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

[Edellaine]

Homework Statement

Show that 2\mathbb{Z} + 5\mathbb{Z} = \mathbb{Z}

Homework Equations

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

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.

 

[Mark44]

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.

 

[Edellaine]

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.

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.

 

[zpconn]

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.

 

[Mark44]

OK, that makes more sense.