◀ ▲ ▶Branches / Geometry / Elements-euclid / Book--9-number-theory-applications / Proof: By Euclid

Proof: By Euclid

(related to Proposition: Prop. 9.12: Elements of Geometric Progression from One Divisible by Prime)

• Let any multitude whatsoever of numbers, $A$, $B$, $C$, $D$, be in (continued) proportion, (starting) from a unit.
• I say that however many prime numbers $D$ is measured by, $A$ will also be measured by the same (prime numbers).

• For let $D$ be measured by some prime number $E$.
• I say that $E$ measures $A$.
• For (suppose it does) not.
• $E$ is prime, and every prime number is prime to every number which it does not measure [Prop. 7.29].
• Thus, $E$ and $A$ are prime to one another.
• And since $E$ measures $D$, let it measure it according to $F$.
• Thus, $E$ has made $D$ (by) multiplying $F$.
• Again, since $A$ measures $D$ according to the units in $C$ [Prop. 9.11 corr.] , $A$ has thus made $D$ (by) multiplying $C$.
• But, in fact, $E$ has also made $D$ (by) multiplying $F$.
• Thus, the (number created) from (multiplying) $A$, $C$ is equal to the (number created) from (multiplying) $E$, $F$.
• Thus, as $A$ is to $E$, (so) $F$ (is) to $C$ [Prop. 7.19].
• And $A$ and $E$ (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], 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].
• Thus, $E$ measures $C$.
• Let it measure it according to $G$.
• Thus, $E$ has made $C$ (by) multiplying $G$.
• But, in fact, via the (proposition) before this, $A$ has also made $C$ (by) multiplying $B$ [Prop. 9.11 corr.] .
• Thus, the (number created) from (multiplying) $A$, $B$ is equal to the (number created) from (multiplying) $E$, $G$.
• Thus, as $A$ is to $E$, (so) $G$ (is) to $B$ [Prop. 7.19].
• And $A$ and $E$ (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], 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].
• Thus, $E$ measures $B$.
• Let it measure it according to $H$.
• Thus, $E$ has made $B$ (by) multiplying $H$.
• But, in fact, $A$ has also made $B$ (by) multiplying itself [Prop. 9.8].
• Thus, the (number created) from (multiplying) $E$, $H$ is equal to the (square) on $A$.
• Thus, as $E$ is to $A$, (so) $A$ (is) to $H$ [Prop. 7.19].
• And $A$ and $E$ are prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], 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].
• Thus, $E$ measures $A$, as the leading (measuring the) leading.
• But, in fact, ($E$) also does not measure ($A$).
• The very thing (is) impossible.
• Thus, $E$ and $A$ are not prime to one another.
• Thus, (they are) composite (to one another).
• And (numbers) composite (to one another) are (both) measured by some [prime] number [Def. 7.14] .
• And since $E$ is assumed (to be) prime, and a prime (number) is not measured by another number (other) than itself [Def. 7.11] , $E$ thus measures (both) $A$ and $E$.
• Hence, $E$ measures $A$.
• And it also measures $D$.
• Thus, $E$ measures (both) $A$ and $D$.
• So, similarly, we can show that however many prime numbers $D$ is measured by, $A$ will also be measured by the same (prime numbers).
• (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"