# Magnitudes with Rational Ratio are Commensurable

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## 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.

### 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 $B$.

## 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:

whatever part the unit is of $D$, the same part is $C$ of $A$.
$\dfrac C A = \dfrac 1 D$

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:

$\dfrac C F = \dfrac 1 E$
$\dfrac A F = \dfrac D E$

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.