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Consider the expression

$$\left\lfloor\frac{2n-1}{2}\right\rfloor\;:$$

does it make sense that this floor function will evaluate to $n$? Or should it be $n-1$?

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  • $\begingroup$ Are we saying anything about n? Is it an integer, real number, etc.? $\endgroup$ – The Great Duck Mar 10 '16 at 5:19
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$\frac{2n-1}{2}=n-\frac{1}{2}$, so the floor is $n-1$.

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  • $\begingroup$ ...of course assuming that $\;n\;$ ranges over the integers, which seems implicit in the OP's choice of that variable name. $\endgroup$ – Marnix Klooster Mar 10 '16 at 5:05

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