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Given two line segments $\overline{AB}$ and $\overline{CD}$, it's always possible to find a third line segment whose length divides evenly into the first two. In modern terminology, if we assign $x = \overline{AB}$ and $y = \overline{CD}$, than the above statement is equivalent to asserting that $x = ay$, where $a \in \mathbb{Q}$.

I'm having difficulty understanding why these two statements are equivalent, mainly because I find the phrasing of the first sentence confusing. If $\overline{CD}$ is a rational multiple of $\overline{AB}$, then why can we always find a third line segment that "evenly" divides into them?

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If $a = \frac p q$, then $x$ and $y$ are both integral multiples of $\frac y q$ by multplying by $p$ and by $q$ respectively. So here, $\frac p q$ is the length of the third length.

This is what meant by “evenly dividing”: You can use the third length to measure both other lengths, it introduces a suitable measure, just like inches measure feet and feet measure yards.

The statement is essentially that the lengths are commensurable and is equivalent to saying that the euclidian algorithm for these lengths works: The third line segment is literally their greatest common divisor, i.e. the greatest length which measures both of the others.


And also, it is not always possible to find such a third length – the diagonal of a square is never commensurable with its sides which was exactly the unexpected discovery that unsettled the Pythagoreans back in their days. See here or here.

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  • $\begingroup$ Good answer, thanks I understand now. $\endgroup$ – St Vincent Apr 18 '14 at 7:06

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