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Let $⌊x⌋$ denote the integer part of $x$ and let $a$ be a real number.

My questions are:

(1) Find sufficient and necessary conditions for which the inequality: $$⌊x⌋<x-a$$ is possible

(2) Solve that inequality with respect to $x$.

My attempt: I can think that when $0<a<1$, the inequalty is possible. But I cannot go further on that.

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    $\begingroup$ With $1400$ reputation, you should know better than to post questions with zero effort shown... $\endgroup$ – 5xum Oct 16 '15 at 17:24
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$\lfloor x\rfloor< x-a$ has a solution if and only if $a<1$.

$\underline{\implies}$

Suppose $\lfloor b\rfloor<b-a$ then $a<b-\lfloor b\rfloor<1$.

$\underline{\impliedby}$

Suppose $a<1$, then for any $x$ such that $a<x-\lfloor x\rfloor<1$ is a solution.

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