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Proof: By Euclid

(related to Proposition: Prop. 10.004: Greatest Common Measure of Three Commensurable Magnitudes)

• Let $A$, $B$, $C$ be the three given commensurable magnitudes.
• So it is required to find the greatest common measure of $A$, $B$, $C$.

• For let the greatest common measure of the two (magnitudes) $A$ and $B$ have been taken [Prop. 10.3], and let it be $D$.
• So $D$ either measures, or [does] not [measure], $C$.
• Let it, first of all, measure ($C$).
• Therefore, since $D$ measures $C$, and it also measures $A$ and $B$, $D$ thus measures $A$, $B$, $C$.
• Thus, $D$ is a common measure of $A$, $B$, $C$.
• And (it is) clear that (it is) also (the) greatest (common measure).
• For no magnitude larger than $D$ measures (both) $A$ and $B$.
• So let $D$ not measure $C$.
• I say, first, that $C$ and $D$ are commensurable.
• For if $A$, $B$, $C$ are commensurable then some magnitude will measure them which will clearly also measure $A$ and $B$.
• Hence, it will also measure $D$, the greatest common measure of $A$ and $B$ [Prop. 10.3 corr.] .
• And it also measures $C$.
• Hence, the aforementioned magnitude will measure (both) $C$ and $D$.
• Thus, $C$ and $D$ are commensurable [Def. 10.1] .
• Therefore, let their greatest common measure have been taken [Prop. 10.3], and let it be $E$.
• Therefore, since $E$ measures $D$, but $D$ measures (both) $A$ and $B$, $E$ will thus also measure $A$ and $B$.
• And it also measures $C$.
• Thus, $E$ measures $A$, $B$, $C$.
• Thus, $E$ is a common measure of $A$, $B$, $C$.
• So I say that (it is) also (the) greatest (common measure).
• For, if possible, let $F$ be some magnitude greater than $E$, and let it measure $A$, $B$, $C$.
• And since $F$ measures $A$, $B$, $C$, it will thus also measure $A$ and $B$, and will (thus) measure the greatest common measure of $A$ and $B$ [Prop. 10.3 corr.] .
• And $D$ is the greatest common measure of $A$ and $B$.
• Thus, $F$ measures $D$.
• And it also measures $C$.
• Thus, $F$ measures (both) $C$ and $D$.
• Thus, $F$ will also measure the greatest common measure of $C$ and $D$ [Prop. 10.3 corr.] .
• And it is $E$.
• Thus, $F$ will measure $E$, the greater (measuring) the lesser.
• The very thing is impossible.
• Thus, some [magnitude] greater than the magnitude $E$ cannot measure $A$, $B$, $C$.
• Thus, if $D$ does not measure $C$ then $E$ is the greatest common measure of $A$, $B$, $C$.
• And if it does measure ($C$) then $D$ itself (is the greatest common measure).
• Thus, the greatest common measure of three given commensurable magnitudes has been found.
• (Which is) the very thing it was required to show.
  ∎

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"