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Let $(a_{n})_{n\geq 1}$ be an arithmetic progression with $r>0$ such as $\vee n\geq 1,\vee x\in (a_{n},a_{n+1})$ we have $\left [ x \right ]=\left [ a_{n} \right ]$. Prove that ∃m$\in N,m\neq 0$ such as $r=\frac{1}{m}$. I figured out that $x\in (a_{n},a_{n}+r)$, but I don't know what to do from here

J. W. Tanner
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omega123
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1 Answers1

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Consider what would happen if there were an $n\geq 1$ such that $a_n < k < a_{n+1}$ where $k$ is an integer. Also consider what the terms of the sequence near $k+1$ might be.

Remember the property $\forall n\geq 1,\forall x\in (a_{n},a_{n+1}), \left [ x \right ]=\left [ a_{n} \right ]$. It will matter somehow.

David K
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