Multiplication of Real Numbers is Right Distributive over Subtraction

Theorem
That is, for any number $a$ and for any integers $m, n$:
 * $m a - n a = \paren {m - n} a$

Proof
Let two magnitudes $AB, CD$ be equimultiples of two magnitudes $E, F$.

Let $AG, CH$ subtracted from them be equimultiples of the same two $E, F$.

We need to show that the remainders $GB, HD$ are either equal to $E, F$ or are equimultiples of them.


 * Euclid-V-6.png

First let $GB = E$.

Let $CK$ be made equal to $F$.

We have that $AG$ is the same multiple of $E$ that $CH$ is of $F$, while $GB = E$ and $KC = F$.

Therefore from the Distributive Laws, $AB$ is the same multiple of $E$ that $KH$ is of $F$.

But, $AB$ is the same multiple of $E$ that $CD$ is of $F$.

Since then, each of the magnitudes $KH, CD$ is the same multiple of $F$.

Therefore $KH = CD$.

Let $CH$ be subtracted from each.

Therefore the remainder $KC$ equals the remainder $HD$.

But $F = KC$, so $HD = F$.

Hence, if $GB = E$ then $HD = F$.

Similarly we can prove that, even if $GB$ is a multiple of $E$, then $HD$ is also the same multiple of $F$.

Also see

 * Multiplication of Real Numbers is Left Distributive over Subtraction


 * Distributive Laws (Arithmetic)