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If we take a number $\lfloor{xe}\rfloor$ and another number $\left\lfloor\lfloor{\frac{xe}{e-1}\rfloor}(e-1)\right\rfloor$, then can we say that they are equal $\forall x\geq 2 \in \mathbb{N}$? If not, when? I think they are pretty equal since $a \lfloor\frac{b}{a}\rfloor + r=b$ where $0\leq r < a$ and here, $a=e-1 < 1.7$. I am not sure if they hold infinitely or not, that's my question.

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  • $\begingroup$ What does "equal infinitely" mean? $\endgroup$ – Kenny Lau Jun 8 '17 at 3:42
  • $\begingroup$ Is it true for $x=e-1$. $\endgroup$ – hamam_Abdallah Jun 8 '17 at 3:43
  • $\begingroup$ Check - what happens when $x=1$? $\endgroup$ – Jaideep Khare Jun 8 '17 at 3:44
  • $\begingroup$ @JaideepKhare it doesn't hold. But check for some more values. $\endgroup$ – Mathejunior Jun 8 '17 at 3:45
  • $\begingroup$ @KennyLau I think OP wants to ask whether is it true $\forall x \in \Mathbb R$. $\endgroup$ – Jaideep Khare Jun 8 '17 at 3:45
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There are many values of $x$ for which this is not true; for instance, $x=101$

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  • $\begingroup$ I strongly doubt this inaccuracy goes away as $x$ gets large: for instance, around $x=110100$ they were different in 5 of the 7 cases I checked. $\endgroup$ – Eric Stucky Jun 8 '17 at 4:19
  • $\begingroup$ As far as finding the cases where there is no inaccuracy, my best answer is good luck. The discrepancy stays bounded above by 3 for all $x$, and so as $x$ gets big you're looking to know the exact value of $1/(e-1)$ to very high accuracy. That doesn't seem so impossible, but it might not be easy to make any uniform statement. $\endgroup$ – Eric Stucky Jun 8 '17 at 4:33

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