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Thread: real analysis

  1. #1
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    real analysis

    $\displaystyle 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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  2. #2
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    Do you know that the positive integers are well ordered?
    Let $\displaystyle K=\left\{k\in \mathbb{Z}^+: x<k\right\}$.
    Let $\displaystyle n$ be the first term in $\displaystyle K$.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Question

    Quote Originally Posted by flower3 View Post
    $\displaystyle 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 $\displaystyle n\in\mathbb{N}$, we have $\displaystyle x<n-1$ and $\displaystyle x\geq n$, which implies $\displaystyle n-1>x\geq n$ or equivalently $\displaystyle -1\geq0$.
    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 $\displaystyle n\in\mathbb{N}$, we have $\displaystyle x<n-1$ and $\displaystyle x\geq n$, which implies $\displaystyle n-1>x\geq n$ or equivalently $\displaystyle -1\geq0$.
    A contradiction.
    The nagation in the above is incorrect.
    $\displaystyle n-1\le x < n$ means $\displaystyle n-1\le x\text{ and }x < n$.
    The negation of that is $\displaystyle n-1> x\text{ or }x \ge n$.
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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.
    $\displaystyle n-1\le x < n$ means $\displaystyle n-1\le x\text{ and }x < n$.
    The negation of that is $\displaystyle n-1> x\text{ or }x \ge n$.
    Oops! How did I do that mistake
    Then, let me correct.
    Suppose that for every $\displaystyle n\in\mathbb{N}$, $\displaystyle x\not\in[n-1,n)$. However, since $\displaystyle x$ is fixed, we can find $\displaystyle m\in\mathbb{N}$ such that $\displaystyle x<m$, i.e., $\displaystyle x\in[0,m)$
    Then, we know that $\displaystyle x\not\in[m-1,m)$, similarly $\displaystyle x\not\in[m-2,m-1)$, and by following similar arguments, we show that $\displaystyle x\not\in[m-m,m-m+1)=[0,1)$.
    This shows that $\displaystyle x$ 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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