Hello everyone. I think i have this right but im' not 100% sure due to the last set, [tex]A_3[/tex] Here is the problem: http://suprfile.com/src/1/3m5fyjg/lastscan.jpg [Broken] Here is my answer: Yes. By the quotient-remainder theorem, every integer n can be represented in exactly one of the three forms. n = 4k or n = 4k+1 or n = 4k+2, for some integer k. This implies that no integer can be in any two of the sets A0, A1, A2, or A3. So A0, A1, A2, and A3 are mutually disjoint. It also implies that every integer is in one of the sets A0, A1, A3, and A4. So [tex]Z = A_0 \bigcup A_1 \bigcup A_2 \bigcup A_3.[/tex] But what makes me question myself is they have [tex]A_3[/tex] such that n = 4k+3, which is just another odd integer....meaning if all integers can be expressed by n = 4k or n = 4k+1 or n = 4k+2, does that mean n = 4k+3 is redudnant and it will contain a duplicated number that n = 4k or n = 4k+1 or n = 4k+2 has already formed? If this is the case then this false because for [tex]Z = A_0 \bigcup A_1 \bigcup A_2 \bigcup A_3.[/tex] to be true, none of the partions are allowed to have duplicated numbers. Any help would be great, thanks!