# Thread: floor function - An Identity

1. ## floor function - An Identity

Prove that $\forall n \in \mathbb{N}: \ \lfloor \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} \rfloor = \lfloor \sqrt{9n+8} \rfloor .$

Source: American Mathematical Monthly (1988)

2. Originally Posted by NonCommAlg
Prove that $\forall n \in \mathbb{N}: \ \lfloor \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} \rfloor = \lfloor \sqrt{9n+8} \rfloor .$

Source: American Mathematical Monthly (1988)
I have a partial solution, using a graphing calculator (I will try to think of another one afterwards). If anyone could add to this, that would be great. Here is my current approach:

Observation: If $\lfloor \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} \rfloor = \lfloor \sqrt{9n+8} \rfloor$ holds $\forall n \in \mathbb{R}$, then it must hold $\forall n \in \mathbb{N}$, since $\mathbb{N} \subset \mathbb{R}$.

To show that $\lfloor \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} \rfloor = \lfloor \sqrt{9n+8} \rfloor$ holds $\forall n \in \mathbb{R}$, we can use the definition of the floor function:
$\lfloor x \rfloor = c \qquad \longrightarrow \qquad x - \epsilon = c, 0 \leq \epsilon < 1$.

Thus, we can re-write the original question in the following format:
$\sqrt{n} + \sqrt{n+1} + \sqrt{n+2} - \epsilon_1 = \sqrt{9n+8} - \epsilon_2 \qquad$ where $0 \leq \epsilon_1, \epsilon_2 < 1$

Combining the two error terms into one and move it to the right hand side of the equation, we obtain:
$\sqrt{n} + \sqrt{n+1} + \sqrt{n+2} = \sqrt{9n+8} + \underbrace{\epsilon_1 - \epsilon_2}_{\epsilon_3}$ where $-1 < \epsilon_3 < 1$

Our search has now become: show that for all $n$, $-1 < \epsilon_3 < 1$ for $\sqrt{n} + \sqrt{n+1} + \sqrt{n+2} - \sqrt{9n+8} = \epsilon_3$ . In other words, if we find the minimum and the maximum of the left hand side and conclude that it's bounded between $-1,1$, we would be done.

To find the extrema of the left hand side, we need to take the first-order derivative and set it equal to zero:
$\frac{\partial \epsilon_3}{\partial n} = n^{-1/2} + (n+1)^{-1/2} + (n+2)^{-1/2} - 9(9n+8)^{-1/2} = 0$

And the solution to this is $n_0 = 3.578$, which, when plugged back into the original equation, gives a maximum value of $\epsilon_3 = 0.05246$, which is indeed less than $1$. Because $n^{-1/2} + (n+1)^{-1/2} + (n+2)^{-1/2}$ is monotonically increasing simply by observation and $- 9(9n+8)^{-1/2}$ is monotonically decreasing by observation, we can be sure that $n_0 = 3.578$ is the only extreme point in $(0,\infty)$.

To find the minimum, we check the edge point $n=0$ because the first-order derivative would not help us there. At $n=0$, we get $\epsilon_3 = 1 - \sqrt{2} = -0.414...$, which is indeed greater than $-1$.

The proof is complete (that is, if you believe that I have found all critical values of $n$ such as $3.578,-0.414$ or whatever).

I was originally going to try to use induction, as the phrase "for all $n$ in the naturals" seemed to suggest, but the square roots really gave me trouble because I could not free up my $k$ and $k+1$ terms; it was a serious mess. Anyone have better luck with this?