Multiplication of Numbers Distributes over Subtraction

Theorem

As Euclid defined it:

If a magnitude be the same multiple that a part subtracted is of a part subtracted, the remainder will also be the same multiple of the remainder that the whole is of the whole.

(The Elements: Book V: Proposition $5$)

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.

(The Elements: Book V: Proposition $6$)

That is, for any numbers $a, b$ and for any integers $m, n$:

$ma - mb = m \left({a - b}\right)$

$ma - na = \left({m - n}\right) a$

Proof

Proof of Proposition 5

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 $CD$.

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 $CD$.

$\blacksquare$

Proof of Proposition 6

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.

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

• Multiplication of Numbers Distributes over Addition

Historical Note

This is Proposition 5 and 6 of Book V of Euclid's The Elements.