Proof: By Euclid
(related to Proposition: 7.03: Greatest Common Divisor of Three Numbers)
- For let the greatest common measure, $D$, of the two (numbers) $A$ and $B$ have been taken [Prop. 7.2].
- So $D$ either measures, or does not measure, $C$.
- First of all, let it measure ($C$).
- And it also measures $A$ and $B$.
- Thus, $D$ measures $A$, $B$, and $C$.
- Thus, $D$ is a common measure of $A$, $B$, and $C$.
- So I say that (it is) also the greatest (common measure).
- For if $D$ is not the greatest common measure of $A$, $B$, and $C$ then some number greater than $D$ will measure the numbers $A$, $B$, and $C$.
- Let it (so) measure ($A$, $B$, and $C$), and let it be $E$.
- Therefore, since $E$ measures $A$, $B$, and $C$, it will thus also measure $A$ and $B$.
- Thus, it will also measure the greatest common measure of $A$ and $B$ [Prop. 7.2 corr.] .
- And $D$ is the greatest common measure of $A$ and $B$.
- Thus, $E$ measures $D$, the greater (measuring) the lesser.
- The very thing is impossible.
- Thus, some number which is greater than $D$ cannot measure the numbers $A$, $B$, and $C$.
- Thus, $D$ is the greatest common measure of $A$, $B$, and $C$.
- So let $D$ not measure $C$.
- I say, first of all, that $C$ and $D$ are not prime to one another.
- For since $A$, $B$, $C$ are not prime to one another, some number will measure them.
- So the (number) measuring $A$, $B$, and $C$ will also measure $A$ and $B$, and it will also measure the greatest common measure, $D$, of $A$ and $B$ [Prop. 7.2 corr.] .
- And it also measures $C$.
- Thus, some number will measure the numbers $D$ and $C$.
- Thus, $D$ and $C$ are not prime to one another.
- Therefore, let their greatest common measure, $E$, have been taken [Prop. 7.2].
- And since $E$ measures $D$, and $D$ measures $A$ and $B$, $E$ thus also measures $A$ and $B$.
- And it also measures $C$.
- Thus, $E$ measures $A$, $B$, and $C$.
- Thus, $E$ is a common measure of $A$, $B$, and $C$.
- So I say that (it is) also the greatest (common measure).
- For if $E$ is not the greatest common measure of $A$, $B$, and $C$ then some number greater than $E$ will measure the numbers $A$, $B$, and $C$.
- Let it (so) measure ($A$, $B$, and $C$), and let it be $F$.
- And since $F$ measures $A$, $B$, and $C$, it also measures $A$ and $B$.
- Thus, it will also measure the greatest common measure of $A$ and $B$ [Prop. 7.2 corr.] .
- And $D$ is the greatest common measure of $A$ and $B$.
- Thus, $F$ measures $D$.
- And it also measures $C$.
- Thus, $F$ measures $D$ and $C$.
- Thus, it will also measure the greatest common measure of $D$ and $C$ [Prop. 7.2 corr.] .
- And $E$ is the greatest common measure of $D$ and $C$.
- Thus, $F$ measures $E$, the greater (measuring) the lesser.
- The very thing is impossible.
- Thus, some number which is greater than $E$ does not measure the numbers $A$, $B$, and $C$.
- Thus, $E$ is the greatest common measure of $A$, $B$, and $C$.
- (Which is) the very thing it was required to show.
∎
Thank you to the contributors under CC BY-SA 4.0!
- Github:
-
- non-Github:
- @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"