# Subtraction of Divisors obeys Distributive Law

## Theorem

In the words of Euclid:

If a number be that part of a number, which a number subtracted is of a number subtracted, the remainder will also be the same part of the remainder that that the whole is of the whole.

In modern algebraic language:

$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a - c = \dfrac 1 n \paren {b - d}$

## Euclid's Proof

Let $AB$ be that aliquot part of the (natural) number $CD$ which $AE$ subtracted is of $CF$ subtracted.

We need to show that the remainder $EB$ is also the same part of the number $CD$ which $AE$ subtracted is of $CF$ subtracted. Whatever part $AE$ is of $CF$, let the same part $EB$ be of $CG$.

Then from Proposition $5$ of Book $\text{VII}$: Divisors obey Distributive Law, whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $AB$ of $GF$.

But whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $AB$ of $CD$, by hypothesis.

Therefore, whatever aliquot part $AB$ is of $GF$, the same aliquot part is it of $CD$ also.

Therefore $GF = CD$.

Let $CF$ be subtracted from each.

Therefore $GC = FD$.

We have that whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $EB$ of $CG$

Therefore whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $EB$ of $FD$.

But whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $AB$ of $CD$.

Therefore the remainder $EB$ is the same aliquot part of the remainder $FD$ that the whole $AB$ is of the whole $CD$.

$\blacksquare$

## Modern Proof

A direct application of the Distributive Property:

 $\ds \frac 1 n b - \frac 1 n d$ $=$ $\ds \frac 1 n b + \frac 1 n \paren {-d}$ $\ds$ $=$ $\ds \frac 1 n \paren {b + \paren {-d} }$ $\ds$ $=$ $\ds \frac 1 n \paren {b - d}$

$\blacksquare$

## Historical Note

This proof is Proposition $7$ of Book $\text{VII}$ of Euclid's The Elements.