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Is this statement valid?

  1. Jul 13, 2013 #1
    |1-0.[itex]\bar{9}[/itex]|=dx?
     
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  3. Jul 13, 2013 #2

    WannabeNewton

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    No. First of all ##0.\bar{9} = 1##. Secondly, as you were told in the other thread, ##dx## is not a real number.
     
  4. Jul 13, 2013 #3

    micromass

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    No. We have that ##1=0.999...##. So ##|1-0.999...|=0##. On the other hand, ##dx## is not a real number.
     
  5. Jul 13, 2013 #4
    See https://www.physicsforums.com/showthread.php?t=507002 [Broken] for a good explanation of why ##0.999...=1##
     
    Last edited by a moderator: May 6, 2017
  6. Jul 14, 2013 #5
    how does 0.999...=1?... .....how?.....

    just to be sure, im not rounding..

    that's like saying i have a piece wise function.. f(x)={2x , 0<x<5}
    {3 , x=5 }
    {2x , 5<x<∞}

    the statement 4.999999....=5 is NOT even saying as x approaches 5... no. This is like saying at 4.999999...... wer're already AT 5.
    how does this make sense??

    and if 4.99999.... = 5 , then this must also be true 5.00000000.......00001 = 5.
    okay so.. now we have three "fives" on the number line... let's work our way outward shall we

    4.999...[itex]\equiv[/itex]5 so now.. 4.999......9998 is also 5.. and 4.999.....9997 is also 5...
    5-[itex]\Sigma[/itex]dxi=5 no matter what the value of i is.. even as it approaches ∞ this is still five.
    and now we have an infinite number of fives on the number line. the implications do not make sense to me.
    if i am wrong with my analysis, i assume i am wrong from the point when i start to work my way outward after the point where 4.999.... is said to be equal to five..
     
    Last edited: Jul 14, 2013
  7. Jul 14, 2013 #6

    pwsnafu

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    See the FAQ HS-Scientist linked to in post #4.

    Also Wikipedia article.
     
  8. Jul 14, 2013 #7

    micromass

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    That is actually the second part of the FAQ, so be sure to read the first part as well: https://www.physicsforums.com/showthread.php?t=507001 [Broken]
     
    Last edited by a moderator: May 6, 2017
  9. Jul 14, 2013 #8

    WannabeNewton

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    You really have to stop treating ##dx## like it's some kind of number. It's fine for hand wavy arguments in subpar physics books but not for what you're interested in here.
     
  10. Jul 14, 2013 #9

    micromass

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    If you want to be rigorous about it, then for any function ##f:\mathbb{R}\rightarrow \mathbb{R}##, you can define

    [tex]df:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}: (p,h)\rightarrow f^\prime(p) h[/tex]

    In particular, we have that

    [tex]dx:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}: (p,h)\rightarrow h[/tex]

    The main reason for this definition is to make the following true:

    [tex]df = \frac{df}{dx} dx[/tex]

    If you want to do physics, then infinitesimals come in very handy. Those infinitesimals are made rigorous by the hyperreal and surreal number system. But they are not denoted as ##dx##.

    In short: you need to stop treating ##dx## as a number, or even worse: as a real number. It isn't, and it's very confusing.
     
  11. Jul 15, 2013 #10
    About 0.999...=1, is representing it as a geometric infinite series a valid proof?
     
  12. Jul 15, 2013 #11

    HallsofIvy

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    Yes, it is probably the simplest strictly valid proof.

    0.9999...= .9+ .09+ .009+ .0009+ ...= .9(1)+ .9(.1)+ .9(.01)+ .9(.001)+ ...= .9(1)+ .9(.1)+ .9(.1^2)+ .9(.1^3)+ ...

    That is a greometric series of the form [itex]\sum ar^n[/itex] with a= .9, r= .1. A geometric series with |r|< 1 (which is the case here: |r|= .1< 1) converges to a/(1- r).

    Here, a/(1- r)= .9/(1- .1)= .9/.9= 1.
     
  13. Jul 16, 2013 #12
    guys what level of math are you all in? ie what's the highest math courses you've taken? i just finished diff eq and can't wait to get to where you guys are at.. things are so confusing for me right now :'(
     
  14. Jul 16, 2013 #13

    WannabeNewton

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    Don't worry, it will all come in due time. Most of the people answering you are doing their PhDs/masters or already have their PhDs so don't fret, you'll learn it all in sooner or later.
     
  15. Jul 16, 2013 #14

    HallsofIvy

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    Yes, at PhD level, but the "geometric series" proof that 0.999...= 1 is one that I learned in PreCalculus.
     
  16. Jul 17, 2013 #15
    I think that ##|1-0.\bar{9}|=0##, which admittedly IS close to 0 if you want to think about it that way, seeing as the distance between the two is 0. :biggrin:

    For now, it's best to treat ##dx## and ##df## as variable holders for notation. Later, in differential geometry, you'll be introduced to the concept of the "exterior derivative," which extends the idea of the differential to a meaningful, mathematically rigorous context. For now, though, just treat it as notation.

    Whereas you and micro seem to do better than most of us without a degree. :tongue:

    It's not about your level of education in school. It's about how much you study by yourself. You can get a lot further by self study than you can by a formal college education, in my opinion.
     
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