Prove that every open interval in real numbers contains rational, irrational and dyadic numbers.

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zastenjkcy · Accepted Answer

Let   a  &amp;lt;  b be any real numbers. By Archimedean property there exists a natural number   n such that   n  (  b  −  a  )  &amp;gt;  1. It follows that       2    n    (  b  −  a  )  &amp;gt;  1, so the interval   (  a  ,  b  ) has length bigger that             1              2        n             and so it contains a dyadic rational points of the form             k              2        n             for some integer   k, because such points are placed at the real line with a distance             1              2        n              &amp;lt;      |    b  −  a      |   between consecutive points. Similarly we can show that the interval   (  a  ,  b  ) contains an rational point of the form             k              2        n              ⋅            1              2             (with a little care assuring that   k  ≠  0).

Prove that every open interval in real numbers contains rational, irrational and dyadic numbers.

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