B counts as being greater than A. CP doesn't work for signed numbers, but if we add, subtract, or multiply by a signed number, it works as you might expect. For example, (-3) mod_{256}*(-2) mod_{256}=253*254=6 mod_{256}. Division is a little trickier and will require its own routine for signed numbers.

So for the proof that 'negative' numbers work for addition, subtraction, and multiplication in mod_{n} works like this:

- Define a negative number in mod
_{n} as -a mod_{n}=0-a mod_{n}=(n mod_{n})-(a mod_{n})=(n-a mod_{n})=n-a. So -a=n-a. Now select any two numbers, a and b:

(At this point, assume that all operations are in mod_{n})

**Addition:** First, the Integer numbers yield -a+b=b-a. We want to see if this holds in **Z**_{n} (the mod_{n} system):
- -a+b=(n-a)+b=b+(n-a)=b+n-a=b-a
- a+(-b)=a+(n-b)=a-b
- a+b = a+b
- (-a)+(-b)=(n-a)+(n-b) = n+n-a-b =-(a-b)

So addition is okay in all 4 cases.

If you are not familiar with mod_{n} notation, basically it is all remainders. 15 mod 4 is the remainder of 15/4, so it is 3. n mod n = 0 since the remainder of n/n is 0. So the remainder of (n+x)/n is the same as the remainder of x/n

**Subtraction:** In the Integers, (-a)-b=-(a+b). We want to see if the four cases hold in **Z**_{n}:
- (-a)-b = (n-a)-b = n-a-b = n-(a+b) = -(a+b).
- (-a)-(-b) = (n-a)-(n-b) = n-a-n+b = b-a
- a-(-b) = a-(n-b) = a-n+b = a+b
- a-b = a-b

**Multiplication:**
- (-a)b = (n-a)b = bn-ba = 0-(ba) = -ba.
- a(-b) = a(n-b) = an-ab = 0-(ab) = -ab=-ba.
- (-a)(-b) = (n-a)(n-b) = n
^{2}-nb-an+ab = n(n-b-a)+ab = 0+ab=ab
- ab=ab

What does this all mean? It means that any addition, subtraction, or multiplication routine that works for unsigned numbers (meaning you are treating them as positives) will work for signed numbers in whatever base you work in. 8-bit math is mod_{256} and 16-bit math is mod_{65536}. Signed division and unsigned division are **not** interchangeable, though.

Does this help at all? I am not sure how familiar you are with any of this terminology, so sorry if non of that made sense .__.