Multiplying euclidean vector with ket vector The commutator of position operator $\mathbf x$ with the translation operator defined as $\mathscr{J}\left(d \mathbf{x}^{\prime}\right)\left|\mathbf{x}^{\prime}\right\rangle=\left|\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right\rangle$ is  $\mathbf {dx'}$ where sakurai says "$\mathbf {dx'}$ is a number multiplied by a identity operator"
So:
$[\mathbf x, \mathscr{J( d\mathbf x'}]= d\mathbf x' \mathbf 1 $
Then
$[\mathbf x, \mathscr{J( \mathbf dx'}] |u \rangle
    = \mathbf{dx'}\mathbf 1|u \rangle =\mathbf {dx'}$ $| u\rangle$
Isn't this number $\mathbf {dx'}$ actually a euclidean vector? And we are multiplying a euclidean vector with a ket vector, which I've not seen how that's defined.
What does $\mathbf {dx'}$ $|u\rangle$ mean then?
 A: The expression
$$[\hat{\mathbf x}, \mathscr T(\mathrm d\mathbf x)] = \mathrm d\mathbf x\  \mathbb I$$
is shorthand for the three equations
$$[\hat x,\mathscr T(\mathrm d\mathbf x) ] = \mathrm dx \ \mathbb I\qquad 
[\hat y,\mathscr T(\mathrm d\mathbf x)] = \mathrm dy\ \mathbb I\qquad 
[\hat z,\mathscr T(\mathrm d\mathbf x)] = \mathrm dz \ \mathbb I$$
or, in index form, $[\hat x_i ,\mathscr T(\mathrm d\mathbf x)] = \mathrm dx_i$.  This can be proven straightforwardly:
$$[\hat x_i ,\mathscr T(\mathrm d\mathbf x)] |\mathbf x\rangle = \hat x_i \mathscr T(\mathrm d\mathbf x)|\mathbf x\rangle - \mathscr T(\mathrm d\mathbf x)\hat x_i |\mathbf x\rangle = (x_i + \mathrm dx_i)|\mathbf x+\mathrm d\mathbf x\rangle - x_i |\mathbf x + \mathrm d\mathbf x\rangle$$
$$= \mathrm dx_i |\mathbf x+\mathrm d\mathbf x\rangle = \mathrm dx_i |\mathbf x\rangle + \mathscr O(|\mathrm d\mathbf x|^2)$$
$$\implies [\hat x_i, \mathscr T(\mathrm d\mathbf x)] = \mathrm dx_i \ \mathbb I + \mathscr O(|\mathrm d\mathbf x|^2)$$

Typically we don't define the multiplication of kets by vectors, so when you see an expression like this it should be interpreted as shorthand (though we should also remember that vector operators have well-defined transformation properties under rotations).
