Show that $[\sqrt{n}]=[\sqrt{n}+\frac{1}{n}]$, for any $n\in N, n\geq 2$

Question

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!

Answer 1

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.

Answer 2

The condition $$k^2\le n\le (k+1)^2-1=k^2+2k$$ is equivalent to $[\sqrt{n}]=k.$ We have $$\sqrt{n}+{1\over n}<\sqrt{k^2+2k}+{1\over k^2}\\ = k+1-(k+1-\sqrt{k^2+2k})+{1\over k^2}\\ =k+1-{1\over \sqrt{k^2+2k}+k+1}+{1\over k^2}\\ < k+1-{1\over 2(k+1)}+{1\over k^2}$$ The last expression is less than $k+1$ as $k^2> 2(k+1)$ for $k\ge 3.$ The cases $k=1,2$ can be verified by hand.

Answer 3

Let $e = a-k$ be the fractional part of $a$, which means $0 \leq e < 1$. Then the problem becomes $$0 \leq e + \frac{1}{(k+e)^2} < 1$$ The LHS is obvious, so we only need to consider the RHS. Now let's consider the condition $n < (k+1)^2$. Since n is integer we can further restrict it to $n \leq (k+1)^2 - 1$. Then we have $$ \begin{align} & &(k+e)^2 &\leq (k+1)^2 - 1 \\ &\Leftrightarrow &e &\leq \sqrt{k^2+2k}-k\\ & & &= 1 + (\sqrt{k^2+2k}-k-1)\\ & & &= 1 - \frac{1}{k+\sqrt{k^2+2k}}\\ & & &< 1 - \frac{1}{3k}\\ \end{align} $$

Hence for $k \geq 3$ $$ e + \frac{1}{(k+e)^2} < 1-\frac{1}{3k}+\frac{1}{k^2} \leq 1$$ Cases for $k < 3$ can be checked by hand

Answer 4

Thinking aloud. $\sqrt n$ is between two integers. $\frac 1n < 1$ so adding $\frac 1n$ to $\sqrt n$ will either a) keep the results between the same two integers (in which case $[\sqrt n + \frac 1n] =[\sqrt n]$), or b) $\frac 1n$ was just enough to push the result above the upper integer. In this case $[\sqrt n + \frac 1n] =[\sqrt n] + 1$ and there is an integer $k$ so that $\sqrt n < k \le \sqrt n + \frac 1n$.

It is now our job to prove $\sqrt n < k \le \sqrt n + \frac 1n$ just simply can never happen (if $n \ge 2$).

$\sqrt n < k$ so $n < k^2$ so $k^2 \ge n+1$ and $k \ge \sqrt {n+1}$

If we can sho $\sqrt{n+1} - \sqrt n \ge \frac 1n$ we will be done

Let $e = \sqrt{n+1} -\sqrt n = \frac 1{\sqrt{n+1} + \sqrt n}>\frac 1{2\sqrt{n+1}}$

We have $\frac 1{2\sqrt{n+1}}\ge \frac 1n \iff n \ge 2\sqrt{n+1} \iff n^2\ge 4n + 4 \iff n^2 -4n + 4 \ge 8\iff n-2 \ge 2\sqrt 2 \iff n\ge 2\sqrt 2+2$ which holds for $n\ge 5$ but we have to show also holds for $n = 2,3,4$.

In these cases we may have $\sqrt{n+1} - \sqrt n <\frac 1n$ but we probably do not have $k =\sqrt{n+1}$.

If $n = 2$ then $1< \sqrt n < 1.5$ and $2-\sqrt 2 > \frac 12 =\frac 12$.

If $n=3$ then as $(\frac 53)^2 = \frac {25}9< 3$ we have $1 < \sqrt 3 < \frac 53$ and $3-\sqrt 3 > \frac 13 = \frac 1n$.

And for $4$ we have $\sqrt 4 = 2$ and $\sqrt 4 + \frac 14 = 4 +\frac 14 < 5$ so we are done.