anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Astrophysics
  • Astrophysics
@ganeshie8
Loser66
  • Loser66
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.
anonymous
  • anonymous
The answer sheet tells me that this statement is true... I'm not sure why.

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anonymous
  • anonymous
Wouldn't the integers (mod m) mean { [0], [1], [2],...,[m-1]}.
zepdrix
  • zepdrix
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 :]
anonymous
  • anonymous
I thought that the integers mod m meant the set of all the equivalence classes mod m and not the actual elements themselves.
zepdrix
  • zepdrix
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:\[\large\rm [1]=\{~1,5,9,...~\}\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
Loser66
  • Loser66
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.

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