If u > 0 is any real number and x < y, show that there exists a rational number r such that x < ru < y

by justMath   Last Updated February 13, 2018 13:20 PM

If $u > 0$ is any real number and $x < y$, show that there exists a rational number $r$ such that $x < ru < y$.

Since $u > 0$ and $x < y$, then we know that $x/u<y/u$.

By the Density Theorem, we know there exists $r \in\mathbb Q$ such that $x/u<r<y/u$.

We finish the proof by multiplying the inequalities by $u$.

Is this enough for the proof?

Tags : real-analysis


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