I've tried to show that:
$$[0,1]\sim([0,1] ∩R-Q)$$
I know from this answer :
$$[0,1]\sim R-Q$$
But how to construct a bijection between R-Q and $([0,1]∩R-Q)$ ?
I think the function would be like $f:R-Q→[0,1]∩R-Q$:
$$f(x) = \cases{ 1/x & \text{if $ ~x∈(R-Q)-[0,1]$} \\ x & \text{if $~x ∈ (0,1)$} }$$
But I think this function is not complete ... could someone help me please to improve this?