1. ## linearity properties

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

Here's how it went:

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

By assumption $\displaystyle m|a-b$ and $\displaystyle m|c-d$.

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

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.
$\displaystyle a+c \equiv b+d$ (mod m)

But what if I went and did
$\displaystyle m|8-5$ and $\displaystyle m|12-3$
Then $\displaystyle m|(8-5)-(12-3)$ so.... $\displaystyle m|12+(8+5-3) \Rightarrow 12 \equiv 10$ (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?

2. Originally Posted by DivideBy0
By the linearity property, $\displaystyle m|(a-b)-(c-d)$. Re-arranging we get $\displaystyle m|(a+c)- (b+d)$
How are you doing your rearrangement?
$\displaystyle (a - b) - (c - d) = (a + d) - (b + c) \neq (a + c) - (b + d)$

$\displaystyle m|(8 - 5)$ and $\displaystyle m|(12 - 3)$

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

-Dan

3. If $\displaystyle a|b$ and $\displaystyle a|c$ then $\displaystyle a| (bx+cy)$ for all $\displaystyle x,y\in \mathbb{Z}$.
In congruence notation we can prove this easily.
We know that $\displaystyle b\equiv 0 (\bmod a)]\$ and $\displaystyle c\equiv 0 (\bmod a)$.
Multiply both side of the congruences by $\displaystyle x$ and $\displaystyle y$ respectively to get,
$\displaystyle bx\equiv 0(\bmod a)$ and $\displaystyle cy\equiv 0(\bmod a)$.
$\displaystyle bx+cy\equiv 0(\bmod a)$.
Thus, $\displaystyle a|(bx+cy)$.

4. Originally Posted by topsquark
How are you doing your rearrangement?
$\displaystyle (a - b) - (c - d) = (a + d) - (b + c) \neq (a + c) - (b + d)$

$\displaystyle m|(8 - 5)$ and $\displaystyle m|(12 - 3)$
$\displaystyle m|((8 - 5) - (12 - 3)) \implies m|(3 - 9) \implies 9 \equiv 3 ~\text{mod(m)}$
Sorry, that was supposed to be $\displaystyle m|(a-b)+(c-d) \Rightarrow m|(a+c)-(b+d)$