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Thread: Modular Arithmetic

  1. #1
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    Modular Arithmetic

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

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

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

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

    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 $\displaystyle 2-26=b12$ where $\displaystyle b$is an integer. So here, $\displaystyle b=-2$.
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    One more example with a different mod that isn't so familiar:

    $\displaystyle r=3(\mod 7)$

    $\displaystyle r-3=b7$

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

    $\displaystyle 10=3(\mod 7)$

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

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

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

    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 $\displaystyle 2-26=b12$ where $\displaystyle b$is an integer. So here, $\displaystyle b=-2$.
    -----------------------------------------------------------------
    Your reasoning here is good.

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

    and

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

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

    $\displaystyle r=3(\mod 7)$

    $\displaystyle r-3=b7$

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

    $\displaystyle 10=3(\mod 7)$

    where b is an integer.
    ------------------------------------------------------------------
    In this case, $\displaystyle r=3\equiv3\pmod{7}$ 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.

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

    and

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



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

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

    Your going through all of those steps I underbraced, where as I'm saying that $\displaystyle 23=11(\mod 12)$ because 23 is the smallest number such that $\displaystyle 23-11=b12$, where $\displaystyle b$ is an integer. In this case, -1. Any number in $\displaystyle 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 $\displaystyle r=a(\mod P)$, I find $\displaystyle r$ such that $\displaystyle r$ is the smallest number such that$\displaystyle r-a=bP$

    So if $\displaystyle 3=3(\mod 7)$, then $\displaystyle 3-3=b7$ therefore $\displaystyle b=0$. So if I have $\displaystyle r=11 (\mod 12)$, it seems that I could just as easily claim that this is equal to 11 since $\displaystyle 11-11=b12$ with $\displaystyle b=0$ 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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