Prove that a number $a$ is rational if and only if there exists a positive integer $k$ such that $[ka]=ka$; likewise for $[k! a]= k! a$

Question

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 $a$ is rational if and only if there exists $k$ such that $[k! a]= k! a$

I cannot see where Farey fractions appear – mau May 04 '14 at 19:49 — mau, May 04 '14 at 19:49
@mau: I have removed "Farey fractions" from the title. – J W Sep 22 '20 at 15:18 — J W, Sep 22 '20 at 15:18

score 3 · Accepted Answer · answered May 04 '14 at 19:17

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.

Prove that a number $a$ is rational if and only if there exists a positive integer $k$ such that $[ka]=ka$; likewise for $[k! a]= k! a$

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