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Archimedian Property and Epsilon-Delta Type Problems

  1. Aug 26, 2015 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    1. Let ε > 0. Determine how large n ∈ ℕ must be to ensure that the given inequality is satisfied, and use the Archimedean Property to establish that such n exist.
    a.) [itex] \frac{1}{n} < \epsilon [/itex]
    b.) [itex] \frac{1}{n^{2}} < \epsilon [/itex]
    c.) [itex] \frac{1}{\sqrt{n}} < \epsilon [/itex]​
    2. Let ε > 0. Find a number δ > 0 small enough so that | a - b | < δ and | c - b | < δ implies | a - c | < ε.
    2. Relevant equations
    The archmidean property says that ∀ε > 0, ∀M > 0, ∃n ∈ ℕ such that n*ε > M

    3. The attempt at a solution
    For part a I multiplied both sides by n, which made the inequality 1 < n*ε, which is a statement of the archimedian property with M = 1, so in order for this to always be true, n = 2, but I run into problems with the reasoning of that, because ε > 0, it says nothing about it being an element of ℕ, so I'm not sure if that's right.
    For C I noticed that squaring both sides, becomes [itex] \frac{1}{n} < \epsilon^{2} [/itex], but redefining ε2 as δ, we have the same statement as part a. But I'm still stuck because I'm not sure if the question wants a numerical answer or what.

    For 2 I know the answer is "We can use any value of δ satisfying the double inequality 0 < δ < [itex] \frac{\epsilon}{2} [/itex]", but I'm not even sure where to start getting this.
     
  2. jcsd
  3. Aug 27, 2015 #2

    andrewkirk

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    For 2 you need to use the triangle inequality ##|a+b|\leq |a|+|b|##.

    For 1 it will help if you re-express your Archimedean property without using any symbols used in your problem, ie
    ∀θ > 0, ∀M > 0, ∃n ∈ ℕ such that nθ > M.
    Then you can do each of a-c by just saying what you put for each of M and θ in this.
    For (a) it's M=1, ##\theta=\epsilon##
    What is it for (b) and (c)?

    I don't understand the worry you express at the end of your first para in your '3. Attempt at a solution'. Can you explain more clearly what it is that concerns you?
     
  4. Aug 27, 2015 #3

    HallsofIvy

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    It would be better to also divide both sides by the positive number [itex]\epsilon[/itex] to get [itex]n> \frac{1}{\epsilon}[/itex]. Now apply the Archimedian property.

    I think it is clear that the answer should be the same type as part A. You can't give a numerical answer because you are not given any numerical data.

    [itex]|a- b|< \delta[/itex] is the same as [itex]-\delta< a- b< \delta[/itex] and [itex||c- b|< \delta[/itex] is the same as [itex]-\delta< c- b< \delta. What is (a- b)- (c- b)?
     
  5. Aug 27, 2015 #4

    B3NR4Y

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    So for (b) M = 1, again, θ = ε, or is it θ = √(ε) ? With the inequality being 1 < n2 ε
    (c) has M =1, again, and θ = ε2 = δ

    The trouble I am having is finding the n, because 2 I think is valid for all values of ε if you assume ε ∈ ℕ, and therefore a whole number, but the problem just says it's greater than 0, so I thought it was any element of ℝ > 0, therefore if ε ≤ ½, n = 2 doesn't work and there is something more so I think the answer would have to be an inequality.

    And for problem two,
    | a - b + c -c | ≤ |a - c| + |b - c|, from the triangle inequality. coincidentally, |a - c| and |b - c| are both less than delta, but | a - c| + |b - c| is greater than or equal to | a - b | so I'm not sure where to put them in the inequality. |a - b| ≤ |a - c| + |b - c| < δ? Not quite sure this is true, or how to prove if it is.
     
  6. Aug 27, 2015 #5

    B3NR4Y

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    Okay, that makes sense. Especially for the second question, I think I've got it.

    The first question I'll still have to work it, it's of course obviously an inequality as a final answer to find n, 0 < 1/ε < n, I think. And such an n has to exist, because of the archimedian principle.
     
  7. Aug 27, 2015 #6

    Zondrina

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    For part 1 a) we can say ##n \varepsilon > 1##, which like you said is the Archimedian property with ##M = 1##. So there exists an ##n## such that ##n \varepsilon > 1##. So choose ##n > \frac{1}{\epsilon}## to guarantee ##n \varepsilon > M## (you stated this prior).

    For b) take the root to obtain ##|n| \sqrt{ \varepsilon } > 1##. Let ##\varepsilon' = \sqrt{ \varepsilon }##. Since ##n > 0##, we obtain ##n \varepsilon' > 1##. Now apply part a) again.

    For c), square both sides and apply b) again.

    2. For this, re-read HallsofIvy's hint.
     
  8. Aug 27, 2015 #7

    B3NR4Y

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    I'm sorry if this is silly, but I really want to understand this.
    For 1, the answers I put are
    a.) The archimedian property says that ∀ε > 0 and ∀M >0 then ∃n∈ℕ : M < nε therefore such an n exists if you let M = 1, and ε=ε
    Specifically 0 < [itex]\frac{1}{\epsilon} [/itex] < n
    b.) Manipulating, we get 1 < [itex] | n | \sqrt{\epsilon} [/itex], the archimedian property says that an n satisfying exists if M = 1, and [itex] \epsilon' = \sqrt{\epsilon} [/itex] therefore 0 < [itex] \frac{1}{\sqrt{\epsilon}} [/itex] < n
    c.) Manipulating, again, we get 1 < ε2n let ε'=ε2, therefore 1 < ε'n. n exists due to the archimedian property, specifically 0 < [itex]\frac{1}{\epsilon^{2}} [/itex] < n
    For 2, -δ < a - b < δ and -δ < c - b < δ, (a-b) - (c-b) is = a - c, which is good because that's what the problem tells us ε is greater than (absolute value wise), but I'm not sure how to get this in a useable form, from the axiom that says if a < b, a + c < b + c we can say that -δ - c + b < a - c < δ - c + b. Now I'm stuck, I don't see how to mess with this and get the absolute values back.
     
  9. Aug 27, 2015 #8

    andrewkirk

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    As I said in the first line of post 2, you need to use the triangle inequality. To avoid confusion between variable names, let's re-write it as ##|u+v|\leq|u|+|v|##.
    Now what do you get if you substitute a-b for u and b-c for v?

    By the way, as a general principle, if you are using a theorem or law in a problem, it's good practice to ensure none of the variable names used in your statement of the law are the same as any of those in your problem, otherwise you'll get confused as to what any given instance of a variable name refers to, and end up with confusing formulas like the 'ε=ε' in your last post. That's why I re-stated the Archimedean Property without using ##\epsilon## above, and why I wrote the triangle inequality without using any of ##a,b,c## in this post. You adopted this principle in post 4 but then relapsed into using ε to refer to two different things in post 7.
     
    Last edited: Aug 27, 2015
  10. Aug 27, 2015 #9

    B3NR4Y

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    Substituting a-b for u and b-c for v, we get | a - c| ≤ | a - b | + | b - c |. Inequalities can add if they're going in the same direction, so since |a - b| < δ, and | b - c | < δ, adding | a - b | + | b - c | < 2δ, which is promising considering I know the answer is 0 < δ < ½ε, but I'm not sure how to get the triangle inequality to fit into this in order to get ε involved.
     
  11. Aug 27, 2015 #10

    andrewkirk

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    You've got | a - c| ≤ | a - b | + | b - c | by using the triangle inequality. You don't need to use it again.

    You also have | a - b | + | b - c | < 2δ

    Now what happens if you set ##\delta=\frac{\epsilon}{2}## and use transitivity of <?
     
  12. Aug 27, 2015 #11

    B3NR4Y

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    Setting [itex] \delta = \frac{\epsilon}{2} [/itex] we get the inequality | a - b | + | b - c | < ε, | a - c | ≤ | a - b | + | b - c | which implies that, | a - c | < ε, as the problem told us to ensure. I think I can see that this means 0 < δ < ½ ε. Thank you.
     
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