Proof: By Euclid

Modern Formulation

(not yet contributed)

• For if $A$ and $B$ are commensurable (magnitudes) then some magnitude will measure them.
• Let it (so) measure (them), and let it be $C$.
• And as many times as $C$ measures $A$, so many units let there be in $D$.
• And as many times as $C$ measures $B$, so many units let there be in $E$.
• Therefore, since $C$ measures $A$ according to the units in $D$, and a unit also measures $D$ according to the units in it, a unit thus measures the number $D$ as many times as the magnitude $C$ (measures) $A$.
• Thus, as $C$ is to $A$, so a unit (is) to $D$1 [Def. 7.20] .
• Thus, inversely, as $A$ (is) to $C$, so $D$ (is) to a unit [Prop. 5.7 corr.] .
• Again, since $C$ measures $B$ according to the units in $E$, and a unit also measures $E$ according to the units in it, a unit thus measures $E$ the same number of times that $C$ (measures) $B$.
• Thus, as $C$ is to $B$, so a unit (is) to $E$ [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 $B$, so the number $D$ (is) to the (number) $E$ [Prop. 5.22].
• Thus, the commensurable magnitudes $A$ and $B$ have to one another the ratio which the number $D$ (has) to the number $E$.
• (Which is) the very thing it was required to show.

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