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How can I go about proving the following problem: Prove that a number a is rational if and only if there exists a positive integer k such that $[ka]=ka$. Prove that a number is rational if and only if there exist k such that $[k! a]= k! a$

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    $\begingroup$ I cannot see where Farey fractions appear $\endgroup$ – mau May 4 '14 at 19:49
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The following chain of reasoning proves both parts of both problems:


If $a$ is rational, then there are integers $p,q$ with $a=p/q$.

If $a=p/q$, then $qa=p$, so $[qa]=[p]=p=qa$.

If $[ka]=ka$, then $ka\in\mathbb{Z}$, then $(ka)(k-1)!\in \mathbb{Z}$, then $[k!a]=k!a$.

If $[k!a]=k!a$, then $k!a=t\in \mathbb{Z}$, then $a=t/k!$, a ratio of two integers, so $a$ is rational.

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