I read this article (images below) about tiling a rectangle ($x \times 1$, $x$ is irrational), and I can't quite get the part where it says "it's easy to see".
I understand that we need to prove that $v(R)=\sum_{i=1}^n v(Q_{i})$.
But why is it simple that $v(R) = v(B)$?
And why is it simple that $v(Q_{i})=\sum v(B)$?
Thanks,
Sam
Consider all the rectangles formed after you extend the edges of the squares $Q_i$, in the figure below I highlight three of them and label them with their side-lengths
$$ B(a_i,b_j): ~\mbox{rectangle with width} ~a_i~ \mbox{and height} ~b_j $$
What is important to realize here is that
$$ \sum_k a_k = a_R = 1 ~~\mbox{and}~~~ \sum_k b_k = b_R = x \tag{1} $$
With this in mind
\begin{eqnarray} v(R) &=& f(a_R)f(b_R) \\ &\stackrel{(1)}{=}& f\left(\sum_k a_k\right)f\left(\sum_l b_l\right) \\ &\stackrel{f~{\rm is linear}}{=}& \left(\sum_k f(a_k)\right) \left(\sum_l f(b_l)\right) \\ &=& \sum_{k,l} f(a_k)f(b_l) \\ &=& \sum_{k,l} v(B(a_k, b_l)) \tag{2} \end{eqnarray}
Or in the notation of the paper
$$ v(R) = \sum_B v(B) \tag{3} $$
Same argument can be applied to one individual square $Q_i$, but now $B$ only runs on the rectangles spanned by $Q_i$