Proving Existence of Integer n for Left Cosets in Q/Z | Homework Solution

  • Thread starter UNChaneul
  • Start date
  • Tags
    Cosets
In summary: Z = ..., -2, -1, 0, 1, 2, ...1/2 + Z = ..., -1 1/2, -1/2, 1/2, 1 1/2, 2 1/2, ...2(1/2 + Z) = ..., -3, -1, 1, 3, 5, ... != Z which makes it seem as though distributivity does not apply when multiplying an additive coset by a constant.And at any rate, it would seem impossible get 0 from 1/2 + Z regardless of what we multiply it by...but 0 is in Z so we have to retrieve that somehow.
  • #1
UNChaneul
18
0

Homework Statement


View [itex]Z[/itex] as a subgroup of the additive group of rational numbers [itex]Q[/itex]. Show that given an element [itex]\bar{x} \in Q/Z[/itex] there exists an integer [itex]n \geq 1[/itex] such that [itex]n \bar{x} = 0[/itex].


Homework Equations





The Attempt at a Solution


As we are working in an additive group, it is clear that any left coset [itex]a+Z \in Q/Z[/itex] is constructed from [itex]a \in Q, 0 \leq a \ll 1[/itex]. This is because otherwise we get a nonzero intersection between cosets, and cosets either have a zero intersection or are completely identical.

Now, the problem I am having understanding is the the existence of an integer [itex]n \geq 1[/itex] such that [itex]n \bar{x} = 0[/itex]. Because, the way I am interpreting it, for any coset in Q/Z there is a positive integer greater to or equal to 1 such that multiplying ANY element in that coset gives 0. I can't seem to picture this being possible, as some cosets will consist only of nonzero rational numbers, and there is no positive integer which when multiplied by any of them would result in 0.

An alternate explanation would be that by 0, they mean 0 in [itex]Q/Z[/itex], which would be [itex]Z[/itex]. Here too I have issues, as [itex]Z[/itex] includes the integer 0, which again leads to the problem mentioned above.

So, what am I misunderstanding about this problem?
 
Physics news on Phys.org
  • #2
The problem you are having is that the number zero is not the zero in the quotient. The additive identity of Q/Z is the coset 0+Z. So 1/2+Z is a cose and if you multiply this coset by 2 you get 1+Z = Z.
 
  • #3
Robert1986 said:
The problem you are having is that the number zero is not the zero in the quotient. The additive identity of Q/Z is the coset 0+Z. So 1/2+Z is a cose and if you multiply this coset by 2 you get 1+Z = Z.

Actually, it's 2Z+1. It's certainly a subset of Z. But it's not really a coset.
 
  • #4
Dick said:
Actually, it's 2Z+1. It's certainly a subset of Z. But it's not really a coset.

I was interpreting 2(1/2 + Z) as (1/2 + Z) + (1/2 + Z) = (1/2 + 1/2) + Z = 1 + Z. When the group is additive, isn't it pretty standard to write 2g for g+g? But the idea is that we add a rational number to itself a certain number of times (ie, multiply it by some integer) and we get an integer, z, so that z+Z = Z which is 0 in the quotient space. So if q is a rational number and q=z_1/z_2 then z_2q = z_1 and z_1 + Z = Z. Also, (z_2)(q+Z) = (q + q + ... + q)+Z = z_1 + Z = Z.
 
  • #5
Robert1986 said:
I was interpreting 2(1/2 + Z) as (1/2 + Z) + (1/2 + Z) = (1/2 + 1/2) + Z = 1 + Z. When the group is additive, isn't it pretty standard to write 2g for g+g? But the idea is that we add a rational number to itself a certain number of times (ie, multiply it by some integer) and we get an integer, z, so that z+Z = Z which is 0 in the quotient space. So if q is a rational number and q=z_1/z_2 then z_2q = z_1 and z_1 + Z = Z. Also, (z_2)(q+Z) = (q + q + ... + q)+Z = z_1 + Z = Z.

Ah, right. g+g. Not 2 times every element of g. Thought I might be missing something...
 
  • #6
Robert1986 said:
I was interpreting 2(1/2 + Z) as (1/2 + Z) + (1/2 + Z) = (1/2 + 1/2) + Z = 1 + Z. When the group is additive, isn't it pretty standard to write 2g for g+g? But the idea is that we add a rational number to itself a certain number of times (ie, multiply it by some integer) and we get an integer, z, so that z+Z = Z which is 0 in the quotient space. So if q is a rational number and q=z_1/z_2 then z_2q = z_1 and z_1 + Z = Z. Also, (z_2)(q+Z) = (q + q + ... + q)+Z = z_1 + Z = Z.

I had thought that, but then it seemed as though multiplication weren't quite distribution with that since

Z = ..., -2, -1, 0, 1, 2, ...
1/2 + Z = ..., -1 1/2, -1/2, 1/2, 1 1/2, 2 1/2, ...
2(1/2 + Z) = ..., -3, -1, 1, 3, 5, ... != Z which makes it seem as though distributivity does not apply when multiplying an additive coset by a constant.

And at any rate it would seem impossible get 0 from 1/2 + Z regardless of what we multiply it by...but 0 is in Z so we have to retrieve that somehow.
 
  • #7
UNChaneul said:
I had thought that, but then it seemed as though multiplication weren't quite distribution with that since

Z = ..., -2, -1, 0, 1, 2, ...
1/2 + Z = ..., -1 1/2, -1/2, 1/2, 1 1/2, 2 1/2, ...
2(1/2 + Z) = ..., -3, -1, 1, 3, 5, ... != Z which makes it seem as though distributivity does not apply when multiplying an additive coset by a constant.

And at any rate it would seem impossible get 0 from 1/2 + Z regardless of what we multiply it by...but 0 is in Z so we have to retrieve that somehow.

I made the same mistake, listen to what Robert1986 is saying. In this problem 2(Z+1/2) means (Z+1/2)+(Z+1/2) NOT 2Z+2(1/2). So yes, it's not a distributive 'multiplication'.
 
  • #8
Dick said:
I made the same mistake, listen to what Robert1986 is saying. In this problem 2(Z+1/2) means (Z+1/2)+(Z+1/2) NOT 2Z+2(1/2). So yes, it's not a distributive 'multiplication'.

Oh, so it's not implying multiplying every element of the set by a constant. Well then that's easy enough to show lol.

I guess I just got confused by what the notation was referring to ~~
 

1. What is the definition of a left coset of Z in Q?

A left coset of Z in Q is a subset of Q, the set of rational numbers, that is formed by multiplying each element of Z, the set of integers, by a fixed element of Q. In other words, it is a set of the form {qz | q∈Q}, where z is an integer.

2. How do you determine the number of left cosets of Z in Q?

The number of left cosets of Z in Q is equal to the number of elements in Q. This is because each element in Q can be multiplied by every element in Z to create a unique left coset. Since Q is an infinite set, there are infinitely many left cosets of Z in Q.

3. Is the set of left cosets of Z in Q a group?

No, the set of left cosets of Z in Q is not a group. In order for a set to be a group, it must satisfy certain properties such as closure, associativity, identity, and inverses. However, the set of left cosets of Z in Q does not satisfy these properties and therefore cannot be considered a group.

4. How are left cosets of Z in Q related to equivalence relations?

Left cosets of Z in Q are closely related to equivalence relations. In fact, the set of left cosets of Z in Q can be seen as the set of equivalence classes of the equivalence relation defined by the subgroup Z in the group Q. Each left coset represents a distinct equivalence class, and all elements within a left coset are considered equivalent.

5. Can left cosets of Z in Q be used to solve problems in number theory?

Yes, left cosets of Z in Q have many applications in number theory. For example, they can be used to prove the fundamental theorem of arithmetic, as well as to find solutions to equations such as the diophantine equation. They also play a significant role in understanding the structure of the rational numbers and their relationship to the integers.

Similar threads

  • Calculus and Beyond Homework Help
Replies
21
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
421
  • Calculus and Beyond Homework Help
Replies
3
Views
503
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
936
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
532
Back
Top