Can "Infinity" ever be more than a mathematical abstraction?

Discussion in 'Physics & Math' started by Seattle, Jun 24, 2018.

  1. arfa brane call me arf Valued Senior Member

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    5,711
    Thanks. I'm trying really hard not to.
    And you communicate like a disparaging, condescending twat. Mostly.

    Y0u come across like someone who thinks they have all the right ideas. That's probably just youthful arrogance. Please don't get the idea I'm trying to give myself some comfort here, I don't really care
     
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  3. someguy1 Registered Senior Member

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    Then why do you keep replying to my posts? I'd be perfectly happy if you stopped.

    You know for several pages you've been throwing out random math phrases you've read on Wikipedia and asking me to put them in context for you. Why are you doing this? To troll me? Guess you got me good.
     
    Last edited: Aug 15, 2018
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  5. arfa brane call me arf Valued Senior Member

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    5,711
    So would I. But this thread isn't about you or me, is it?
     
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  7. someguy1 Registered Senior Member

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    Hey man I'm sorry if I upset you. Or anyone. You keep asking these questions and I'm answering the best I can but I can't tell where they're coming from. At some point I lost patience. My apologies. Time for a forum break. Peace.
     
  8. arfa brane call me arf Valued Senior Member

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    5,711
    The quotes aren't from Wikipedia and I haven't had the least expectation that you will put them in context for me. Nor have I asked you or anyone else to do that. Your arrogance is a given, it seems. You can't help yourself.

    To the rest:

    The ideas I've been trying to explore go like this: the real line is continuous, so is physical motion. With calculus you can define a set of instantaneous positions and velocities for a classical object with motion. Calculus usually defines the derivative in terms of limits, the epsilon-delta method uses real numbers, not infinitesimals (mainly because there's only one infinitesimal real). But the derivative can be defined in other ways; if you want to invoke infinitesimals you "need" a map from them to the standard reals.

    Also, when you see a textbook saying dy/dx means an infinitesimal change in y and in x in the reals, they don't mean actual infinitesimals but very small real numbers (one supposes).


    What calculus seems to say is, it doesn't matter if the real line is after all a subset of the hyperreal line, or if we call one or the other an abstraction of reality. Zeno's problems with continuity of motion seem to have more than one solution. And Zeno was just saying there's a problem if we assume real (actual) distances are infinitely divisible. (that is, into 1/2 the distance, then 1/2 the remaining distance, ad infinitum), so he thought, an object in motion has to pass through an infinite number of "checkpoints" in a finite time. He got that wrong, but why?

    Is it because the real line isn't, after all, infinitely divisible? Or is it because an object's motion really isn't "from point to point"?
     
    Last edited: Aug 16, 2018
  9. iceaura Valued Senior Member

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    I did not. I singled out and removed from the discussion a difficulty, or confusion, that results from using inappropriate physical metaphors - such as "holes" - to refer to mathematical objects and features.
    There are no physical holes, or anything corresponding to physical holes, in the computable reals. There are infinities - limits, such as the limit of the series 1+1/2+1/4 - - -. It is equal to "2".
    In particular, it does not claim that the misleadingly labeled noncomputable "holes" in the computable real number line correspond to, abstract, any physical reality whatsoever.
    And as far as we have any evidence, they don't.
    If you don't have the number, you don't have what you are approximating. Your model is missing - say - the diameter of a circle. If you don't need the diameter of a circle to model your physical reality, then ok - you don't need pi. If you do need the diameter of a circle, you do need pi.
     
    Last edited: Aug 16, 2018
  10. Throbber Registered Senior Member

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    Not neccessarily. The diameter is one centremetre. Pi is not needed.
     
  11. Throbber Registered Senior Member

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    172
    Incidentally if every number could equal the same number, does it continue for infinity?

    n-n=0

    In this way ANY number can equal zero.
     
  12. Write4U Valued Senior Member

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    hehe...I see it that (n1) can be any number and be approximate to (n2) with a difference of zero, between the measurement of each.
     
  13. Throbber Registered Senior Member

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    172
    So mathematics should be a mess! The Laws should not exist in this case!
     
  14. Michael 345 Looking for Bali in Nov Valued Senior Member

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    Is it because the real line isn't, after all, infinitely divisible?

    In THEORY yes
    Practice NO

    I would contend Planck length would be a answer your looking for

    http://www.physlink.com/education/askexperts/ae281.cfm

    Or is it because an object's motion really isn't "from point to point"?

    If you are thinking about the runner not passing the tortoise

    In THEORY yes but only if you restrict the distance to a finite length

    If you restrict the distance, as a example, to 100 metres, again in THEORY, runner does not pass tortoise (see also Planck length)

    Allow unlimited distance tortoise looses in a very short period or distance

    Please Register or Log in to view the hidden image!

     
  15. QuarkHead Remedial Math Student Valued Senior Member

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    I don't think this quite right. Of course, I may have misunderstood this exchange, but try this.

    Take the Real Line \(R^1\) (this btw is a topological construct). Assume for a contradiction that it can be arbitrarily partitioned into 2 disjoint sets, \(L\) and \(U\) such that \(L\cap U = \emptyset\).

    Now every element in \(L\) is a lower bound for every element in \(U\), and every element in \(U\) is an upper bound for every element in \(L\). But since \(R^1\) is equipped with a a total order, there is a greatest element in \(L\) which is the greatest upper bound for \(L\) and therefore in \(U\).

    Likewise there is a least element in \(U\) is the least lower bound for \(U\), and therefore is in \(L\).

    Therefore my assumption fails, this partition into disjoint sets cannot exist, which implies that the Real line is "continuous" in the colloquial sense of the word

    t
     
  16. iceaura Valued Senior Member

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    26,906
    Without pi, you have no circle to go with your linear distance - your one centimeter is not a diameter.
    The condition that every element of one is a bound for the other does not imply that every bound for the one is an element of the other.
     
  17. QuarkHead Remedial Math Student Valued Senior Member

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    1,555
    Do you not understand the process of "proof by contradiction"?
     
  18. iceaura Valued Senior Member

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    Each appearance of the word "therefore", in your proof, is an error.
     
  19. QuarkHead Remedial Math Student Valued Senior Member

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    1,555
    As I said earlier, you do not understand proof by contradiction. My "therefores" where part of the assumption I proved wrong. That is how this sort of proof works.
     
  20. someguy1 Registered Senior Member

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    Like \(L = \mathbb Q\) and \(U = \mathbb R \setminus \mathbb Q\)? Perhaps you left out a vital assumption.


    Surely false with the example I just gave. Methinks you forgot to say:

    ... And every element of \(L\) is strictly less than every element of \(R\). Right? That's the part you left out.

    So, let \(L = (-\infty, 0]\) and \(R = (0, +\infty)\). That satisfies your hypotheses, including the one you forgot to mention.

    True.


    \(0\) is indeed the greatest element in \(L\), but it's not in \(U\).

    No, there is no least element in \(L\). You're wrong here.

    It's your "proof" that fails, since even AFTER stating the assumption you failed to mention, the most obvious example falsifies it.

    Perhaps you forgot to mention that the sets are assumed to be open, in which case you could work out a proof (but definitely not this one) that the reals are connected. Not "continuous," which has no meaning in this context.

    @QH, How many unstated assumptions are the readers supposed to supply? Even after assuming all your missing assumptions, the proof's still wrong. That is, if the readers figure out that (1) everything in L is less than everything in R; and (2) L and R are open sets; then you STILL haven't given a proof that the reals are connected.
     
  21. iceaura Valued Senior Member

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    They are mistakes in deduction.
     
    Last edited: Aug 23, 2018
  22. QuarkHead Remedial Math Student Valued Senior Member

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    1,555
    I challenge you to find where I ever claimed that


    Then you are being disingenuous. I made 2 assumptions - that not many here have a grounding in topology, and that nonetheless they would be happy with the intuitive notion that the Real Line is simply all the Real numbers arranged in the usual order as some sort of "line".

    The "partition" I described, as you will know very well, is called a Dedekind Cut
     
  23. iceaura Valued Senior Member

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    26,906
    Someguy mentioned a problem with your description, I mentioned a problem with your reasoning, here's a closer look:
    Let's accept that under advisement, despite someguy's correct objection (it's intuitively ok, once past the linguistic hitch of bounded elements)
    1} is false as it stands - there are many such sets LL that have no greatest element.
    2} is false - every element of UU is an upper bound of L, and UU has no greatest element - and does not make sense in your argument, where you seem to need a least upper bound for LL.
    3} is false as it stands, since many sets LL as described (the ones that have a greatest element) do contain their {least} upper bound - it isn't in UU.
     

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