Proposition: Prop. 10.042: Binomial Straight Line is Divisible into Terms Uniquely

Euclid's Formulation

A binomial (straight line) can be divided into its (component) terms at one point only. * Let $AB$ be a binomial (straight line) which has been divided into its (component) terms at $C$. * $AC$ and $CB$ are thus rational (straight lines which are) commensurable in square only [Prop. 10.36]. * I say that $AB$ cannot be divided at another point into two rational (straight lines which are) commensurable in square only.

fig042e

Modern Formulation

In other words, \[\alpha + \sqrt{\beta} = \gamma + \sqrt{\delta}\] has only one solution: i.e., \[\gamma=\alpha\quad\text{ and }\quad\delta=\beta,\] where \(\alpha,\beta,\gamma,\delta\) denote positive rational numbers. Likewise, \[\sqrt{\alpha} + \sqrt{\beta} =\sqrt{\gamma}+\sqrt{\delta}\] has only one solution: i.e., \[\gamma=\alpha\quad\text{ and }\quad\delta=\beta,\] or, equivalently, \[\gamma=\beta\quad\text{ and }\quad\delta=\alpha.\]

Notes

This proposition corresponds to [Prop. 10.79], with plus signs instead of minus signs.

Proofs: 1

Proofs: 1 2
Propositions: 3


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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