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Math Help - Proof involving inequalities

  1. #1
    Senior Member Pinkk's Avatar
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    Proof involving inequalities

    Let a,b \in \mathbb{R}. If  a \le b_{1} for every b_{1} > b, then a \le b. I'm fairly certain I have to apply some of the ordered field properties but I am not sure how. Thanks!
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Suppose a>b; let a-b=\epsilon. Let b_1=b+\epsilon/2. Then b_1 >b so by assumption a \leq b_1. So a \leq b+\epsilon/2 =b+\frac{a-b}{2} = \frac{a+b}{2} < a, which is absurd.
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  3. #3
    Senior Member Pinkk's Avatar
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    So the contradiction you arrived at is a > a, if I'm not mistaken. Thanks!
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    That is correct!

    There may be a more abstract approach but this is pretty straightforward.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    Let a,b \in \mathbb{R}. If  a \le b_{1} for every b_{1} > b, then a \le b. I'm fairly certain I have to apply some of the ordered field properties but I am not sure how. Thanks!
    http://www.mathhelpforum.com/math-he...her-proof.html. Look familiar? Clearly, if it is true for every b_1>b then it is true for the above post.
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