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Math Help - Division Algorithm...

  1. #1
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    Post Division Algorithm...

    I need some help in division algorithm...
    please help!

    Q1) prove that a+c is in equivalence relation with b+d (mod n)
    (i.e.) a+c=b+d(mod n)

    Q2) prove that ac=bd(mod n)..

    thanks in advance!
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  2. #2
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    I got a little but of idea for the second one...
    if a = b(mod n) and c = d(mod n), then ac = bd(mod n).

    So, by using the definition of the modulus:

    a = b + nk and c = d + nl for some integers we'll call k and l.

    So by multiplying both of these conditions we then get:-

    ac = (b + nk)(d + nl)
    = bd + nkd + nlb + n^2kl
    = bd + n(kd + lb + nkl)

    which means that ac = bd(mod n) since (kd + lb + nkl) is just some
    integer.

    I hope this is right!

    I did the similar operation for 1st but I am still not sure on how to resolve it?
    coz I keep getting the some random terms in there...


    Thanks for the any help!
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  3. #3
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    Quote Originally Posted by Vedicmaths View Post
    I got a little but of idea for the second one...
    if a = b(mod n) and c = d(mod n), then ac = bd(mod n).

    So, by using the definition of the modulus:

    a = b + nk and c = d + nl for some integers we'll call k and l.

    So by multiplying both of these conditions we then get:-

    ac = (b + nk)(d + nl)
    = bd + nkd + nlb + n^2kl
    = bd + n(kd + lb + nkl)

    which means that ac = bd(mod n) since (kd + lb + nkl) is just some
    integer.

    I hope this is right!
    Yes thats right...

    I did the similar operation for 1st but I am still not sure on how to resolve it?
    coz I keep getting the some random terms in there...

    Thanks for the any help!
    Use the same thing for the first:

    "a = b + nk and c = d + nl for some integers we'll call k and l"

    a+c = b+d + n(k+l)

    "which means that a+c = b+d(mod n) since (k + l) is just some
    integer."
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