rest of your post is wildly off the mark. You keep posting an erroneous proof and I keep refuting it with an example. You say that if you delete the first element from the tail of a linear order (of the reals), the rest of the set has no smallest element. But that's false. "0, 1, everything" refutes it. You say ok take the next deleted tail. "0, 1, 2, everything" refutes it. You say ok do it countably many times. Then "1, 2, 3, 4, ... 0, everything" refutes it. That's the ordinal [math]\omega + 1[/math]. You can keep going to "odds, evens, everything." That's [math]\omega + \omega[/math]. I can just keep walking through the countable ordinals. You claim a deleted tail has no first element but clearly these examples refute your claim. That's all that's going on here. You have a faulty proof and I keep showing it's faulty.
  11. You're addressing me but you've indicated a disinclination to hear any more from me. I prefer not to play that game. For the record I'm done here unless something new and/or interesting gets said. I'll leave it with this. If -- I'm not saying there is, but if -- there were a well-order on the reals, it would look just like the well-order on the natural numbers: one element after another. The only difference is that you'd have to periodically take limits. A limit ordinal is an ordinal that doesn't have an immediate predecessor. [math]\omega[/math] is a limit ordinal. It's the upward limit of 0, 1, 2, 3, ... Technically it's implemented as the set-theoretic union of all the preceding ordinals. Alternatively, it's the set of all the preceding ordinals. Those two are the same since ordinals are transitive sets. So if you take all the countable ordinals, and take their upward limit (as their union, or by taking the set of all the countable ordinals) you have an ordinal that can't be one of the countable ordinals, so it's an uncountable ordinal. I completely agree with you that the idea of an uncountable well-ordered set is mind boggling. But that's not a proof against it. Rather, it's another one of those counterintuitive things in math that we have to just "get used to," as John von Neumann said. You have an intuition that there's no uncountable well-ordered set, and you think your intuition is a proof. But you haven't got a proof. Only an intuition which turns out to be false.
  12. LOLOL All the best. But since you asked I'll explain. The 1, 2, 3, ... example is ... an example. An illustration of how to conceptualize well-ordered sets. You remove the first element and there's another next element. You remove that one and there's another. It's perfectly true that we can't visualize an uncountable well-ordered set. But its existence can be proved even without the axiom of choice. You have an intuition that there are "too many" reals to well-order, but you haven't got a proof. https://en.wikipedia.org/wiki/First_uncountable_ordinal
  13. I agree with that. You keep making the same logic error, I keep correcting it. But what if we run the argument on the natural numbers in their usual order? Doesn't every deleted tail still have a smallest element? And where's your proof that this can't happen with an arbitrary linear order on the reals?
  14. How does ignoring the counterexamples to your argument help you convince anyone? The examples I'm giving you falsify your argument.
  15. I already explained this. It would go better if you'd read my posts. Run your argument on the natural numbers in their usual order 0, 1, 2, 3, ... Take the tail 14, 15, 16, 17, ... Remove 14 to get 15, 16, 17, ... What's the smallest element? 15. Remove 15 to get 16, 17, 18, 19, ... What's the smallest element? 16 This is exactly what would happen if you have a well-order for the reals. You haven't shown this can't happen. You're still confusing the usual order on the reals with an arbitrary linear order. It doesn't even have to be a well-order. Say your order is 14, 15, everything else. Then you pick the tail S = 14, 15, everything else. You delete 14. Now you have 15, everything else. The smallest element is 15. But what if it was 14, 15, 16, everything else. Same thing. No matter how many times you remove an element, there's a linear order that breaks your proof.
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