i need help to test the following:

“prove that every open interval in real numbers contains rational, irrational and dyadic numbers.”

I had tried defining an isometry on a set where the property was fulfilled but my teacher told me that it must be through the Archimedean property, I would be very grateful if you could help me.


Let $a<b$ be any real numbers. By Archimedean property there exists a natural number $n$ such that $n(b-a)>1$. It follows that $2^n(b-a)>1$, so the interval $(a,b)$ has length bigger that $\tfrac {1}{2^n}$ and so it contains a dyadic rational points of the form $\tfrac {k}{2^n}$ for some integer $k$, because such points are placed at the real line with a distance $\tfrac {1}{2^n}<|b-a|$ between consecutive points. Similarly we can show that the interval $(a,b)$ contains an rational point of the form $\tfrac {k}{2^n}\cdot \tfrac{1}{\sqrt 2}$ (with a little care assuring that $k\ne 0$).

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