## Tuesday, September 8, 2015

### Prove that: $⌊√n+1/(√n+√(n+2))⌋=⌊√n⌋$, for all $n\in N$.

Prove that: $\left\lfloor{\sqrt{n}+\dfrac{1}{\sqrt{n}+\sqrt{n+2}}}\right\rfloor= \left\lfloor{\sqrt{n}}\right\rfloor$, for all $n\in N$.

My solution:

Step 1:

First, note that the expression inside the floor function on the left can be rewritten such that we have:

$\sqrt{n}+\dfrac{1}{\sqrt{n}+\sqrt{n+2}}$

$=\sqrt{n}+\dfrac{1}{\sqrt{n}+\sqrt{n+2}}\cdot \dfrac{\sqrt{n}-\sqrt{n+2}}{\sqrt{n}-\sqrt{n+2}}$

$=\sqrt{n}+\dfrac{\sqrt{n}-\sqrt{n+2}}{n-(n+2)}$

$=\sqrt{n}-\dfrac{\sqrt{n}-\sqrt{n+2}}{2}$

$=\sqrt{n}+\dfrac{\sqrt{n+2}-\sqrt{n}}{2}$

$=\dfrac{\sqrt{n}+\sqrt{n+2}}{2}$

Step 2:

Next, note that for all $n\in N$, $4\sqrt{n}\gt -2$ is always true. Algebraically manipulating it such that we get:

$n+4\sqrt{n}+4\gt -2+n+4$

$(\sqrt{n}+2)^2\gt n+2$

$\sqrt{n}+2\gt \sqrt{n+2}$

Therefore we have:

$\sqrt{n}+\sqrt{n}+2\gt \sqrt{n}+\sqrt{n+2}$, which is

$2\sqrt{n}+2\gt \sqrt{n}+\sqrt{n+2}$

Step 3:

Note that we can set:

$\sqrt{n}+\sqrt{n}\lt \sqrt{n}+\sqrt{n+2}\lt 2\sqrt{n}+2$

which is

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

$\sqrt{n}\lt \dfrac{\sqrt{n}+\sqrt{n+2}}{2}\lt \sqrt{n}+1$

Thus, we can conclude at this juncture that $\left\lfloor{\dfrac{\sqrt{n}+\sqrt{n+2}}{2}}\right\rfloor=\left\lfloor{\sqrt{n}+\dfrac{1}{\sqrt{n}+\sqrt{n+2}}}\right\rfloor=\left\lfloor{\sqrt{n}}\right\rfloor$ and we're hence done.