Mathematics Asked by npdfe on November 26, 2021
I have the following problem.
Let $A$ be the set of rational numbers which have the form ${dfrac{k}{2^n}}$ for some $ k in mathbb{Z} $ and $ n in mathbb{N} $ . Prove that if $p$ and $q$ are rational number with $p < q$ then there exist $ r in A$ such that $p le r le q $
I think I should starts with $p = dfrac ab$ and $q = dfrac cd$, with $a,b,c,d$ in integers, and $b,d neq 0$. But not sure what to do next. Maybe I need to use axioms?
Hint: First notice that if $q-pgeq 1$ then you can just place an integer in between and integers are in $A.$ Then notice that if $p>1$ you can just substract $lfloor p rfloor$ and so without loss of generality you just need to show this for $p<qin [0,1)$ Now play a little game in which you divide the interval by $2$ and if the two numbers are separated you are done and if they are not then you divide the subinterval in which you are. Use the archimedian property to show that there exists an $n$ such that $frac{1}{2^n}<frac{1}{q-p}$ and so the game has to end at some point.
Answered by Phicar on November 26, 2021
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