Magnitudes Proportional Compounded are Proportional Separated
Theorem
In the words of Euclid:
- If magnitudes are proportional componendo, they will also be proportional separando.
(The Elements: Book $\text{V}$: Proposition $17$)
That is:
- $a : b = c : d \implies \left({a - b}\right) : b = \left({c - d}\right) : d$
Proof
Let $AB, BE, CD, DF$ be magnitudes which are proportional componendo, so that $AB : BE = CD : DF$.
We need to show that they are proportional separando, that is, that $AE : EB = CF : DF$.
Let equimultiples $GH, HK, LM, MN$ be taken of $AE, EB, CF, FD$.
Let $KO, NP$ be other arbitrary equimultiples of $EB, FD$.
We have that $GH$ is the same multiple of $AE$ that $HK$ is of $EB$.
So from Multiplication of Numbers is Left Distributive over Addition $GH$ is the same multiple of $AE$ that $GK$ is of $AB$.
But $GH$ is the same multiple of $AE$ that $LM$ is of $CF$.
Therefore $GK$ is the same multiple of $AB$ that $LM$ is of $CF$.
Again, we have that $LM$ is the same multiple of $CF$ that $MN$ is of $FD$.
So from Multiplication of Numbers is Left Distributive over Addition $LM$ is the same multiple of $CF$ that $LN$ is of $CD$.
Therefore $GK, LN$ are equimultiples of $AB, CD$.
Again, we have that $HK$ is the same multiple of $EB$ that $MN$ is of $FD$.
Also, $KO$ is the same multiple of $EB$ that $NP$ is of $FD$.
So from Multiplication of Numbers is Right Distributive over Addition the sum $HO$ is also the same multiple of $EB$ that $MP$ is of $FD$.
We also have that $AB : BE = CD : DF$.
Also, we have that $GK, LN$ are equimultiples of $AB, CD$.
Also, we have that $EB, FD$ are equimultiples of $HO, MP$.
Therefore:
- $GK > HO \implies LN > MP$
- $GK = HO \implies LN = MP$
- $GK < HO \implies LN < MP$
Suppose $GK > HO$.
If we subtract $HK$ from each, we have that $GH > KO$.
But $GK > HO \implies LN > MP$.
Therefore $LN > MP$.
So if $MN$ is subtracted from each, we have that $LM > NP$.
So:
- $GH > KO \implies LM > NP$
Similarly we can prove that:
- $GH = KO \implies LM = NP$
- $GH < KO \implies LM < NP$
Also, we have that:
- $GH, LM$ are equimultiples of $AE, CF$
- $KO, NP$ are equimultiples of $EB, FD$
Therefore $AE : EB = CF : FD$.
$\blacksquare$
Historical Note
This proof is Proposition $17$ of Book $\text{V}$ of Euclid's The Elements.
It is the converse of Proposition $18$ of Book $\text{V} $: Magnitudes Proportional Separated are Proportional Compounded.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{V}$. Propositions