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Math Help - Floor function proof (probably easy)

  1. #1
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    Floor function proof (probably easy)

    I want to know if my attempt is anywhere near being correct...


    I'm supposed to prove that floor(n + m) = n + floor(m) where n is an integer and m is a real number. I'll write this as [n + m] = n + [m].


    [n + m] can be rewritten as floor( ( n + [m] ) + ( m - [m] ) ).

    By definition, [m] is the largest integer not greater than m. Therefore the absolute value of ( m - [m] ) is less than 1.

    n is an integer and [m] is an integer, so ( n + [m] ) is also an integer because the set of integers is closed under addition.

    So floor( ( n + [m] ) + ( m - [m] ) ) or floor(n + m) is equal to the integer ( n + [m] ), because this is the largest integer not greater than (n + m).

    Therefore [n + m] = n + [m].


    Thank you!
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  2. #2
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    Re: Floor function proof (probably easy)

    Hi,
    In proofs involving floor, the following characterization is often useful. Namely, for x a real,
    \lfloor x\rfloor is the unique integer n with n\leq x<n+1.

    For your problem, let j=\lfloor m\rfloor. Then clearly j+n is an integer with
    j+n\leq m+n<j+n+1. QED.

    Similarly, ceiling(x) = \lceil x\rceil, is the unique integer with \lceil x\rceil\geq x>\lceil x\rceil-1
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