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Math Help - Modular Arithmetic

  1. #1
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    Modular Arithmetic

    I'm having a hard time understanding modular numbers and aritmetic.

    If I have  r=a(\mod P), then r-a=bP where b is an integer. So if I want to find 23 in \mod 12, I write r=23(\mod 12) which should turn out to be 11 right? So 23 is equal to 11 in \mod 12.

    So if I want to find:
    r=26(\mod 12)

    I need to find r such that:
    r-26=b12

    where b is an integer and r is the smallest possible positve number satisfying the equation. Is this correct? 26 hours would bring us to 2 o-clock since 2 is the smallest positive integer such that  2-26=b12 where bis an integer. So here, b=-2.
    -----------------------------------------------------------------
    One more example with a different mod that isn't so familiar:

    r=3(\mod 7)

    r-3=b7

    So the smallest possible number seems to be 10 with b=1.

    10=3(\mod 7)

    where b is an integer.
    ------------------------------------------------------------------

    Is my reasoning correct?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by adkinsjr View Post
    I'm having a hard time understanding modular numbers and aritmetic.

    If I have  r=a(\mod P), then r-a=bP where b is an integer. So if I want to find 23 in \mod 12, I write r=23(\mod 12) which should turn out to be 11 right? So 23 is equal to 11 in \mod 12.

    So if I want to find:
    r=26(\mod 12)

    I need to find r such that:
    r-26=b12

    where b is an integer and r is the smallest possible positve number satisfying the equation. Is this correct? 26 hours would bring us to 2 o-clock since 2 is the smallest positive integer such that  2-26=b12 where bis an integer. So here, b=-2.
    -----------------------------------------------------------------
    Your reasoning here is good.

    23\equiv 11\pmod{12} since 23\pmod{12}\equiv11+12\pmod{12}\equiv11+0\pmod{12}  \equiv11\pmod{12}

    and

    26\equiv 2\pmod{12} since 26\equiv 24+2\pmod{12}\equiv 2(12)+2\pmod{12}\equiv 0+2\pmod{12}\equiv 2\pmod{12}

    One more example with a different mod that isn't so familiar:

    r=3(\mod 7)

    r-3=b7

    So the smallest possible number seems to be 10 with b=1.

    10=3(\mod 7)

    where b is an integer.
    ------------------------------------------------------------------
    In this case, r=3\equiv3\pmod{7} is the smallest positive value. Can you try to reason out why?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Your reasoning here is good.

    23\equiv 11\pmod{12} since 23\pmod{12}\equiv11+12\pmod{12}\equiv11+0\pmod{12}  \equiv11\pmod{12}

    and

    26\equiv 2\pmod{12} since 26\equiv 24+2\pmod{12}\equiv 2(12)+2\pmod{12}\equiv 0+2\pmod{12}\equiv 2\pmod{12}



    In this case, r=3\equiv3\pmod{7} is the smallest positive value. Can you try to reason out why?
    I'm not understanding this at all:

    23\equiv 11\pmod{12} since \underbrace{23\pmod{12}\equiv11+12\pmod{12}\equiv1  1+0\pmod{12}\equiv11\pmod{12}}

    Your going through all of those steps I underbraced, where as I'm saying that 23=11(\mod 12) because 23 is the smallest number such that 23-11=b12, where b is an integer. In this case, -1. Any number in 0\leq x\leq23\ will not give an integral multiple of twelve when I subtrace 11.

    In this case, is the smallest positive value. Can you try to reason out why?
    No, my reasoning wouldn't work here then. For r=a(\mod P), I find r such that r is the smallest number such that r-a=bP

    So if 3=3(\mod 7), then 3-3=b7 therefore b=0. So if I have r=11 (\mod 12), it seems that I could just as easily claim that this is equal to 11 since 11-11=b12 with b=0 again.
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  4. #4
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    I think you've got your definition backwards. You say "23= 11 (mod 12) because 23 is the smallest number such that 23- 11= b12, where b is an integer." No, 23= 11 (mod 12) because 11 is the smallest number such that 23- 11 is a multiple of 12. And, of course, 11= 11 (mod 12) because 11 is the smallest number such that 11- 11 is a multiple of 12.
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