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Math Help - linearity properties

  1. #1
    Senior Member DivideBy0's Avatar
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    linearity properties

    A proof about congruences got me wondering about the linearity property:

    a|b \wedge a|c \Rightarrow a|bm+cn for m,n \in R

    Here's how it went:

    If a \equiv b (mod m) and c \equiv d (mod m), then a \pm c \equiv b \pm d (mod m)

    By assumption m|a-b and m|c-d.

    By the linearity property, m|(a-b)-(c-d). Re-arranging we get m|(a+c)- (b+d)

    Can you just rearrange that easily??? Sure the total sum (a+c)-(b+d) stays the same, and it also does when you convert it back to mods.
    a+c \equiv b+d (mod m)

    But what if I went and did
    m|8-5 and m|12-3
    Then m|(8-5)-(12-3) so.... m|12+(8+5-3) \Rightarrow 12 \equiv 10 (mod m) ???
    Which is not true.
    I've rearranged it so that there are two terms, x = 12 and y = 8+5-3. That's what happened in the proof. But my version doesn't work.

    What the associativity rules and restrictions on this arithmetic?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    By the linearity property, m|(a-b)-(c-d). Re-arranging we get m|(a+c)- (b+d)
    How are you doing your rearrangement?
    (a - b) - (c - d) = (a + d) - (b + c) \neq (a + c) - (b + d)

    Then your example reads:
    m|(8 - 5) and m|(12 - 3)

    So
    m|((8 - 5) - (12 - 3)) \implies m|(3 - 9) \implies 9 \equiv 3 ~\text{mod(m)}
    which is more or less what your initial statements imply.

    -Dan
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  3. #3
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    If a|b and a|c then a| (bx+cy) for all x,y\in \mathbb{Z}.
    In congruence notation we can prove this easily.
    We know that b\equiv 0 (\bmod a)]\ and c\equiv 0 (\bmod a).
    Multiply both side of the congruences by x and y respectively to get,
    bx\equiv 0(\bmod a) and cy\equiv 0(\bmod a).
    Add them together,
    bx+cy\equiv 0(\bmod a).
    Thus, a|(bx+cy).
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  4. #4
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by topsquark View Post
    How are you doing your rearrangement?
    (a - b) - (c - d) = (a + d) - (b + c) \neq (a + c) - (b + d)

    Then your example reads:
    m|(8 - 5) and m|(12 - 3)

    So
    m|((8 - 5) - (12 - 3)) \implies m|(3 - 9) \implies 9 \equiv 3 ~\text{mod(m)}
    which is more or less what your initial statements imply.

    -Dan
    Sorry, that was supposed to be m|(a-b)+(c-d) \Rightarrow m|(a+c)-(b+d)

    As for my example, I think I got confused with the arithmetic somehow...
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