Lát sàn (Tiling)
A favorite topic of mine is linear algebra. It keeps surprising me with wonderful and useful facts. Among others, I found this gem recently:
Theorem. Let be a rectangular in the place with edge lengths and . It is clear that if is rational, then can be partitioned into edge parallel squares. Prove that the inverse also holds, namely that if can be partitioned into edge parallel (not necessarily equal)
squares, then is rational.
Sketch of proof. Assume that is partitioned into squares with sides in a particular way. Looking at the sides and the positions of
the squares, we can write up a system of linear equations, each of which has the form either , or
or . Collecting all these equations, we end up with , where , has integer entries, and the coordinates of are or .
Assume that there is a unique solution. Then each can be written as , for some rational (since has integer coefficients). Plugging these back to the original equations and subtracting or from the right hand side if necessary, we obtain equations of the form , where are linear combinations (with integer coefficients) of some . If there is some such that , then is rational and we are done.
If for all , then this means that the given partition works for all , regardless the values of and . On the
other hand, by considering the areas, we have . The right hand side can be rewritten as a quadratic form in
and as . But it is
impossible that such a form equals for all . A contradiction.
To conclude the proof, it suffices to show that there is a unique solution. We leave it as an exercise. For more about this theorem and other interesting results, I highly recommend
PROBLEMS AND THEOREMS IN LINEAR ALGEBRA by V. Prasolov.