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

Theorem
That is, for any numbers $a, b$ and for any integer $m$:
 * $ma - mb = m \left({a - b}\right)$

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 $CD$.


 * Euclid-V-5.png

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 Multiplication of Numbers Distributes over Addition, $AE$ 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 $CD$.