In the original question, I'm asked to prove that the quotient group $\mathbb{R}/\mathbb{Q}$ is torsion-free.
According to the book that this question is from the definition of the torsion-free group is: "An abelian group where all its members except the identity element have infinite order is called torsion-free group."
So the obvious strategy for the proof is showing that all members have infinite order or assuming that this group have an element with finite order and getting a contradiction.
But I have trouble understating what this group "looks" like and what its members are like.
Is there some visual representation of this group or relatively simple description that will help me understand this quotient group?
Best Answer
The group $\mathbb{R}$ is a vector space over $\mathbb{Q}$ and $\mathbb{Q}$ is a subspace thereof.
Every vector space over $\mathbb{Q}$ is torsion-free.
By the way (assuming the axiom of choice), $\mathbb{R}/\mathbb{Q}\cong\mathbb{R}$, because the two vector spaces have the same dimension over $\mathbb{Q}$.
Direct proof
Suppose $x+\mathbb{Q}$ is torsion. This means that $nx\in\mathbb{Q}$, for some integer $n>0$. Thus $x\in\mathbb{Q}$.