Question

Prove or Disprove: If m and n are distinct integers such that m,n >= 2 then the integers (mod m) and the integers (mod n) are disjoint.

Answer 1

@ganeshie8

Answer 2

To me, we disprove it, just take n = 2, m = 4, then set of integers =0 mod 4 and set of integers =0 mod 2 are joint.
\(\{2k | k \in \mathbb Z\} \cap \{4l|l \in \mathbb Z\}\neq \emptyset \)
Ex, \(4 \in \) both sets.

Answer 3

The answer sheet tells me that this statement is true... I'm not sure why.

Answer 4

Wouldn't the integers (mod m) mean { [0], [1], [2],...,[m-1]}.

Answer 5

Yes, which is why I don't understand how it could possibly be true.\[\large\rm \mathbb Z_m=\{~0,1,2,3,...,m-1~\}\]\[\large\rm \mathbb Z_n=\{~0,1,2,3,...,n-1~\}\]

So if you take like... m=4 and n=5,\[\large\rm \mathbb Z_4=\{~0,1,2,3~\}\]\[\large\rm \mathbb Z_5=\{~0,1,2,3,4~\}\]Clearly they share some stuff...


I guess maybe I'm not understanding what they mean by disjoint.
Does that refer to the elements? Err blah.. my math brain not working today :]

Answer 6

I thought that the integers mod m meant the set of all the equivalence classes mod m and not the actual elements themselves.

Answer 7

Hmm ya, you're probably right :) Sorry I'm a lil rusty.
So then do we want to show that they share no equivalence class then?
Is that what we're looking for with disjoint perhaps?

Like for \(\large\rm \mathbb Z_5\) we would say that:\[\large\rm [1]=\{~1,6,11,...~\}\]But in \(\large\rm \mathbb Z_4\) we instead have:\[\large\rm [1]=\{~1,5,9,...~\}\]

Answer 8

Something along those lines, I assume they want a way to show that no two equivalence classes are the same for all m and n.

Answer 9

So for Z (mod 5) and Z (mod 4), they do not share any common elements such that there are no two equivalence classes that have the same elements in it.

Answer 10

Since m,n are arbitrary, if you take m = s*n, then pretty sure the equivalent classes are same, like this \([3] (mod 15) \equiv [1] (mod 3)\), clearly, 3 is not 15.