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Math Help - Another proof

  1. #1
    Senior Member Pinkk's Avatar
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    Another proof

    Let a,b \in \mathbb{R} Show that if a \le b + \frac{1}{n} for all n\in \mathbb{N}, then a \le b.

    I have no idea how to go about this. Any help would be appreciated.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    Let a,b \in \mathbb{R} Show that if a \le b + \frac{1}{n} for all n\in \mathbb{N}, then a \le b.

    I have no idea how to go about this. Any help would be appreciated.
    Suppose that a>b\implies a-b>0. By the Archimedean principle there exists some n'\in\mathbb{N} such that \frac{1}{n'}<a-b. Therefore a\le b+\frac{1}{n'}\implies a-b\le\frac{1}{n'}<a-b\implies a-b<a-b contradiction.
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  3. #3
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    Quote Originally Posted by Pinkk View Post
    Let a,b \in \mathbb{R} Show that if a \le b + \frac{1}{n} for all n\in \mathbb{N}, then a \le b.

    I have no idea how to go about this. Any help would be appreciated.

    Suppose a>b and put \epsilon=a-b>0. As \frac{1}{n}\xrightarrow [n\to\infty] {}0, there exists n\in \mathbb{N} s.t. \frac{1}{n}<\epsilon.
    Get now your contradiction...

    Tonio
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