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  1. #1
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    real analysis

     Let \ x \in R \ with \ x>0 \  . show \ that \ there \ is \ n \in N  such \  that \  n-1 \leq x < n .
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    Do you know that the positive integers are well ordered?
    Let K=\left\{k\in \mathbb{Z}^+: x<k\right\}.
    Let n be the first term in K.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Question

    Quote Originally Posted by flower3 View Post
     Let \ x \in R \ with \ x>0 \  . show \ that \ there \ is \ n \in N  such \  that \  n-1 \leq x < n .
    Suppose the contrary that such number does not exist, then for any n\in\mathbb{N}, we have x<n-1 and x\geq n, which implies n-1>x\geq n or equivalently -1\geq0.
    A contradiction.
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  4. #4
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    Quote Originally Posted by bkarpuz View Post
    Suppose the contrary that such number does not exist, then for any n\in\mathbb{N}, we have x<n-1 and x\geq n, which implies n-1>x\geq n or equivalently -1\geq0.
    A contradiction.
    The nagation in the above is incorrect.
    n-1\le x < n means n-1\le x\text{ and }x < n.
    The negation of that is n-1> x\text{ or }x \ge n.
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  5. #5
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Plato View Post
    The nagation in the above is incorrect.
    n-1\le x < n means n-1\le x\text{ and }x < n.
    The negation of that is n-1> x\text{ or }x \ge n.
    Oops! How did I do that mistake
    Then, let me correct.
    Suppose that for every n\in\mathbb{N}, x\not\in[n-1,n). However, since x is fixed, we can find m\in\mathbb{N} such that x<m, i.e., x\in[0,m)
    Then, we know that x\not\in[m-1,m), similarly x\not\in[m-2,m-1), and by following similar arguments, we show that x\not\in[m-m,m-m+1)=[0,1).
    This shows that x must be negative, and thus is a contradiction.

    But I have to say that this is not a nice proof.

    Thanks to Plato for pointing out my mistake.
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