TITLE: The cardinality of $\mathbb{R}/\mathbb Q$
QUESTION [11 upvotes]: How to prove the cardinality of $\mathbb{R}/ \mathbb Q$ is equal to the cardinality of $\mathbb{R}$

REPLY [7 votes]: I don't see what I would have expected to be the "standard answer" to this question, so let me leave it in the hopes it will be helpful to someone.

Proposition: Let $G$ be an infinite group, and let $H$ be a subgroup with $\# H < \# G$.  Then $\# G/H = \# G$.

Proof: Let $\{g_i\}_{i \in G/H}$ be a system of coset representatives for $H$ in $G$: then every element $x$ in $G$ can be written as $x =  g_{i_x} h_x$ for unique $h_x \in H$ and $i_x \in G/H$.  (Note that there is no canonical system of coset representatives: getting one is an archetypical use of the Axiom of Choice.)  Thus we have defined a bijection from $G$ to $G/H \times H$, so $\# G = \# G/H \cdot \# H$.  Since $\# G$ is infinite, so must be at least one of $\# G/H$, $\# H$, and then standard cardinal arithmetic (again AC gets used...) gives that 
$\# G = \# G/H \cdot \# H = \max(\#G/H, \# H)$.  
Since we've assumed $\# H < \# G$, we conclude $\# G = \#G/H$.  
This applies in particular with $G = \mathbb{R}$, $H = \mathbb{Q}$ to give $\# \mathbb{R}/\mathbb{Q} = \# \mathbb{R} = 2^{\aleph_0}$.