---
book: 10
number: 16
id: "X.16"
kind: "theorem"
uses: []
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.16"
license: "CC-BY-SA-4.0"
---

# X.16

*If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable*.

## Proof

For let the two incommensurable magnitudes *AB*, *BC* be added together; I say that the whole *AC* is also incommensurable with each of the magnitudes *AB*, *BC*.

For, if *CA*, *AB* are not incommensurable, some magnitude will measure them.

Let it measure them, if possible, and let it be *D*.

Since then *D* measures *CA*, *AB*, therefore it will also measure the remainder *BC*.

But it measures *AB* also; therefore *D* measures *AB*, *BC*.

Therefore *AB*, *BC* are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.

Therefore no magnitude will measure *CA*, *AB*; therefore *CA*, *AB* are incommensurable. [[book-10/definitions#Definition 1|X. Def. 1]]

Similarly we can prove that *AC*, *CB* are also incommensurable.

Therefore *AC* is incommensurable with each of the magnitudes *AB*, *BC*.

Next, let *AC* be incommensurable with one of the magnitudes *AB*, *BC*.

First, let it be incommensurable with *AB*; I say that *AB*, *BC* are also incommensurable.

For, if they are commensurable, some magnitude will measure them.

Let it measure them, and let it be *D*.

Since then *D* measures *AB*, *BC*. therefore it will also measure the whole *AC*.

But it measures *AB* also; therefore *D* measures *CA*, *AB*.

Therefore *CA*, *AB* are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.

Therefore no magnitude will measure *AB*, *BC*; therefore *AB*, *BC* are incommensurable. [[book-10/definitions#Definition 1|X. Def. 1]]

Therefore etc.

Lemma. *If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application.*

For let there be applied to the straight line *AB* the parallelogram *AD* deficient by the square figure *DB*; I say that *AD* is equal to the rectangle contained by *AC*, *CB*.

This is indeed at once manifest; for, since *DB* is a square, *DC* is equal to *CB*; and *AD* is the rectangle *AC*, *CD*, that is, the rectangle *AC*, *CB*.

Therefore etc.