722. Find the common measure of segments of length $\frac{4}{15}$ and $\frac{8}{21}$.
$\triangleright$ We can take a segment of length $\frac{1}{15 \cdot 21}=\frac{1}{315}$ as a common measure. It is clear that

$$
\frac{4}{15}=\frac{4 \cdot 21}{15 \cdot 21} \quad \text { and } \quad \frac{8}{21}=\frac{15 \cdot 8}{15 \cdot 21}
$$

so the proposed segment fits $4 \cdot 21$ times in the first segment and $15 \cdot 8$ times in the second. $\triangleleft$

723 (Continuation). Are there longer common measures?

$\triangleright$ Yes. For example, we can notice that $15=3 \cdot 5$ and $21=3 \cdot 7$, so the fractions can be rewritten as $\frac{4}{15}=\frac{4 \cdot 7}{3 \cdot 5 \cdot 7}$ and $\frac{8}{21}=\frac{8 \cdot 5}{3 \cdot 5 \cdot 7}$, so a segment of length
$\frac{1}{3 \cdot 5 \cdot 7}=\frac{1}{105}$ will be a common measure. It is three times longer than the previously proposed one and fits $4 \cdot 7=28$ times in the first segment and $8 \cdot 5=40$ times in the second. $\triangleleft$

724 (Continuation). Is there an even longer common measure?

Yes. Notice that both numbers 28 and 40 are multiples of 4, so if we take a segment four times longer, it will fit 7 times in the first and 10 times in the second. $\triangleleft$

Not any two segments have a common measure. This is related to the existence of irrational numbers (see the discussion on p. 159). For example, the side of a square and its diagonal do not have a common measure. Indeed, if in some units of measurement the square had an integer side $n$ and an integer diagonal $m$, then $m^{2}=n^{2}+n^{2}=2 n^{2}$ by the Pythagorean theorem, and $(m / n)^{2}=2$, but no fraction with integer numerator and denominator gives 2 when squared. Another, geometric proof of the incommensurability of the side and diagonal of a square we will discuss in this section after the Euclidean algorithm - a method for finding a common measure of two segments (if it exists).

Segments that have a common measure are called commensurable.