Prove that $\Bbb{R^3}\setminus\Bbb{Q^3}$ is path connected in the euclidian metric space $\Bbb{E^3}$. I understand the concept of one coordinate moving while the rest don't change, however I can't make up the  exact mapping that would prove this. Can anyone give me the concrete mapping?
 A: Take $(a_1,a_2,a_3),(b_1,b_2,b_3)\in\Bbb R^3\setminus\Bbb Q^3$. Assume, for instance, that $a_1,b_3\notin\Bbb Q$. Then consider$$\begin{array}{rccc}\gamma\colon&[0,2]&\longrightarrow&\Bbb R^3\\&t&\mapsto&\begin{cases}\bigl(a_1,tb_2+(1-t)a_2,tb_3+(1-t)a_3\bigr)&\text{ if }t\in[0,1]\\\bigl((t-1)b_1+(2-t)a_1,b_2,b_3\bigr)&\text{ if }t\in[1,2].\end{cases}\end{array}$$Since $a_1,b_3\notin\Bbb Q$, $(\forall t\in[0,2]):\gamma(t)\in\Bbb R^3\setminus\Bbb Q^3$.
A: Let $(a,b,c)$, $(d,e,f)$ be two points in $\mathbb{R}^3\setminus\mathbb{Q}^3$. Then at least one of $a,b,c$ is not rational. By symmetry, we may assume WLOG that it is $a$.
If $d$ is not rational, we can follow the following path which is entirely contained in  $\mathbb{R}^3\setminus\mathbb{Q}^3$:
$$(a,b,c) \to (a,\sqrt2,c)\to(d,\sqrt2,c)\to(d,e,f)$$
where $P\to Q$ signifies a straight line path from $P$ to $Q$.
If $e$ is not rational, we follow:
$$(a,b,c) \to (a,e,c) \to (d,e,f)$$
If $f$ is not rational, we follow:
$$(a,b,c) \to (a,b,f) \to (d,e,f)$$
