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Math Help - Proof involving the Archimedean property.

  1. #1
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    Proof involving the Archimedean property.

    Given any x \in R prove that there exists a unique n \in Z such that n - 1 \leq x < n.

    Proof: Suppose x \in R. By the Archimedean property, there is an n \in Z such that n > x. Since N is well-defined, we have that n - 1 \in N and n - 1 < n. Then it follows that n - 1 \leq x < n for x \in R.

    I feel like I've jumped to conclusions, but to my non-math-genius brain, I feel like this is correct. Can someone prod me along?

    Last edited by mjlaz; September 5th 2010 at 05:50 PM.
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  2. #2
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    Quote Originally Posted by mjlaz View Post
    Given any x \in R prove that there exists a unique n \in Z such that n - 1 \leq x < n.

    Proof: Suppose x \in R. By the Archimedean property, there is an n \in Z such that n > x. Since N is well-defined, we have that n - 1 \in N and n - 1 < n. Then it follows that n - 1 \leq x < n for x \in R.

    I feel like I've jumped to conclusions, but to my non-math-genius brain, I feel like this is correct. Can someone prod me along?

    Your error is in asserting that n-1< x. The Archimedean property only tells you that there is an integer larger than x. it does not tell you that that integer is anywhere near x. For example, if the x= 1.5, the Archimedean property might be telling you that the number 1,000,000 is larger than 1.5.

    You will also need to use the "well ordering property" of the positive integers: every non-empty set of positive integers contains a smallest member. Apply that to the set of all integers larger than x which the Archimedean property tells you is non-empty.
    Last edited by HallsofIvy; September 7th 2010 at 09:05 AM.
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    After visiting the professor today, I made a little ground. How about this variation?

    Proof: Let x \in R. By the Archimedean property, there exists some n > x, n \in Z. Define S = { y \in Z : y > x}. Since S is bounded below by x, then by the well-ordering principle of integers, there is a least element in S, namely n = min(S). Since n \in S, then n > x. However, n - 1 \notin S and so n - 1 \leq x < n.

    I haven't proved the uniqueness component.
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  4. #4
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    Quote Originally Posted by mjlaz View Post
    I haven't proved the uniqueness component.
    Is it not the case that a least element (from well ordering) is unique?
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    Allow me to give you a different proof.
    You need the completeness axiom.
    We need to have proved that the set of integers, \mathbb{Z}, is not bounded.

    If a\in \mathbb{R} the define \mathcal{S}=\{n\in \mathbb{Z}:n\le a\}.
    Clearly we can say that b=\sup(\mathcal{S}).
    Also \left( {\exists c \in \mathcal{S}} \right)\left[ {b - 1 < c \leqslant b \leqslant a} \right].
    From which we get b < c + 1 \leqslant b + 1 which means that  c + 1 \notin \mathcal{S}\, \Rightarrow \,a < c + 1\, \Rightarrow \,c \leqslant a < c + 1.
    That completes the proof.
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