Magnitudes with Rational Ratio are Commensurable
Theorem
In the words of Euclid:
- If two magnitudes have to one another the ratio which a number has to a number, the magnitudes will be commensurable.
(The Elements: Book $\text{X}$: Proposition $6$)
Porism
In the words of Euclid:
- From this it is manifest that, if there be two numbers, as $D$, $E$, and a straight line, as $A$, it is possible to make a straight line [$F$] such that the given straight line is to it as the number $D$ to the number $E$.
And, if a mean proportional be also taken between $A$, $F$, as $B$,
as $A$ is to $F$, so will the square on $A$ be to the square on $B$, that is, as the first is to the third, so is the figure on the first to that which is similar and similarly described on the second.
But, as $A$ is to $F$, so is the number $D$ to the number $E$; therefore it has been contrived that, as the number $D$ is to the number $E$, so also is the figure on the straight line $A$ to the figure on the straight line $B$.
(The Elements: Book $\text{X}$: Proposition $6$ : Porism)
Proof
Let $A$ and $B$ be magnitudes which have to one another the ratio which the number $D$ has to the number $E$.
Let $A$ be divided into as many equal parts as there are units in $D$.
Let $C$ be equal to one of those parts.
Let $F$ be made up of as many magnitudes equal to $C$ as there are units in $E$.
Since:
- there are in $A$ as many magnitudes equal to $C$ as there are units in $D$
then:
From Ratios of Fractions in Lowest Terms:
- $\dfrac C A = \dfrac 1 D$
and so from the porism to Ratios of Equal Magnitudes:
- $\dfrac A C = \dfrac D 1 = D$
Since:
- there are in $F$ as many magnitudes equal to $C$ as there are units in $E$
then from Ratios of Fractions in Lowest Terms:
- $\dfrac C F = \dfrac 1 E$
From Equality of Ratios Ex Aequali:
- $\dfrac A F = \dfrac D E$
But:
- $\dfrac D E = \dfrac A B$
and so from Equality of Ratios is Transitive:
- $\dfrac A B = \dfrac A F$
Therefore $A$ has the same ratio to each of the magnitudes $B$ and $F$.
Therefore from Magnitudes with Same Ratios are Equal:
- $B = F$
But $C$ measures $F$.
Therefore $C$ also measures $B$.
Further, $C$ also measures $A$.
Therefore $C$ measures both $A$ and $B$.
Therefore $A$ is commensurable with $B$.
$\blacksquare$
Historical Note
This proof is Proposition $6$ of Book $\text{X}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions