Multiplication of Real Numbers is Right Distributive over Subtraction

Theorem

 * If two magnitudes be equimultiples of two magnitudes, and any magnitudes subtracted from them be equimultiples of the same, the remainders also are either equal to the same of equimultiples of them.

That is, for any number $a$ and for any integers $m, n$:
 * $ma - na = \left({m - n}\right) a$

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 Multiplication of Numbers Distributes over Addition, $AB$ is the same multiple of $E$ that $KH$ is of $F$.

But by hypothesis $AB$ 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 Numbers is Left Distributive over Subtraction


 * Multiplication of Numbers Distributes over Addition