Multiplication of Numbers is Left Distributive over Addition

Theorem
That is, if $ma, mb, mc$ etc. be any equimultiples of $a, b, c$ etc., then:
 * $ma + mb + mc + \cdots = m \left({a + b + c + \cdots }\right)$

Proof
Let any number of magnitudes whatever $AB, CD$ be respectively equimultiples of any magnitudes $E, F$ equal in multitude.

Then we are to show that whatever multiple $AB$ is of $E$, then that multiple will $AB + CD$ be of $E + F$.


 * Euclid-V-1.png

Since $AB$ is the same multiple of $E$ that $CD$ is of $F$, as many magnitudes as there are in $AB$ equal to $E$, so many also are there in $CD$ equal to $F$.

Let $AB$ be divided into the magnitudes $AG, GB$ equal to $E$, and $CH, HD$ equal to $F$.

Then the multitude of the magnitudes $AG, GB$ will be equal to the multitude of the magnitudes $CH, HD$.

Since $AG = E$ and $CH = F$ it follows that $AG = E$ and $AG + CH = E + F$.

For the same reason, $GB = E$ and $GB + HD = E + F$.

Therefore, as many magnitudes as there are in $AB$ equal to $E$, so many also are there in $AB + CD$ equal to $E + F$.

Also see

 * Real Multiplication Distributes over Addition
 * Multiplication of Numbers is Right Distributive over Addition