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Thread: Bounding a sum of real functions with rationals

  1. #1
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    Bounding a sum of real functions with rationals

    I am stuck with this problem

    Show that for real functions $\displaystyle f$ and $\displaystyle g$ such that $\displaystyle f+g<x$, there exists rationals $\displaystyle r$ and $\displaystyle s$ such that $\displaystyle r+s<x, f<r$ and $\displaystyle g<s$

    Would using the Archimedean principle help?

    Let $\displaystyle \alpha \in \mathbb{Q}$ and $\displaystyle \alpha>0$.
    By the Archimedean principle, there exists $\displaystyle k \in \mathbb{Z}$ such that
    $\displaystyle k\alpha \leq x < (k+1)\alpha$

    Is this in the right direction?


    Thanks in advance
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  2. #2
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    Re: Bounding a sum of real functions with rationals

    Note that $\displaystyle f<x-g$. So, there exists rational $\displaystyle r$, such that $\displaystyle f<r<x-g$... And now think how to obtain second rational
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  3. #3
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    Re: Bounding a sum of real functions with rationals

    "Show that for real functions $\displaystyle f$ and $\displaystyle g$ such that $\displaystyle f+g<x$, there exists rationals $\displaystyle r$ and $\displaystyle s$ such that $\displaystyle r+s<x, f<r$ and $\displaystyle g<s$"

    let A =lubf, B=lubg

    If A and B are rational and belong to f and g respectiveley, you are finished.

    Otherwise you can pick rational numbers r and s greater than A and B respectiveley and arbitrarily close so that r+s < x
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