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 $R$ be a rectangular in the place with edge lengths $a$ and $b$. It is clear that if $a/b$ is rational, then $R$ can be partitioned into edge parallel squares. Prove that the inverse also holds, namely that if $R$ can be partitioned into edge parallel (not necessarily equal)
squares, then $a/b$ is rational.

Sketch of proof. Assume that $R$ is partitioned into squares with sides $x_1, \dots, x_n$ 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 $c_1x_1+ \dots+c_n x_n =a$, or $c_1x_1+\dots + c_n x_n =b$
or $x_i=x_j$. Collecting all these equations, we end up with $Ax= h$, where $x=(x_1, \dots, x_n)$, $A$ has integer entries, and the coordinates of $h$ are $a,b$ or $0$.

Assume that there is a unique solution. Then each $x_i$ can be written as $\alpha_i a +\beta _i b$, for some rational $\alpha_i, \beta_i$ (since $A$ has integer coefficients). Plugging these back to the original equations and subtracting $a$ or $b$ from the right hand side if necessary, we obtain equations of the form $p_j a + q_j b=0$, where $p_j, q_j$ are linear combinations (with integer coefficients) of some $\alpha_i, \beta_i$. If there is some $j$ such that $p_j, q_j \neq 0$, then $a/b= -q_j/p_j$ is rational and we are done.

If $p_j=q_j=0$ for all $j$, then this means that the given partition works for all $R$, regardless the values of $a$ and $b$. On the
other hand, by considering the areas, we have $\sum_{i=1}^n x_i^2 =ab$. The right hand side can be rewritten as a quadratic form in
$a$ and $b$ as $(\sum_{i=1}^n \alpha_i^2)a^2 + (\sum_{i=1}^n \beta_i^2)b^2 + 2 \sum_{i=1}^n \alpha_i \beta_i ab$. But it is
impossible that such a form equals $ab$ for all $a,b >0$. 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.

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4 phản hồi

Hai thằng cu nhà anh có được gen giỏi toán của bố không???

2. chung no choi may tinh rat tai !!!