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ElementsX 6©

translated by Henry Mendell (Cal. State U., L.A.)

Return to Elements X, introduction

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Prop. 6: If two magnitudes have a ratio to one another that an arithmos has to an arithmos, the magnitudes will be commensurable.

Corollary: Given two arithmoi and a straight-line, and similar figures similarly constructed from the line and a straight-line mean in-ratio between the straight-line and another straight-line in the ratio of the arithmoi will also be in the ratio of the two arithmoi.

(diagram 1) For let two magnitudes, A, B, have a ratio to each other that an arithmos, D, has to an arithmos, E. I say that magnitudes A, B are commensurable.

(diagram 2) For as many as are units in in D, let A be divided into so-many equals, (diagram 3) and let G be equal to one of them. (diagram 4) But as many as are units in E, let Z be composed from so-many magnitudes equal to G. And so, since as many as are units in D so-many are also magnitudes in A equal to G, G is also the same part of A that the unit is a part of D. (diagram 5) Therefore, it is: as G to A so the unit to D. But the unit measures arithmos D. Therefore, G also measures A.Analysis

1. A : B = d : e

2. Let A/d = G

3. Let G*e = Z (to show that Z = B)

3. Hence, G measures Z

5. G : A = u : d (let u be the unit)

6. Hence, G measures A

7. A : G = d : u (inverse of 6)

8. G : Z = u : e (3)

9. A : Z = d : e (7, 8 ex aequali)

10. Hence, A : B = A : Z (1, 9)

11. B = Z

12. G measures B

13. A and B are commensurable (6, 13)

1. This is an odd step in the argument, since G was constructed as equal to a measure of A, that is, A being divided according to the units in D. We need that statement that G measures A, but why not infer it from the construction?

Provided (Corollary):

From this it is obvious that if there are two arithmoi, as D, E, and a straight-line, as A, it is possible to make it as arithmos D to arithmos E, so the straight-line to a straight-line. But if a mean in-ratio of A, Z is also taken, as B, it will be: as A to Z so will the figure from A to the figure from B, that is as the first to the third, so the figure from the first to the figure from the second, whatâ€™s similar and similarly inscribed up [VI, cor. 2, cf. C def. 9]. But as A to Z so is arithmos D to arithmos E. Therefore, as arithmos D also has come-about to arithmos E, so the figure from straight-line A to the figure from straight-line B, just what it was required to show.

In other words, Figure(A) : Figure(B) = A : Z = D : E, an arithmos to an arithmos

Note that A and B may be incommensurable (in length only), even thought the figures will be commensurable.

In the context in Book X, the expression 'that from a line' normally indicates a square drawn on the line. However, the theorem and corollary alluded to concern any similar polygons ('the figure (eidos) from...'), though they mention as well squares and triangles. Since there is a little ambiguity in the text, it seemed better to leave the generality in the translation. In the diagram, I have obviously used squares.

The corollary guarantees that we can readily get commensurable squares.

Differently the 6th

For let two magnitudes, A, B, have to each other a ratio that arithmos G has to arithmos D. I say that the magnitudes are commensurable.

For as-many-as are units in G, let A be divided into so-many equals, and there be an equal to one of them, E. Therefore, it is: as the unit to arithmos G, E to A. But it is also: as G to D, A to B. Therefore, ex aequali, it is: as the unit to D, E to B. But the unit also measures D. Therefore, E also measures B.. But E also measures A, since the unit also does G. Therefore, E measures each of A, B. Therefore, A, B are commensurable, and E is their common measure, just what it was required to show.