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Proposition: 5.05: Multiplication of Real Numbers is Left Distributive over Subtraction

(Proposition 5 from Book 5 of Euclid's “Elements”)

If a magnitude is the same multiple of a magnitude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively). * For let the magnitude $AB$ be the same multiple of the magnitude $CD$ that the (part) taken away $AE$ (is) of the (part) taken away $CF$ (respectively). * I say that the remainder $EB$ will also be the same multiple of the remainder $FD$ as that which the whole $AB$ (is) of the whole $CD$ (respectively).

Modern Formulation

If we are given two positive real numbers \(\alpha\), \(\beta\),¹ and the following multiples of aliquot parts \(m\ge 1\), \(n\ge 1\): \[\alpha m=\beta m,\quad \alpha n=\beta n\quad\quad( * )\] then adding both equations gives us \[\alpha(m-n)=\beta m-\beta n.\]

General Modern Formulation

See distributivity law for real numbers.

Table of Contents

Proofs: 1

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"

Adapted from CC BY-SA 3.0 Sources:

1. Prime.mover and others: "Pr∞fWiki", https://proofwiki.org/wiki/Main_Page, 2016

Footnotes