x~y <=> $$ x\quad \equiv \quad y\quad mod\quad m $$
habe daraus erstmal x = y + n m gemacht stimmt das?
reflexiv: x~x => x mod m = x mod m RICHTIG
symmetrie: x~y => y~x
x = y + n m => m = (x-y)/n
y = x + l m => y = x + l ((x-y)/n) | -x | * n
n(y-x) = l(x-y) | : (x-y)
n * 1 = l = n
GILT
transitivität: x~y, y~z => x~z
x = y + n m => y = x-nm
y = z + o m => x = z + o m + m n
=>
x = z + p m
z + o m + m n = z + p m | -z | :m
o+n = p
q.e.d??
