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Modular operation: Addition

  1. Jun 4, 2012 #1
    Hello all,

    I was wondering if someone can explain to me a step in a proof given to me by my professor in regards to a modular operation theorem.

    Addition theorem: Given three integers x, y, d (d > 0), (x+y)%d = (x%d + y%d) %d

    Let x = q(1)d + r(1) and y = q(2)d + r(2).
    We have (x+y)%d = (q(1)d + r(1) + q(2)d + r(2)) %d
    = (r(1) + r(2)) %d
    Therefore: (x+y)%d = (x%d + y%d) %d

    I dont get how my professor jumped from (q(1)d + r(1) + q(2)d + r(2))%d to (r(1) + r(2))%d.

    Is there a specific reason for why we just ignore the product of q(1)d and q(2)d?

    Thank you in advance.
  2. jcsd
  3. Jun 4, 2012 #2


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    The Big Idea is that d is zero modulo d, so q(1)d is zero because you're multiplying q(1) by zero.

    As for the technical detail, doesn't the equality follow directly from the definition of x%d? If you don't think so, then please state what definition you are using, and apply that definition to the two sides of that equation.
  4. Jun 4, 2012 #3
    d has to be greater than zero in our given. so i dont believe q(1)d and q(2)d are negated because d is zero.

  5. Jun 4, 2012 #4


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    d is zero modulo d.
  6. Jun 4, 2012 #5
    oh i get it! ty ty!
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