7.4. A square was divided into two rectangles. It turned out that the perimeters of both rectangles are integers. Can we conclude that the perimeter of the original square is also an integer?
7.4. A square was cut into two rectangles. It turned out that the perimeters of both rectangles are integers. Can we conclude that the perimeter of the original square is also an integer?

Answer: no, we cannot. Hint: Let's provide such an example: a square with a side of 5/6 is cut into two rectangles of length 5/6 and widths 1/6 and 4/6. Then the perimeters of the rectangles are $2\left(\frac{1}{6}+\frac{5}{6}\right)=2$ and $2\left(\frac{4}{6}+\frac{5}{6}\right)=3$. Comment. We can arrive at a similar example as follows: let $a$ be the side of the square, $x$ be the width of one of the rectangles. Then $(a-x)$ is the width of the other rectangle, and their perimeters are $2(a+x)$ and $2(2a-x)$. The sum of these perimeters is $6a$, so $a$ has the form $a=\frac{n}{6}$, where $n$ is a natural number. Therefore, $a+x=\frac{m}{2}$ ($m$ is a natural number). For $n=5, m=2$, we get our example.