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Proving a floor function

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Ok, So ive tried to prove this statement unsuccessfully.
    how would you guys do it? (assume [] means floor)

    [x-1][x+1]+1 = ([x])^2
     
  2. jcsd
  3. Nov 6, 2012 #2

    HallsofIvy

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    Let x= a+ r for some integer a and r between 0 and 1. [x-1]= a-1, [x+ 1]= a+1, and [x]= a.
     
  4. Nov 6, 2012 #3
    I'll assume were working with [itex]x \in \mathbb{R}[/itex].

    Just to be clear of the definition of the floor function, it is the function [itex]\left\lfloor \right\rfloor : \mathbb{R} \longrightarrow \mathbb{Z}[/itex], with the mapping [itex]x \in \mathbb{R} \longmapsto \mathrm{max} \left\{ y \in \mathbb{Z} : y \leq x \right\}[/itex], i.e. the floor of a real number is the maximum integer in that set.

    Notice that you have a few cases, from different numbers giving different results for the floor function:
    1. [itex]x[/itex] being an integer
    2. [itex]x[/itex] not being an integer, but greater than zero
    3. [itex]x[/itex] not being an integer, but less than zero

    The first case requires only a correct application of the definition of the floor function and some algebraic manipulations; just start with the left-hand side of the equation and work your way to the right-hand side (same for the other two cases), or the other way.

    The second and third cases are more difficult. As already suggested, write [itex]x[/itex] in the way [itex]x = \alpha + \beta[/itex], where [itex]\alpha \in \mathbb{Z}[/itex], [itex]\beta \in \mathbb{Q} - \mathbb{Z}[/itex], i.e. [itex]\beta[/itex] to be a rational number that is not an integer. For the second case, what additional conditions would and must you apply to [itex]\alpha[/itex] and [itex]\beta[/itex]? Similarly for the third case? Hint: [itex]2.7 = 2 + 0.7[/itex].
     
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