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Why modulo m1 and modulo m2 implies modulo [m1, m2]

  1. Sep 17, 2010 #1
    Why "modulo m1" and "modulo m2" implies "modulo [m1, m2]"

    If [tex]a \equiv r[/tex] (mod m1) and [tex]a \equiv r[/tex] (mod m2) then [tex]a \equiv r[/tex] (mod [m1, m2]), where [a, b] is the least common multiple of a and b.

    I have tried to prove that.

    Assume that
    [tex][m_{1}, m_{2}] = l_{1}m_{1} = l_{2}m_{2}[/tex]

    and

    [tex]a = k_{1}m_{1} + r[/tex]
    [tex]a = k_{2}m_{2} + r[/tex]

    Then

    [tex]al_{1} = k_{1}l_{1}m_{1} + rl_{1}[/tex]
    [tex]al_{2} = k_{2}l_{2}m_{2} + rl_{2}[/tex]

    thus

    [tex]a(l_{1} - l_{2}) = [m_{1}, m_{2}](k_{1} - k_{2}) + r(l_{1} - l_{2})[/tex]

    and

    [tex]a = [m_{1}, m_{2}] {(k_{1} - k_{2}) \over (l_{1} - l_{2})} + r[/tex]

    In order to

    [tex]a \equiv r\ (mod [m_{1}, m_{2}])[/tex], or [tex]a = K[m_{1}, m_{2}] + r[/tex]

    we have to prove that

    [tex](l_{1} - l_{2})\; | \; (k_{1} - k_{2})[/tex]

    But how?
     
  2. jcsd
  3. Sep 17, 2010 #2

    Gokul43201

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    Re: Why "modulo m1" and "modulo m2" implies "modulo [m1, m2]"

    This doesn't directly answer your question, but wouldn't the proof be a lot easier if you started off with m1|(a-r) and m2|(a-r) ... ?
     
  4. Jan 3, 2011 #3
    Re: Why "modulo m1" and "modulo m2" implies "modulo [m1, m2]"

    You are right, the proof is actually very simple:

    If m1 | (a-r) and m2 | (a-r) then a-r is a common multiple of m1 and m2. Therefore it is divisible by the least common multiple, [m1, m2]. Thus [m1, m2] | (a-r), or [tex]a[/tex] [tex] \equiv r[/tex] (mod [m1, m2])
     
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