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    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)
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    Member Last_Singularity's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    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?
    Last edited by Last_Singularity; January 3rd 2009 at 08:56 AM.
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