# Bounding a sum of real functions with rationals

• Oct 14th 2011, 03:46 AM
akolman
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?

• Oct 14th 2011, 04:40 AM
Georgii
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
• Oct 14th 2011, 06:57 AM
Hartlw
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