Multiplication of Real Numbers is Left Distributive over Subtraction/Proof 1

Theorem

In the words of Euclid:

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

(The Elements: Book $\text{V}$: Proposition $5$)

That is, for any numbers $a, b$ and for any integer $m$:

$m a - m b = m \paren {a - b}$

Proof

Let the magnitude $AB$ be the same multiple of the magnitude $CD$ that the part $AE$ subtracted is of the part $CF$ subtracted.

We need to show that the remainder $EB$ is also the same multiple of the remainder $FD$ that the whole $AB$ is of the whole $CD$.

Whatever multiple $AE$ is of $CF$, let $EB$ be made that multiple of $CG$.

We have that $AE$ is the same multiple of $CF$ that $AB$ is of $GC$.

So from Distributive Property, $AE$ is the same multiple of $CF$ that $AB$ is of $GF$.

By by assumption, $AE$ is the same multiple of $CF$ that $AB$ is of $CD$.

Therefore $AB$ is the same multiple of each of the magnitudes $GF, CD$.

Therefore $GF = CD$.

Let $CF$ be subtracted from each.

Then the remainder $GC$ is equal to the remainder $FD$.

Since:

$AE$ is the same multiple of $CF$ that $EB$ is of $GC$

$GC = DF$

it follows that $AE$ is the same multiple of $CF$ that $EB$ is of $CD$.

That is, the remainder $EB$ will be the same multiple of the remainder $FD$ that the whole $AB$ is of the whole $CD$.

$\blacksquare$

Historical Note

This proof is Proposition $5$ of Book $\text{V}$ 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 $\text{V}$. Propositions