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

Proof: By Euclid

(related to Proposition: Prop. 9.13: Divisibility of Elements of Geometric Progression from One where First Element is Prime)

• Let any multitude whatsoever of numbers, $A$, $B$, $C$, $D$, be in continued proportion, (starting) from a unit.
• And let the (number) after the unit, $A$, be prime.
• I say that the greatest of them, $D$, will be measured by no other (numbers) except $A$, $B$, $C$.

• For, if possible, let it be measured by $E$, and let $E$ not be the same as one of $A$, $B$, $C$.
• So it is clear that $E$ is not prime.
• For if $E$ is prime, and measures $D$, then it will also measure $A$, (despite $A$) being prime (and) not being the same as it [Prop. 9.12].
• The very thing is impossible.
• Thus, $E$ is not prime.
• Thus, (it is) composite.
• And every composite number is measured by some prime number [Prop. 7.31].
• Thus, $E$ is measured by some prime number.
• So I say that it will be measured by no other prime number than $A$.
• For if $E$ is measured by another (prime number), and $E$ measures $D$, then this (prime number) will thus also measure $D$.
• Hence, it will also measure $A$, (despite $A$) being prime (and) not being the same as it [Prop. 9.12].
• The very thing is impossible.
• Thus, $A$ measures $E$.
• And since $E$ measures $D$, let it measure it according to $F$.
• I say that $F$ is not the same as one of $A$, $B$, $C$.
• For if $F$ is the same as one of $A$, $B$, $C$, and measures $D$ according to $E$, then one of $A$, $B$, $C$ thus also measures $D$ according to $E$.
• But one of $A$, $B$, $C$ (only) measures $D$ according to some (one) of $A$, $B$, $C$ [Prop. 9.11].
• And thus $E$ is the same as one of $A$, $B$, $C$.
• The very opposite thing was assumed.
• Thus, $F$ is not the same as one of $A$, $B$, $C$.
• Similarly, we can show that $F$ is measured by $A$, (by) again showing that $F$ is not prime.
• For if ($F$ is prime), and measures $D$, then it will also measure $A$, (despite $A$) being prime (and) not being the same as it [Prop. 9.12].
• The very thing is impossible.
• Thus, $F$ is not prime.
• Thus, (it is) composite.
• And every composite number is measured by some prime number [Prop. 7.31].
• Thus, $F$ is measured by some prime number.
• So I say that it will be measured by no other prime number than $A$.
• For if some other prime (number) measures $F$, and $F$ measures $D$, then this (prime number) will thus also measure $D$.
• Hence, it will also measure $A$, (despite $A$) being prime (and) not being the same as it [Prop. 9.12].
• The very thing is impossible.
• Thus, $A$ measures $F$.
• And since $E$ measures $D$ according to $F$, $E$ has thus made $D$ (by) multiplying $F$.
• But, in fact, $A$ has also made $D$ (by) multiplying $C$ [Prop. 9.11 corr.] .
• Thus, the (number created) from (multiplying) $A$, $C$ is equal to the (number created) from (multiplying) $E$, $F$.
• Thus, proportionally, as $A$ is to $E$, so $F$ (is) to $C$ [Prop. 7.19].
• And $A$ measures $E$.
• Thus, $F$ also measures $C$.
• Let it measure it according to $G$.
• So, similarly, we can show that $G$ is not the same as one of $A$, $B$, and that it is measured by $A$.
• And since $F$ measures $C$ according to $G$, $F$ has thus made $C$ (by) multiplying $G$.
• But, in fact, $A$ 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) $F$, $G$.
• Thus, proportionally, as $A$ (is) to $F$, so $G$ (is) to $B$ [Prop. 7.19].
• And $A$ measures $F$.
• Thus, $G$ also measures $B$.
• Let it measure it according to $H$.
• So, similarly, we can show that $H$ is not the same as $A$.
• And since $G$ measures $B$ according to $H$, $G$ has thus 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) $H$, $G$ is equal to the square on $A$.
• Thus, as $H$ is to $A$, (so) $A$ (is) to $G$ [Prop. 7.19].
• And $A$ measures $G$.
• Thus, $H$ also measures $A$, (despite $A$) being prime (and) not being the same as it.
• The very thing (is) absurd.
• Thus, the greatest (number) $D$ cannot be measured by another (number) except (one of) $A$, $B$, $C$.
• (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"