Proof: By Euclid

For let the least number, $G$, measured by (both) $B$ and $C$ have be taken [Prop. 7.34].
And as many times as $B$ measures $G$, so many times let $A$ also measure $H$.
And as many times as $C$ measures $G$, so many times let $D$ also measure $K$.
And $E$ either measures, or does not measure, $K$.
Let it, first of all, measure ($K$).
And as many times as $E$ measures $K$, so many times let $F$ also measure $L$.
And since $A$ measures $H$ the same number of times that $B$ also (measures) $G$, thus as $A$ is to $B$, so $H$ (is) to $G$ [Def. 7.20] , [Prop. 7.13].
And so, for the same (reasons), as $C$ (is) to $D$, so $G$ (is) to $K$, and, further, as $E$ (is) to $F$, so $K$ (is) to $L$.
Thus, $H$, $G$, $K$, $L$ are in continued proportion in the ratio of $A$ to $B$, and of $C$ to $D$, and, further, of $E$ to $F$.
So I say that (they are) also the least (numbers in continued proportion in these ratios).
For if $H$, $G$, $K$, $L$ are not the least numbers in continued proportion in the ratios of $A$ to $B$, and of $C$ to $D$, and of $E$ to $F$, let $N$, $O$, $M$, $P$ be (the least such numbers).
And since as $A$ is to $B$, so $N$ (is) to $O$, and $A$ and $B$ are the least (numbers which have the same ratio as them), and the least (numbers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser - that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20], $B$ thus measures $O$.
So, for the same (reasons), $C$ also measures $O$.
Thus, $B$ and $C$ (both) measure $O$.
Thus, the least number measured by (both) $B$ and $C$ will also measure $O$ [Prop. 7.35].
And $G$ (is) the least number measured by (both) $B$ and $C$.
Thus, $G$ measures $O$, the greater (measuring) the lesser.
The very thing is impossible.
Thus, there cannot be any numbers less than $H$, $G$, $K$, $L$ (which are) continuously (proportional) in the ratio of $A$ to $B$, and of $C$ to $D$, and, further, of $E$ to $F$.
So let $E$ not measure $K$.
And let the least number, $M$, measured by (both) $E$ and $K$ have been taken [Prop. 7.34].
And as many times as $K$ measures $M$, so many times let $H$, $G$ also measure $N$, $O$, respectively.
And as many times as $E$ measures $M$, so many times let $F$ also measure $P$.
Since $H$ measures $N$ the same number of times as $G$ (measures) $O$, thus as $H$ is to $G$, so $N$ (is) to $O$ [Def. 7.20] , [Prop. 7.13].
And as $H$ (is) to $G$, so $A$ (is) to $B$.
And thus as $A$ (is) to $B$, so $N$ (is) to $O$.
And so, for the same (reasons), as $C$ (is) to $D$, so $O$ (is) to $M$.
Again, since $E$ measures $M$ the same number of times as $F$ (measures) $P$, thus as $E$ is to $F$, so $M$ (is) to $P$ [Def. 7.20] , [Prop. 7.13].
Thus, $N$, $O$, $M$, $P$ are in continued proportion in the ratios of $A$ to $B$, and of $C$ to $D$, and, further, of $E$ to $F$.
So I say that (they are) also the least (numbers) in the ratios of $A$ $B$, $C$ $D$, $E$ $F$.
For if not, then there will be some numbers less than $N$, $O$, $M$, $P$ (which are) in continued proportion in the ratios of $A$ $B$, $C$ $D$, $E$ $F$.
Let them be $Q$, $R$, $S$, $T$.
And since as $Q$ is to $R$, so $A$ (is) to $B$, and $A$ and $B$ (are) the least (numbers having the same ratio as them), and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20], $B$ thus measures $R$.
So, for the same (reasons), $C$ also measures $R$.
Thus, $B$ and $C$ (both) measure $R$.
Thus, the least (number) measured by (both) $B$ and $C$ will also measure $R$ [Prop. 7.35].
And $G$ is the least number measured by (both) $B$ and $C$.
Thus, $G$ measures $R$.
And as $G$ is to $R$, so $K$ (is) to $S$.
Thus, $K$ also measures $S$ [Def. 7.20] .
And $E$ also measures $S$ [Prop. 7.20].
Thus, $E$ and $K$ (both) measure $S$.
Thus, the least (number) measured by (both) $E$ and $K$ will also measure $S$ [Prop. 7.35].
And $M$ is the least (number) measured by (both) $E$ and $K$.
Thus, $M$ measures $S$, the greater (measuring) the lesser.
The very thing is impossible.
Thus there cannot be any numbers less than $N$, $O$, $M$, $P$ (which are) in continued proportion in the ratios of $A$ to $B$, and of $C$ to $D$, and, further, of $E$ to $F$.
Thus, $N$, $O$, $M$, $P$ are the least (numbers) in continued proportion in the ratios of $A$ $B$, $C$ $D$, $E$ $F$.
(Which is) the very thing it was required to show.
∎

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References

Adapted from (subject to copyright, with kind permission)

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

◀ ▲ ▶Branches / Geometry / Elements-euclid / Book--8-continued-proportion / Proof: By Euclid

Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)