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    floor function - An Identity

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


    Source: American Mathematical Monthly (1988)
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    Member Last_Singularity's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Prove that $\displaystyle \forall n \in \mathbb{N}: \ \lfloor \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} \rfloor = \lfloor \sqrt{9n+8} \rfloor .$


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

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

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

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

    Our search has now become: show that for all $\displaystyle n$, $\displaystyle -1 < \epsilon_3 < 1$ for $\displaystyle \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} - \sqrt{9n+8} = \epsilon_3$ . In other words, if we find the minimum and the maximum of the left hand side and conclude that it's bounded between $\displaystyle -1,1$, 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:
    $\displaystyle \frac{\partial \epsilon_3}{\partial n} = n^{-1/2} + (n+1)^{-1/2} + (n+2)^{-1/2} - 9(9n+8)^{-1/2} = 0$

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

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

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

    I was originally going to try to use induction, as the phrase "for all $\displaystyle n$ in the naturals" seemed to suggest, but the square roots really gave me trouble because I could not free up my $\displaystyle k$ and $\displaystyle k+1$ terms; it was a serious mess. Anyone have better luck with this?
    Last edited by Last_Singularity; Jan 3rd 2009 at 07:56 AM.
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