Can You Prove the Floor Function Relationship for Positive Integers?

[juantheron]

For any positive integer $n\;,$ prove that$\sqrt{4n+1}<\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}$.

Hence or otherwise, prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+1}}\right\rfloor$

for any positive integer $n$.

 

[kaliprasad]

For any positive integer $n\;,$ prove that$\sqrt{4n+1}<\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}$.

Hence or otherwise, prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+1}}\right\rfloor$

for any positive integer $n$.

Spoiler

we realize that $n(n+1) = (n+\dfrac{1}{2})^2 - \dfrac{1}{4}$

so $\sqrt{n(n+1)}\lt(n+\dfrac{1}{2})$

clearly $n\lt\sqrt{n(n+1)}$



so

we have

$(\sqrt{n} + \sqrt{n+1})^2 = n + n+ 1 + 2 \sqrt{n(n+1)}$

= $2n +1 + 2 \sqrt{n(n+1)}$

>$2n+ 1 + 2 n$ or > $4n+ 1$

and = $2n +1 + 2 \sqrt{n(n+1)}$ < $2n +1 + 2 (n + \dfrac{1}{2})$

or < (4n + 2)

so $(4n+1)\lt(\sqrt{n} + \sqrt{n+1})^2\lt(4n+2)$

because 4n+2 is not a perfect square

we have $\lfloor(\sqrt{4n+1}\rfloor = \lfloor(\sqrt{4n+2}\rfloor$

and as $ (\sqrt{n} + \sqrt{n+1})^2$ is between 4n + 1 and 4n +2 we have

$\lfloor(\sqrt{4n+1}\rfloor = \lfloor(\sqrt{4n+2}\rfloor= \lfloor(\sqrt{n} +\sqrt{n+1}\rfloor$

 

[juantheron]

Thank U kali. Nice solution

My solution is same as you.

Spoiler

Left inequality :: $\displaystyle R = \frac{\sqrt{n} + \sqrt{n+1}}{\sqrt{4n + 1}}$

Squaring the fraction gives:

$\displaystyle R^2 = \frac{2n + 1 + 2\sqrt{n^2 + n}}{4n + 1}$

$\displaystyle R^2 = \frac{2n + 1}{4n + 1} + \frac{2\sqrt{n^2 + n}}{4n + 1}$

Replacing $n^2 + n$ with $n^2$ (which is smaller) gives:

$\displaystyle R^2 > \frac{2n + 1}{4n + 1} + \frac{2n}{4n + 1}$

$\displaystyle R^2 > \frac{4n + 1}{4n + 1}$

$R^2 > 1$

Therefore, $\displaystyle \sqrt{n} + \sqrt{n+1} > \sqrt{4n + 1}$

Right inequality :: Now, let $\displaystyle R = \frac{\sqrt{n} + \sqrt{n+1}}{\sqrt{4n + 2}}$. Again, squaring yields:

$\displaystyle R^2 = \frac{2n + 1 + 2\sqrt{(n)(n+1)}}{4n + 2}$

This time, replace $n(n+1)$ with $(n+\frac{1}{2})^2$, which is bigger (the factors have the same sum, and the biggest product for two numbers with the same sum is when the factors are the same - or just expand $n(n+1) - (n + \frac{1}{2})^2$ to get a convenient negative constant). This gives:

$\displaystyle R^2 < \frac{2n + 1 + 2n + 1}{4n + 2}$

$\displaystyle R^2 < \frac{4n + 2}{4n + 2}$

$R^2 < 1$

Therefore, $\displaystyle \sqrt{n} + \sqrt{n + 1} < \sqrt{4n + 2}$.

Lemma: the numbers $\sqrt{4n + 1}$ and $\sqrt{4n + 2}$ do not have an integer "strictly" in between them. That is to say, there is no integer $k$ such that $\sqrt{4n + 1} < k < \sqrt{4n + 2}$.

Semi-Proof: The two numbers are the square roots of consecutive positive integers, so they are two neighboring numbers in the list $\sqrt{1}, \sqrt{2}, \sqrt{3}, ...$. Every positive integer is on this list, but clearly no two integers are next to each other on this list, since the integers in this list start off "close" and only appear less often as you go up the list.

Thus, there is some integer $k$ for which $k \le \sqrt{4n + 1} < \sqrt{4n + 2} \le k + 1$. Putting our other expression in, we have:

$k \le \sqrt{4n + 1} < \sqrt{n} + \sqrt{n+1} < \sqrt{4n + 2} \le k + 1$

By definition, $\displaystyle \lfloor \sqrt{4n + 1}\rfloor = k$, and $\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = k$, so they have the same floor.