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I How do you move floors and ceilings in discrete math?

  1. Sep 2, 2016 #1
    The title more accurately should have been "How do you cancel floors and ceilings in discrete functions"

    For instance,

    ##\frac{log{\frac{3x}{-6(z)}}}{8t} < 1##

    If I wanted to get rid of the log, I'd just raise the expression by base 10.

    ##\frac{(\frac{3x}{-6(z)})}{10^{8t}} < 10^1##

    But what happens if there's a roof for discrete functions?

    ##\frac{\lceil{log \frac{10x}{4y}}\rceil}{8z} < 1##

    How do I handle this?

    -----------------------------------------------------------------------------------

    EDIT NOTICE: the expressions above have been fixed into its their proper inequalities
     
    Last edited: Sep 2, 2016
  2. jcsd
  3. Sep 2, 2016 #2

    haruspex

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    i rather hope you would not do that, since it is wrong.
     
  4. Sep 2, 2016 #3
    oops, sorry, here's the other side

    ##\frac{log(stuff)}{8t} < 1 ##

    ##\frac{stuff}{10^{8t}} < 10 ##
     
  5. Sep 2, 2016 #4

    haruspex

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    Still wrong.

    ##\frac{\log(stuff)}{8t} < 1 ##
    ##\log(stuff)<8t##
    ##stuff<10^{8t}##
    For your ceil question, it might help if you state the entire problem.
     
  6. Sep 3, 2016 #5

    mfb

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    Assuming z>0, z>0 leads to a very similar case:
    ##\lceil{stuff\rceil} < 8z##
    What is the largest value of stuff that satisfies the inequality? Once you found that, you can continue with the usual approaches.
     
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