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## 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?