Show that $[\sqrt{n}]=[\sqrt{n}+\frac{1}{n}]$, for any $n\in N, n\geq 2$
I let $a=\sqrt{n}$, and we know that $k\leq a < k+1$, where $k\in N$.
From now all we have to do is to show that $k \leq \frac{1}{a^2}+a<k+1$
I tried processing the first inequality but got to nothing useful. I hope one of you can help me! Thank you!
Best Answer
We proceed by contradiction. Suppose $\sqrt{n} + \frac{1}{n} \geq k+1$, so that $\sqrt{n} \geq k+1-\frac{1}{k^2}$. Then we know that $\sqrt{n}$ must lie in the interval $[k+1-\frac{1}{k^2},k+1)$, hence $n$ lies in the interval $$\left[(k+1-\frac{1}{k^2})^2,(k+1)^2\right) = \left[(k+1)^2 + \frac{1}{k^4} - \frac{2(k+1)}{k^2},(k+1)^2\right).$$ We aim to show that there is no integer in this interval, by showing that $\frac{2(k+1)}{k^2} - \frac{1}{k^4} < 1$. Rearranging, we need to show that $$k^4 - 2k^3 - 2k^2 + 1 > 0$$ which is true for $k \geq 4$ (since $k^4 > 4k^3 > 2k^3 + 2k^2$). This inequality is also true for $k=3$ by substitution. As for $k = 1,2$, one may check manually that no such $n$ works.