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Abstract Algebra: Show that 2Z + 5Z = Z

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that [tex]2\mathbb{Z} + 5\mathbb{Z} = \mathbb{Z}[/tex]

    2. Relevant equations

    where 2Z + 5Z = {a+b | a in 2Z and b in 5Z} = Z

    3. The attempt at a solution
    For any n in Z, we can write

    n= (5-4)n = 5n +(-4)n = 5n + 2(-2n)
    And since 5n is in 5Z and 2(-2n) is in 2Z, we can form Z from any combination of elements of 2Z and 5Z.

    What they're asking me to prove makes sense intuitively but I'm not sure how to write it. Thanks for the help.
    Last edited: Apr 13, 2010
  2. jcsd
  3. Apr 13, 2010 #2


    Staff: Mentor

    Maybe I'm looking at this the wrong way, but it doesn't seem true to me. I'm assuming that 2Z = {..., -4, -2, 0, 2, 4, 6, ...} and that 5Z = {..., -10, -5, 0, 5, 10, 15, ...}

    I'm also assuming that + means "union."

    If my assumptions are reasonable, every element of 2Z + 5Z is in Z, but there are a lot of elements in Z that aren't in 2Z + 5Z, such as 3, 7, 9, 11, 13, 17, and so on.
  4. Apr 13, 2010 #3
    The TA said that we're supposed to approach the problem as to show 2Z + 5Z = {a+b | a in 2Z and b in 5Z} = Z. I forgot to paste that into #2. Other than that, you have 2Z and 5Z right. I'm sorry. I was in a rush to get to class and I didn't take my time typing out the post.

    I understand where you're going though. In an earlier problem we were asked to show that 2Z U 5Z is not a subring of Z and I used a counterexample based on your reply to show that.
  5. Apr 13, 2010 #4
    The result is true.

    The ideals of the ring of integers can be characterized very precisely and simply. Find this characterization.

    Now observe that 2 and 5 are relatively prime.
  6. Apr 13, 2010 #5


    Staff: Mentor

    OK, that makes more sense.
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