503. Find the area of a rectangle with sides $2 / 3$ and $3 / 5$.
$\triangle$ Such a rectangle is composed of $2 \times 3$ rectangles of size $1 / 3 \times 1 / 5$, so its area is $6 / 15 = (2 / 3) \cdot (3 / 5). \triangleleft$

![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-154.jpg?height=180&width=183&top_left_y=424&top_left_x=1485)

$\|$ The area of a rectangle with sides $a$ and $b$ is $a b$. In particular, the area of a square with side $a$ is $a^{2}$.

In the previous problem, we essentially proved this for the case when the sides of the rectangle are rational (measured by fractions with integer numerators and denominators). For the general case, this can be proven by enclosing the rectangle between a smaller and a larger rectangle with rational sides.

Note also that this problem explains why the product $a \cdot a$ is called "a squared" or "a square."