Monday, December 21, 2015

Prove that if $x$ and $y$ are irrational numbers, there exists an irrational number $z$ such that $y < z < x$



My teacher proposed this question a few days ago along with the similar case for rational numbers. I've already figured out the proof for rational numbers (just prove that their arithmetic mean is rational), but I'm not really sure where to start with this proof. I guess I'm stuck because it isn't as easy to represent irrational numbers as it is for rational numbers. Also, the arithmetic mean of two irrational numbers isn't necessarily irrational, so that approach doesn't work either. Could someone help me get started?


Answer



If $\frac{x+y}{2}$ is irrational, there you go. If not, then $q=\frac{x+y}{2}$ is rational and what can we say about $\frac{q+x}{2}$?


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