# 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.

### 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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