I'm trying to follow miniature 12 in Thirty-three Miniatures: Mathematical and Algorithmic Applications of Linear Algebra where Matousek proves that it isn't possible to tile a rectangle the ratio of whose side are irrational with a square. Towards the end of the proof, he defines a linear function, $f(x)$. Consider now a rectangle, $R$ with sides $a$ and $b$. He defines the function $v(R)=f(a)f(b)$. It is then claimed that if a set of smaller rectangles, $Q_i$ (with side $a_i$ and $b_i$) tile the rectangle, we must have:
$$v(R) = \sum_i v(Q_i)$$
And this is because the function $f$ is linear. To me, this seems obviously off. Let's consider that the function $f$ is $f(x)=x$. This is a linear function. Now, the equation above suggests:
$$a b = \sum_i a_i b_i $$
But because of the requirement of tiling, we also have:
$$a=\sum_i a_i$$ and $$b = \sum_i b_i$$
This obviously isn't correct since there will be cross-terms. What am I missing?
Per comments, this was a bad miss on my part. I assumed that for a tiling, we required $a=\sum_i a_i$ and $b = \sum_i b_i$. This is certainly not the case as the simple counter-example below shows. The sum of the sides of the four squares that tile the larger square is 4 while the side length of the larger square is only 2.
However, I'm still stuck on the proof itself at another place which I've asked a different question for: Product of linear function applied to the two sides of a rectangle is supposed to equate to the sum across its tiles.