# Proof: By Euclid

• For let the two magnitudes $A$ and $B$ have to one another the ratio which the number $D$ (has) to the number $E$.
• I say that the magnitudes $A$ and $B$ are commensurable. • For, as many units as there are in $D$, let $A$ have been divided into so many equal (divisions).

• And let $C$ be equal to one of them.
• And as many units as there are in $E$, let $F$ be the sum of so many magnitudes equal to $C$.
• Therefore, since as many units as there are in $D$, so many magnitudes equal to $C$ are also in $A$, therefore whichever part a unit is of $D$, $C$ is also the same part of $A$.
• Thus, as $C$ is to $A$, so a unit (is) to $D$ [Def. 7.20] .
• And a unit measures the number $D$.
• Thus, $C$ also measures $A$.
• And since as $C$ is to $A$, so a unit (is) to the [number] $D$, thus, inversely, as $A$ (is) to $C$, so the number $D$ (is) to a unit [Prop. 5.7 corr.] .
• Again, since as many units as there are in $E$, so many (magnitudes) equal to $C$ are also in $F$, thus as $C$ is to $F$, so a unit (is) to the [number] $E$ [Def. 7.20] .
• And it was also shown that as $A$ (is) to $C$, so $D$ (is) to a unit.
• Thus, via equality, as $A$ is to $F$, so $D$ (is) to $E$ [Prop. 5.22].
• But, as $D$ (is) to $E$, so $A$ is to $B$.
• And thus as $A$ (is) to $B$, so (it) also is to $F$ [Prop. 5.11].
• Thus, $A$ has the same ratio to each of $B$ and $F$.
• Thus, $B$ is equal to $F$ [Prop. 5.9].
• And $C$ measures $F$.
• Thus, it also measures $B$.
• But, in fact, (it) also (measures) $A$.
• Thus, $C$ measures (both) $A$ and $B$.
• Thus, $A$ is commensurable with $B$ [Def. 10.1] .
• Thus, if two magnitudes ... to one another, and so on ....

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