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Theory of numbers (Division)

  1. Jul 8, 2012 #1
    Well the standard division theorem says,

    a = bq +r
    0 <= r < b
    after that we were introduced with r = b - r[itex]\acute{}[/itex]
    r[itex]\acute{}[/itex] having the same domain as that of r
    after that the theorem changes to
    a = b(q+1) - r[itex]\acute{}[/itex]

    Solving it with r[itex]\acute{}[/itex]= b-r, gives us the standard equation, but what does it imply?

    also two conditions in r and r[itex]\acute{}[/itex] where given -
    r>=1/2b and r[itex]\acute{}[/itex]<=1/2b
    and then defining Q and R

    a = bQ +R where |R| <=1/2b

    What does all this mean?
    Examples, please?
  2. jcsd
  3. Jul 8, 2012 #2


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    Science Advisor

    Hey Kartik.

    It just means you have a different definition for the quotient and the remainder.

    It's better to use the original definition, and the reason for this is that all of the rigorous proofs are based on what is known as the well ordering principle and for the normal definition that is used, the proofs are standardized.

    The other thing is that the natural definition of the modulus makes more sense when its defined with the normal bq + r instead of your own, because all the congruence results and arithmetic are naturally suited to this definition.

    Is there any reason why you wish to use your definition over the standard one?
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