Subtraction of Divisors obeys Distributive Law/Proof 2

Theorem

In the words of Euclid:

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

(The Elements: Book $\text{VII}$: Proposition $7$)

In modern algebraic language:

$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a - c = \dfrac 1 n \paren {b - d}$

Proof

A direct application of the Distributive Property:

\(\ds \frac 1 n b - \frac 1 n d\)

\(=\)

\(\ds \frac 1 n b + \frac 1 n \paren {-d}\)

\(\ds \)

\(=\)

\(\ds \frac 1 n \paren {b + \paren {-d} }\)

\(\ds \)

\(=\)

\(\ds \frac 1 n \paren {b - d}\)

$\blacksquare$

Historical Note

This proof is Proposition $7$ of Book $\text{VII}$ of Euclid's The Elements.