Product of linear function applied to the two sides of a rectangle is supposed to equate to the sum across its tiles. Background
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. In order to do this, he applies a linear function, $f$ that maps a set $V$ to the set of real numbers, $\Bbb R$. I don't think its important for now what $V$ is, but you can always get the definition from the PDF linked above. This function is to have the following properties:

*

*$f(1)=1$

*$f(x)=-1$, where $x$ is some irrational number greater than zero.

Now for a rectangle $A$ with sides $a$ and $b$, a new function is defined:
$$v(A) = f(a)f(b)$$
Now, if the rectangle $A$ is partitioned into smaller rectangles, $B_i$ that completely tile it, it is claimed that:
$$v(A) = \sum_i v(B_i) \tag{1}$$

Where I'm stuck
Let's consider first that $f(y)=y$. Then, $v(A)$ just becomes the area of the rectangle and equation (1) above obviously holds pretty much by definition (if the smaller rectangles tile the larger one, the sum of their areas should equal that of the larger one).
Now, let's consider $f(y) = \mu y$. This is equivalent to making $v(A)$ the area of the rectangle in some different units. So, equation (1) above still holds by the same area argument.
Finally, we make $f$ as general as possible while still keeping it linear. This gives us: $f(y) = \mu y + \nu$. And at this point, the result seems to no longer hold. Consider a very simple case of a square of side $2$ tiled by four squares each of side $1$ (figure below). Also, consider the function $f(y)=y+1$. For the larger square, we get:
$$v(A) = (2+1)(2+1)=9$$
And for each of the small squares $B_i$ we get:
$$v(B_i) = (1+1)(1+1) = 4$$
And there are four such squares, so $\sum v(B_i) = 16 \neq 9 = v(A)$.
Hence, the result seems to fail to hold. Now, maybe Matousek is only considering functions of the form $f(y)=\mu y$. But the two requirements in the background section won't be satisfied necessarily if we add that additional requirement.


I think maybe you and the author are using two different meaning of the word, "linear".
You are using it in the analytic geometry sense of $f(x)=ax+b$.
The author is using it in the linear algebra sense of $f(x+y)=f(x)+f(y)$, and $f(\alpha x)=\alpha f(x)$ for all vectors $x$ and all scalars $\alpha$. And in this application, we're considering the reals as a vector space over the rationals, so $x$ is real and $\alpha$ is restricted to being rational.
