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Find the greatest real number c such that $\{c\sqrt2\}\ge\frac{c}{n}$ for all positive integers n.

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Are you sure $n$ is required? It reduces to just $\{ c \sqrt{2} \geq \frac{c}{n}$. Is there supposed to be an $n$ in the LHS? – Calvin Lin May 28 '13 at 16:23
oops, it should be {nsqrt(2)} instead of {csqrt(2)} on the RHS – user79889 May 28 '13 at 18:05

It suffices to show it for $\{ c \sqrt{2} \} \geq c $.

Since the LHS is between 0 and 1, it tells us that the RHS is at most 1. Hence, $c$ can be any real value in $[-\infty, \frac{\sqrt{2}}{2})$. There is no greatest real number, but a supremum exists.

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