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New member = lotsa ?'s

  1. Jun 24, 2004 #1
    i know i've only been on for couple of months, but i was always afraid of asking for help. thanks to this site, i now understand all of this theory stuff!


    prove that [alpha] + [alpha + 1/3] + [alpha + 2/3] = [3 alpha] for all reals alpha.

    prove [alpha] + [beta] <= [alpha + beta] alpha/beta are real.

    i honestly do not know what to do for these, could anyone provide a beacon of light???
  2. jcsd
  3. Jun 24, 2004 #2
    And what does [alpha] mean? Can't be the absolute value of alpha, since none of those statements are true then.
  4. Jun 24, 2004 #3

    matt grime

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    i suspect it means floor. the second on requires no proof being quite obvious, the first one, hmm, i can't think of anything other than a brute force way of doing it; someone must be able to spot a nice argument for it. In case 1+1=1 is wondering, the brute force method is to let alpha = n+r where r is a number in the interval [0,1) there are then three cases to consider, r in [0,1/3), [1/3,2/3) and [2/3,1) and so that [alpha]=n.
    Last edited: Jun 24, 2004
  5. Jun 24, 2004 #4


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    1. Consider x - [x] in [0,1/3) , [1/3, 2/3) and [2/3, 1) and prove separately for each case. Perhaps this is what matt means by 'brute force'...

    2. The "no proof" for this could be done by writing alpha = N + n, where N is an integer and n is in [0,1).

    EDIT : Didn't read matt's post completely - this is redundant.
    Last edited: Jun 24, 2004
  6. Jun 24, 2004 #5
    sorry all as i said i am new and i still need to do A LOT of clarifying. the [] means the greatest integer. as far as the brute force, that is the least painful way of doing this? i appreciate all of this! yes i do understand the brute force method.
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