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This isn't true, right?

$$k\left\lfloor\frac n {2k}\right\rfloor\leq \left\lfloor\frac n k\right\rfloor$$

$2<k\leq \left\lfloor\dfrac {n-1} 2\right\rfloor$, $n>4$, $k,n$ are coprime.

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  • $\begingroup$ Those $k$ and $n$ do not meet the conditions provided $\endgroup$ – user155385 Jul 1 '14 at 21:49
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As with your previous question one check is what happens at the numbers grow. If $k$ is as large as possible (for a given $n$) the right side will be $2$. The left side will be about $\frac n2$, which can be large. Take, for example, $n=101,k=50$. Then the left is $50$, the right is $2$. Often it helps just to ignore the floor function for a preliminary investigation.

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Let $n=15$ and $k=4$. Not that $15>4$, $2<4<\left\lfloor\dfrac{15-1}{2}\right\rfloor=7$, and that $gcd(4,15)=1$

Now

$$k\left\lfloor\dfrac {2n} k\right\rfloor=4\left\lfloor\dfrac {30} 4\right\rfloor=28$$

and

$$\left\lfloor\dfrac n k\right\rfloor=\left\lfloor\dfrac {15}4\right\rfloor=3$$

Clearly $28>3$, so this provides a counterexample.

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  • $\begingroup$ SHOOT! I meant to put the 2 in the denominator, sorry! $\endgroup$ – user2154420 Jul 1 '14 at 21:57

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