Area of the lattice generated from $(n, n\sqrt{2} \mod 1)$ I plotted $\Big\{ (n, n \sqrt{2} \, \mathrm{mod} \,1) \;\Big|  -50 \leq n \leq 50 \Big\}$ and even though the $n \sqrt{2}$ is a line, the pattern that emerges is a lattice.  What is the basis of this lattice?  And the area of the fundamental region?  
While these things can be numerically computed.  I would like to have a systematic way of doing this.
The basis seems to be $\big\{(5, 5\sqrt{2}-7), (-7, -7\sqrt{2} + 10) \big\}$.  The area generated by this lattice should be:
$$ \left|\begin{array}{rr} 5 & -7 \\ 5 \sqrt{2} - 7 & - 7 \sqrt{2} + 10\end{array} \right| = (5)( - 7 \sqrt{2} + 10) - (-7)(5 \sqrt{2} - 7)
= 5\cdot 10 - (-7)(-7) = 1$$
Is the area of the fundamental parallelogram of this lattice always $1$?  Even if $\alpha = \sqrt{3}$ or another number?

 A: For now I'm going to answer only the question regarding this particular example: you have correctly identified a fundamental domain, and its area is indeed 1. But this needs to be formulated as an actual mathematical statement using lattices in Lie groups. 
Let me do this formulation and give the proof, in a few steps.
Step 1: We may consider $\Gamma = \{(n,n \sqrt{2} \,\text{mod}\, 1) \bigm| n \in \mathbb{Z}\}$ as a lattice in the Lie group $\mathbb{R} \times S^1$, where "$n \sqrt{2} \,\text{mod}\, 1$" stands for $\exp(2 \pi i \sqrt{2}) \in S^1$; and where by a "lattice" I mean a discrete cocompact subgroup. The Lie group $\mathbb{R} \times S^1$ is, from a topological perspective, a bi-infinite cylinder. The finite subcylinder $[0,1] \times S^1$ is a fundamental domain for the lattice $\Gamma$. If we give $\mathbb{R}$ its ordinary Lebesgue measure and $S^1$ the measure $\text{radians}/2\pi$, then the area of the fundamental domain $[0,1] \times S^1$ equals~$1$.
Step 2: Consider the universal covering map/homomorphism $\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times S^1$ defined by $(x,y) \mapsto (x,\exp(2 \pi i y))$. Let $\widetilde \Gamma \subset \mathbb{R} \times \mathbb{R}$ be the inverse image of $\Gamma$ under this homomorphism. Then $\widetilde\Gamma$ is the lattice in the same sense, a discrete cocompact subgroup of $\mathbb{R} \times \mathbb{R}$. Furthermore, $\widetilde\Gamma$ has, as a basis, the vectors $(0,1)$, $(1,\sqrt{2})$. The picture you have drawn is the intersection 
$$\widetilde\Gamma \cap [-50,50] \times [0,1]
$$
The lattice $\widetilde\Gamma$ has, as a fundamental domain of area~$1$, the parallelogram spanned with vertices $(0,0)$, $(0,1)$, $(1,\sqrt{2})$, $(1,1+\sqrt{2})$.
So, what this comes down to is to demonstrate that the lattice $\widetilde\Gamma$ has a parallelogram shaped fundamental domain of area $1$ completely contained in $[-50,50] \times [0,1]$.
Step 3:
There is an isomorphism $\widetilde\Gamma \approx \mathbb{Z}^2$ whose inverse is given by the linear transformation
$$T(m,n) = m(1,\sqrt{2}) + n(0,1)
$$
and this linear transformation has determinant $1$.
Also, any pair of vectors in $\mathbb{Z}^2$ which form a $2 \times 2$ matrix of determinant $\pm 1$ forms a basis for $\mathbb{Z}^2$.
Combining these, any pair of vectors in $\widetilde\Gamma$ which form a $2 \times 2$ matrix of determinant $\pm 1$ forms a basis for $\widetilde\Gamma$. 
And that's exactly what your calculations show for the vectors $(5, 5\sqrt{2}-7), (-7, -7\sqrt{2} + 10)$.
