Proposition: 5.19: Proportional Magnitudes have Proportional Remainders
Euclid's Formulation
If as the whole is to the whole so the (part) taken away is to the (part) taken away then the remainder to the remainder will also be as the whole (is) to the whole.
- For let the whole $AB$ be to the whole $CD$ as the (part) taken away $AE$ (is) to the (part) taken away $CF$.
- I say that the remainder $EB$ to the remainder $FD$ will also be as the whole $AB$ (is) to the whole $CD$.
Modern Formulation
In modern notation, this proposition reads that if \[\frac\alpha\beta=\frac\gamma\delta,\] then \[\frac\alpha\beta=\frac{\alpha-\gamma}{\beta-\delta},\]
for all positive real numbers \(\alpha,\beta,\gamma,\delta\) with \(\alpha > \gamma\) and \(\beta > \delta\).
Table of Contents
Proofs: 1 Corollaries: 1
Mentioned in:
Proofs: 1 2 3
Sections: 4
Thank you to the contributors under CC BY-SA 4.0! 
- Github:
-
- non-Github:
- @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Adapted from CC BY-SA 3.0 Sources:
- Prime.mover and others: "Pr∞fWiki", https://proofwiki.org/wiki/Main_Page, 2016
