Power of $(\sqrt{2} +1)$ and floor function - Printable Version +- OMath! (http://math.elinkage.net) +-- Forum: Math Forums (/forumdisplay.php?fid=4) +--- Forum: Inequalities (/forumdisplay.php?fid=8) +--- Thread: Power of $(\sqrt{2} +1)$ and floor function (/showthread.php?tid=128) Power of $(\sqrt{2} +1)$ and floor function - elim - 01-07-2011 07:24 PM Prove that $\lfloor (\sqrt{2} + 1)^{2n} \rfloor + 1 - (\sqrt{2} + 1)^{2n} \geq \frac{1}{2}$ and $(\sqrt{2} + 1)^{2n-1} - \lfloor (\sqrt{2} + 1)^{2n-1} \rfloor \leq \frac{1}{2} , \forall n \in \mathbb{N}$