# Magnitudes Proportional Separated are Proportional Compounded

## Theorem

In the words of Euclid:

*If magnitudes be proportional*separando*, they will also be proportional*componendo*.*

(*The Elements*: Book $\text{V}$: Proposition $18$)

That is:

- $a : b = c : d \implies \paren {a + b} : b = \paren {c + d} : d$

## Proof

Let $AE, EB, CF, FD$ be magnitudes which are proportional *separando*, that is:

- $AE : EB = CF : FD$

We need to show that they are also proportional *componendo*, that is:

- $AB : BE = CD : FD$

Suppose $CD : DF \ne AB : BE$.

Then as $AB : BE$ so will $CD$ be to some magnitude less than $DF$ or greater.

Suppose that it be in that ratio to a less magnitude $DG$.

Then since $AB : BE = CD : DG$ it follows from Magnitudes Proportional Compounded are Proportional Separated that:

- $AE : EB = CG : GD$

But by hypothesis:

- $AE : EB = CF : FD$

So by Equality of Ratios is Transitive we have that:

- $CG : GD = CF : FD$

But $CG > CF$.

Therefore $GD > FD$ from Relative Sizes of Components of Ratios.

But it is also less, which is impossible.

Therefore as $AB$ is to $BE$, so is not $CD$ to a lesser magnitude than $FD$.

Similarly we can show that neither is it in that ratio to a greater.

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $18$ of Book $\text{V}$ of Euclid's *The Elements*.

It is the converse of Proposition $17$ of Book $\text{V} $: Magnitudes Proportional Compounded are Proportional Separated.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{V}$. Propositions