Pick a natural number m>0. As discussed in class, every natural number n can be uniquely expanded as n=b*m+r, where b is an integer and
. Consider the relation 
with domain 
, range [0,m-1] and elements (n,r) where r is the remainder of n mod m. Notice that 
is an equivalence relation, for which each of the equivalence classes
consists of numbers congruent to each other mod m,i.e. for 
consists of all numbers with remainder r when divided by m. These equivalence classes form a partition into m pieces; call this partition 
(the upside down capital pi means disjoint union). Notice that that problems 17 and 18 in section 4.6 of the textbook, discuss a way of getting a new partition 
from two old ones P,Q : the new partition has sets that are non-null intersections of the pieces of the old one. Given two integers x and y, we say that the partition P separates x and y, if [x]≠[y]. For example x and y are separated mod 2 if one is even and the other odd, but not separated if they are both even or both odd.2) To prove:
a)
(Note: your classmate Chelsey Anderson points out that this problem can be solved using the Chinese Remainder theorem)b) If
separates x and y then so does 
for every integer k>0
.
, then x,y are separated in every with 
d) if
then x,y are separated in for the least m such that 
