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Are there any infinitesimals in R?

  1. Jul 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that infinitesimals are not a subset of R.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Well, I had two ideas about how to prove this but I'm really not sure about either. Proof 1 was the first idea I had but I think it's probably wrong since it has to do with the limit of a sequence. Proof 2 was my other idea but I think it seems fishy so it's probably wrong. Anyway, here goes nothing . . .

    Proof 1: Suppose that (∃ε)(εR) defined by 0 < ε < 1/nnN. Clearly ε satisfies the definition of an infinitesimal.

    Since ε < 1/n this defines a sequence {xn} = 1/n. We can prove that this sequence converges to zero as n becomes arbitrarily large by using the definition of convergence. Therefore,

    (∀ϵ)(ϵ > 0)(∃N)(NN)(∀n)(if n > N then 0 < |{xn} - 0| < ϵ).

    Clearly, if (ϵ > 0), then the must be some (NN) such that, 1/N < ϵ. The fact that n > N implies that 0 < {xn} = 1/n < 1/N < ϵ. Since this condition is true, we have that, if n > N then 0 < |{xn} - 0| < ϵ.

    To complete the proof, we need only note that since 0 < ε < {xn}, if n > N then 0 < |ε| <|{xn}| < ϵ. This implies that ε = 0 and consequently, infinitesimals are not a subset of R. Q.E.D.


    Proof 2: Suppose that (∃ε)(εR) defined by 0 < ε < 1/nnN. Clearly ε satisfies the definition of an infinitesimal. Since ε ≠ 0, ε-1R. Because 0 < ε < 1/nnN, this implies that ε-1 > nnN. However, this contradicts the Archimedean property of R and consequently infinitesimal numbers cannot be a subset of R. Q.E.D.


    I hope this is the right forum for this. Please pardon any poor wording, I'm not used to formatting things for the computer. Thanks!
     
  2. jcsd
  3. Jul 11, 2009 #2

    HallsofIvy

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    Re: Infinitesimals

    I think that can be done more simply as a "proof by contradiction". If an infinitesmal, [itex]\epsilon[/itex] were a member of R, it would have to be 0, positive or negative. It cannot be 0 by definition. If [itex]\epsilon> 0[/itex] then [itex]1/\epsilon[/itex] would be a positive real number. By the Archimedean property (which essentially is what says infinitesmals cannot be real numbers) there exist a positive integer N such that [itex]N> 1/\epsilon[/itex] from which [itex]1/N< \epsilon[/itex], a contradiction. I'll leave the last case, that [itex]\epsilon[/itex] is a negative real number, to you.
     
  4. Jul 11, 2009 #3

    Hurkyl

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    Re: Infinitesimals

    I thought proof 1 was fine (other than neglecting the possibility of negative infinitessimals). It's the same argument -- the Archmedean principle is essentially the statement that "1/n --> 0 as n --> infinity", and his invokation of the squeeze theorem works out to essentially the same thing as your argument by contradiction.

    Oh wait, nevermind -- I just noticed he didn't actually invoke the squeeze theorem, he just duplicated its proof in his argument.

    Isn't his proof 2 the same thing as your argument, HoI?
     
    Last edited: Jul 11, 2009
  5. Jul 11, 2009 #4

    HallsofIvy

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    Re: Infinitesimals

    Pretty much, yeah. But mine was prettier!

    (There were a number of special symbols my reader could not interpret so I was not clear what he was saying.)
     
  6. Jul 11, 2009 #5
    Re: Infinitesimals

    Thanks for the replies guys! I'm really glad that each approach I used will work with some modifications.

    Hurkyl: Thanks for pointing out that I needed to consider negative infinitesimals. I always seem to forget some important part of a proof.

    HallofIvy: Your proof by contradiction was certainly prettier than mine! Thanks for the input!
     
  7. Jul 11, 2009 #6

    Hurkyl

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    Re: Infinitesimals

    avec_holl: have you worked out how to redo proof #1 in terms of the squeeze theorem? I think it's a good idea -- you were clearly thinking along those lines, and it would be a good exercise to figure out how to work efficiently with those thoughts.
     
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