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Thread: Proof involving inequalities

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

    Let $\displaystyle a,b \in \mathbb{R}$. If $\displaystyle a \le b_{1}$ for every $\displaystyle b_{1} > b$, then $\displaystyle a \le b$. 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 $\displaystyle a>b$; let $\displaystyle a-b=\epsilon$. Let $\displaystyle b_1=b+\epsilon/2$. Then $\displaystyle b_1 >b$ so by assumption $\displaystyle a \leq b_1$. So $\displaystyle a \leq b+\epsilon/2 =b+\frac{a-b}{2} = \frac{a+b}{2} < a$, 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 $\displaystyle a > a$, 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 $\displaystyle a,b \in \mathbb{R}$. If $\displaystyle a \le b_{1}$ for every $\displaystyle b_{1} > b$, then $\displaystyle a \le b$. 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 $\displaystyle b_1>b$ then it is true for the above post.
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