Subtraction of Multiples of Divisors obeys Distributive Law/Proof 1
Theorem
In the words of Euclid:
- If a number be the same parts of a number that a number subtracted is of a number subtracted, the remainder will also be the same parts of the remainder that the whole is of the whole.
(The Elements: Book $\text{VII}$: Proposition $8$)
In modern algebraic language:
- $a = \dfrac m n b, c = \dfrac m n d \implies a - c = \dfrac m n \paren {b - d}$
Proof
Let the (natural) number $AB$ be the same aliquant part of the (natural) number $CD$ that $AE$ subtracted is of $CF$ subtracted.
We need to show that $EB$ is also the same aliquant part of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
Let $GH = AB$.
Then whatever aliquant part $GH$ is of $CD$, the same aliquant part also is $AE$ of $CF$.
Let $GH$ be divided into the aliquant parts of $CD$, namely $GK + KH$, and $AE$ into the aliquant parts of $CF$, namely $AL + LE$.
Thus the multitude of $GK, KH$ will be equal to the multitude of $AL, LE$.
We have that whatever aliquot part $GK$ is of $CD$, the same aliquot part also is $AL$ of $CF$.
We also have that $CD > CF$.
Therefore $GK > AL$.
Now let $GM = AL$.
Then whatever aliquot part $GK$ is of $CD$, the same aliquot part also is $GM$ of $CF$.
Therefore from Subtraction of Divisors obeys Distributive Law the remainder $MK$ is of the same aliquot part of the remainder $FD$ that the whole $GK$ is of the whole $CD$.
Again, we have that whatever aliquot part $KH$ is of $CD$, the same aliquot part also is $EL$ of $CF$.
We also have that $CD > CF$.
Therefore $HK > EL$.
Let $KN = EL$.
Then whatever aliquot part $KH$ is of $CD$, the same aliquot part also is $KN$ of $CF$.
Therefore from Subtraction of Divisors obeys Distributive Law the remainder $NH$ is of the same aliquot part of the remainder $FD$ that the whole $KH$ is of the whole $CD$.
But $MK$ was proved to be the same aliquot part of $FD$ that $GK$ is of $CD$.
Therefore $MK + NH$ is the same aliquant part of $DF$ that $HG$ is of $CD$.
But $MK + NH = EB$ and $HG = BA$.
Therefore $EB$ is the same aliquant part of $FD$ that $AB$ is of $CD$.
$\blacksquare$
Historical Note
This proof is Proposition $8$ of Book $\text{VII}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VII}$. Propositions