# linearity properties

• Sep 23rd 2007, 08:03 AM
DivideBy0
linearity properties

$a|b \wedge a|c \Rightarrow a|bm+cn$ 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?
• Sep 23rd 2007, 08:25 AM
topsquark
Quote:

Originally Posted by DivideBy0
By the linearity property, $m|(a-b)-(c-d)$. Re-arranging we get $m|(a+c)- (b+d)$

How are you doing your rearrangement?
$(a - b) - (c - d) = (a + d) - (b + c) \neq (a + c) - (b + d)$

$m|(8 - 5)$ 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
• Sep 23rd 2007, 08:53 AM
ThePerfectHacker
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 $cy\equiv 0(\bmod a)$.
$bx+cy\equiv 0(\bmod a)$.
Thus, $a|(bx+cy)$.
• Sep 23rd 2007, 09:57 AM
DivideBy0
Quote:

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

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