Multiplication of Real Numbers is Right Distributive over Subtraction

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Theorem

In the words of Euclid:

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 or equimultiples of them.

(The Elements: Book $\text{V}$: Proposition $6$)


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 $F$.

$\blacksquare$


Also see


Historical Note

This theorem is Proposition $6$ of Book $\text{V}$ of Euclid's The Elements.


Sources