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Math Help - short proof

  1. #1
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    short proof

    how would i show that the set S = {cubed root(n+1) - cubed root(m) such that n,m are in N the natural numbers} is dense in the real numbers? the book says to use the fact that (cubed root(n+1) - cubed root(n)) is less than epsilon if n is greater than n0. i'm just in an elementary proof course. could anyone show me how to prove this? i would love some help.
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  2. #2
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    The idea is that if you let m=n then \sqrt[3]{n+1} - \sqrt[3]{n} will be so small if n is huge.

    To show this use the following hint: a^3 - b^3 = (a-b)(a^2+ab+b^2) and rationalize the numerator using the difference of two cubes formula.
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  3. #3
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    how do you get it in the form to use the cubed root formula and why do you neutralize it?

    yeah so basically the limit as it approaches infinity is zero, but how does that help me?

    i don't know, all we've done in class is prove that the rational numbers are dense in the real numbers and the irrational numbers are dense in the real numbers. i don't see how to prove the density of this function based on what we did.
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  4. #4
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    can someone just help me a little bit? i have a whole whack of these type of problems to do and i just have no idea what i'm doing. any further help would be appreciated.

    thanks, cindy
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  5. #5
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    \frac{\sqrt[3]{n+1} - \sqrt[3]{n}}{1}\cdot \frac{\sqrt[3]{n+1}^2 + \sqrt[3]{n+1}\sqrt[3]{n}+\sqrt[3]{n}^2}{\sqrt[3]{n+1}^2 + \sqrt[3]{n+1}\sqrt[3]{n}+\sqrt[3]{n}^2} =  \frac{n+1-1}{\sqrt[3]{n+1}^2 + \sqrt[3]{n+1}\sqrt[3]{n}+\sqrt[3]{n}^2} \leq \frac{1}{3\sqrt[3]{n^2}}

    So choose n > \frac{1}{\sqrt{27\epsilon^3}}.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    So choose n >\frac{1}{\sqrt{27\epsilon^3}}.
    Can you tell me what that has to do with the set being dense is the reals?
    Frankly, I doubt that the stated problem is true.
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  7. #7
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    how would you do it plato?

    thanks for the help theperfecthacker.
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  8. #8
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    Quote Originally Posted by Plato View Post
    Can you tell me what that has to do with the set being dense is the reals?
    Frankly, I doubt that the stated problem is true.
    All I am trying to show is that this difference can be made as little as one wants. That is it. I let the poster figure out what to do with that.
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    All I am trying to show is that this difference can be made as little as one wants. That is it. I let the poster figure out what to do with that.
    But of course that is clearly true. But does that have anything to do with the problem? Does it?
    The question is to show that the given set is dense is the reals.
    I have seen many such problems. BUT not this one.
    Thinking about it for sometime, I now doubt that it is true.
    To be dense, given a real number some element of the set must be “near” to that number.
    Consider –7?
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