wtf

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  1. Studiot is British. "On the game" is quite the insult. I believe you called him a prostitute or a pimp. Studiot please confirm or clarify my understanding.
  2. I actually have your back on that last point. Einstein was told that hundreds of physicists disagreed with his theory. He said: "If my theory is wrong, it would only take one." I agree with you that it doesn't matter who agrees with whom. If the whole of established mathematics believes one thing, and you believe another, and you have a proof, then you are right and they are all wrong. So ... have you got a proof?
  3. > I quite well understand this. Of course they are mutually exclusive. Also, its hard to imagine how any set that is dense with respect to < or about any ordering could be ordered in such a way that it can be enumerated by 1,2,3,... There are just too many numbers that still pop up in between. Ah! But I just explained that. We know there's a bijection between the naturals and the rationals. So to put a well-order on the rationals, we just enumerate them q1, s2, q3, ... where qn is f(n) where f is the bijection. Going the other way, we may put a dense order on the natural numbers by simply using the inverse of that bijection, call ig g. So g(q) is some natural number. So order the naturals n, m according to whether g-inverse(n) and g-inverse(m) are < or > to each other in the rationals. That's why this example is so important. It shows how we may reorder a given set to have strikingly different properties. >Can one show that any set dense with respect to one ordering is therefore dense for all possible orderings? On the contrary we can prove that false. Any set may be well-ordered. That's the well-ordering theorem. Given any dense set you just well-order it. That's the point of the entire thread, right? And of course a MUCH SIMPLER case is that the rationals, which are densely ordered, may nevertheless be well-ordered using any bijection with the naturals. And we do not need to rely on the well-ordering theorem for this fact. That's important. The well-ordering theorem requires the axiom of choice. But establishing a bijection between the naturals and the rationals does not. So your claim is falsified by THE MOST OBVIOUS EXAMPLE. That's another reason this example is so helpful. It's like the well-ordering theorem but without having to invoke the awesome power of the well-ordering theorem! That's why it's of such great interest.
  4. In this discussion it's important to note that the rationals are dense in their usual order; and that the naturals are well-ordered in their usual order. Consider that we may use any bijection between the naturals and the rationals to induce the order of one onto the other. In that way we may impose a dense order on the naturals; or a well-order on the rationals. Another way to look at this is that the naturals and the rationals are actuall the same exact set. Their order properties depend only on the order we impose on them. If we impose the rational order, our set looks like the rationals. If we impose the naturals, our set looks like the naturals. But it's the same underlying set. I'm not claiming this as a "fact," as if there could even be any facts about abstract entities such as sets. Rather, I'm presenting this as a point of view that gives insight into the nature of ordered sets. It's also worth noting (again) that the properties of density and well-order are mutually exclusive. An ordered set may be one or the other, or neither. But it may not be both.
  5. Thread title is "a possible proof." I haven't followed the thread in any detail.
  6. I read the conclusion. You never even claim to have solved the problem.
  7. Cardinality and Bijection of finite sets

    I'm sure Numberphile explained that this is an INFORMAL way of looking at something much more subtle. You need to use the analytic continuation of the Riemann zeta fuction via a process called zeta function regularization, to show that zeta(-1) = -1/12. You can CASUALLY think of zeta(-1) as sort of looking like the formal expression 1 + 2 + 3 + ... but it actually isn't. That series diverges to infinity. It's sad that so many people have gotten confused by this video. Numberphile is usually reputable but they unleashed a lot of confusion online about this issue. https://blogs.scientificamerican.com/roots-of-unity/does-123-really-equal-112/ https://plus.maths.org/content/infinity-or-just-112
  8. Astonishing. * I told you at least three times that I invited you to look at the example of the naturals since they would help you gain insight into the problem you're attacking. That you continually fail to comprehend this point is mystifying to me. I say again: The way you attack a hard problem is to start by analyzing simpler instances of similar problems. If you don't want to get this, so be it. * You say I am "incorrect" that the real numbers can be well ordered; but that they can be bijected to the rationals, which are NOT well-ordered. You say that is incorrect. Please provide a proof or explanation. * Being well-ordered and being dense are mutually exclusive. I thought I pointed this out several times. I thought you UNDERSTOOD this several times. As an exercise -- again, to help you build intuition -- I suggested that you write down a formal proof. I reiterate that suggestion. Since you reactivated this thread you haven't provided a proof or even any additional argument.
  9. Cardinality and Bijection of finite sets

    I don't understand this point at all. Let's look at a simpler case. We have two sets A and B. A = {1, 2, 3, 4, 5} and B = (2, 4, 6, 8, 10}. It seems perfectly clear that we can biject the elements of A to those of B, despite the fact that the numbers in B are larger than some of the ones in A. And it's true that some of the elements of B are not in A. Why does this trouble you?
  10. Continuity and uncountability

    When you evaluate that string using standard positional notation, how do you do it?
  11. definition of derivative

    That'll keep him busy.
  12. Disproof of Riemann Hypothesis

    I'm only following sporadically but this is what I'm confused about from the beginning. * If [math]\widehat{0} = 0[/math] as has been recently acknowledged by OP; and * If correspondingly, [math]\widehat{\infty} = \infty[/math]; and * If the symbol [math]\infty[/math] is identical to [math]+ \infty[/math] in the extended real numbers; which I believe the OP acknowledged a while back; * Then the expression [math]\widehat{\infty} - b[/math] is undefined. It can't be sensibly defined. The exposition so far has not defined it. Has the OP addressed this issue yet? Also, I'm still confused about the neighborhood of infinity. In the real or extended real numbers, we can certainly call the set of reals larger than some fixed real, a neighborhood of infinity. And if we consider the set of reals whose absolute value is greater than some fixed real,then that would be a neighborhood of infinity in the space where you identify plus and minus infinity of the extended reals. In other words it's a circle, but with the point diametrically opposed 0 called infinity. It's a regular circle but with a funny metric. The two-dimensional analog of that construction shows how to regard the complex numbers as a sphere. If you take the complex plane you can add a single point called [math]\infty[/math], which can be visualized as the north pole of a sphere. This is called the Riemann sphere, https://en.wikipedia.org/wiki/Riemann_sphere It's like going "out to infinity" in every direction and sewing together the boundary at infinity to make a sphere. Since there's no order on the complex numbers, there's only one point at infinity. Another name for this construction is the one-point compactification of the plane. If you know what a compact set is, you know the plane's not compact. But using this construction you can add a single point and make it a compact set. If you draw a circle on the sphere centered at the north pole, you could legitimately call that a neighborhood of infinity of the complex numbers. It's the same as if you consider the set of all complex numbers whose modulus, or absolute value, is greater than some fixed nonnegative real number. Now as I understand it, the OP does not mean any of those things. I still don't understand what "the" neighborhood of infinity is; and I also don't understand why OP calls it "the" instead of "a", since in the contexts I mentioned, there are many neighborhoods of infinity.
  13. definition of derivative

    I do not believe this is historically accurate. Leibniz used infinitesimals directly. Newton referred to the "ultimate ratio" of delta-y over delta-x as both become closer and closer to zero. Newton's idea has more in common with the modern viewpoint than Leibniz's. That's my understanding. There is historical debate around this topic.
  14. Disproof of Riemann Hypothesis

    But this couldn't be. You are defining the real numbers as cuts in ... what? In the real number line. Where do those real numbers come form before you have defined them? Your definition is circular. Of course the answer is that the real numbers are cuts in the rationals. You should demand your money back from the university that sold you a course in real analysis, since you clearly didn't learn anything.
  15. Disproof of Riemann Hypothesis

    It's true. I'm not a member of the empty set. Why don't you just explain yourself?