Issues with Miller indices I have read that to find the intercepts of a plane represented by Miller indices is to take the reciprocal of each index. However, I noticed that does not give the original plane if we had to reduce to smallest integers. 
For example, a plane that intercepts x,y, and z at 1, 2, 3 respectively would have Miller indices  (6,3,2). However the reciprocal of that would not give the intercepts of the original plane? so when does this rule apply?
Another issue is that when a plane intercepts contain fractions. For example, plane that intercepts x,y,and z at (1/2), 2, 3 respectively, according to my book is that we need to just take the reciprocal without reducing to smallest integer so we get (2 1/2 1/3) as our Miller indices. However, this contains fractions and Miller indices should not??
Am I missing something here?
 A: So you are talking about a square lattice but let's do this for arbitrary lattices.
You have lattice basis vectors $\vec v_{1,2,3}$ and you have a dual lattice $\vec u_i =\alpha_i \epsilon_{ijk} \vec v_j\times\vec v_j$ where $\alpha_i$ is chosen to normalize $\vec u_i\cdot\vec v_j =\delta_{ij},$ some insert a factor of $2\pi$ here or so.
We want the plane that intersects at points $\{a\vec v_1, b\vec v_2,c\vec v_3\}$ for some $(a,b,c)$. This means that say $a\vec v_1-c\vec v_3$ and $b\vec v_2-c\vec v_3$ lie in the plane, so a normal vector is given as,
$$\begin{align}
\vec n &= (a\vec v_1-c\vec v_3)\times(b\vec v_2-c\vec v_3)\\
&= ab \vec v_1 \times \vec v_2 + bc \vec v_2 \times \vec v_3 + ac \vec v_3 \times \vec v_1
\end{align}$$
And so the idea is that after dividing by $abc$ we have the normal
$$\frac1{abc}\vec n =\frac1{\alpha_1 a} \vec u_1+\frac1{\alpha_2 b} \vec u_2 + 
\frac1{\alpha_3 c} \vec u_3.$$
Now in the square Cartesian lattice $\alpha_{1,2,3}=1$ and this reduces to its familiar form. But in the arbitrary lattice, it is not so much worse.
Now, notice that it was kind of arbitrary that we divided by $abc,$ we could have divided by half of that or a third of that or double that, whatever is convenient. This leads to an ambiguity that always exists on Miller indices, the normal vector to a plane is only defined within a multiplicative constant. And on the flip side, given the Miller indices $(hk\ell)$, you get a whole family of parallel planes, not just the one that intersects the lattice axes at $1/(\alpha_1 h), 1/(\alpha_2 k),1/(\alpha_3 \ell)$ but also the one that intersects these lattice axes at twice those points, or half those points.
So it's always a parallel plane to the one that you're looking for, but it's not always the exact same plane that you're looking for. Put another way the plane is always the points $x_1\vec v_1+x_2\vec v_2 +x_3\vec v_3$ where $$\alpha_1 h x_1 + \alpha_2 k x_2 + \alpha_3 \ell x_3 = C$$for indeterminate $C$, and you will not be able to determine that $C$ until you find any single point on the plane.
This also tells you that if you want to “integralize” (2 ½ ⅓) the right way is just to multiply by 6 to get (12 3 2). That scale factor was not well defined in the first place, so it is fine to choose it arbitrarily.
