Return to Vignettes of Ancient Mathematics
The text used is the edition of Tannery (1893), but I have also consulted the translation of ver Eecke (1959) and the paraphrase of Heath (1910).
To read this file, the font must be set to Uniicode, i.e., UTF-8
For notes on translation, go to the introduction to Book I.
In fact, let it be prescribed to divide 16 into two squares. And let the 1st be assigned ΔΥ 1, therefore the other will be 16 ΔΥ. It will be required, therefore, for 16 ΔΥ to be equal to a ⬜.
I form the ⬜ from however many ’s Μ’s that are the side of 16 . Let it be 2 4. Therefore, the ⬜ will be ΔΥ 4 16 16. These are equal to 16 ΔΥ. Let a common be added, the deficiency and similars from similars.
Therefore, ΔΥ 5 are equal to 16, and becomes 16 of fifths.
One will be 26425´, another 14425´, and the two added make 40025´, or 15, and each is a square.
In fact let it again be required to divide square 16 into two squares. Let again the side of the 1st be assigned 1, that of the other however many ’s of as many as the side of the divided. In fact, let it be 2 4.
Therefore, the ⬜’s will be, one ΔΥ 1 and the other ΔΥ 4 16 16. Remaining, I want the two added equal to 16.
Therefore, ΔΥ 5 16 16 are equal to 16. And becomes 165´.
The sd. of the 1st will be 165´; therefore, it will be 25625´. An the sd. of the 2nd 125´; therefore, it will be 14425. And the demonstration is obvious.
Let it be required with 13, which is composed from squares 4 and 9, to divide it again into two other squares.
Let there be taken of the mentioned squares the sds, 2, 3, and let there be assigned the sds of the prescribed squares, one 1 2, the other of as many it may be of as many as is the side of the remaining. Let it be 2 3. And the squares become, one ΔΥ 1 4 4, the other ΔΥ 4 9 12.
Remaining is to make the two added 13. But the two added make ΔΥ 5 13 8. These are equal to 13. and becomes 85. For the actualities: I assigned the sd of the 1st, 1 2. It will be 18Μ5. The sd of the 2nd I assigned 2 3. It will be the side of one of them. And the ⬜’s themselves will be, one 32425´, which is of one. And the two added make 32525´, which brings together the prescribed 13.
In fact, let it be prescribed that the excess of them is 60.
Let it be assigned one whose side is 1 and the other whose side is 1 and however many ’s you want, merely so that the ⬜ from the ’s does not exceed the excess that’s given, nor, truly, that it be equal. For, when one form is left equal to one form, the problem will be established.
Let it be 1 3. Therefore the squares will be ΔΥ 1 and ΔΥ 1 6 9. And the excess of them is 6 9. These are equal to 60. And becomes 8 𐅵. The side of the 1st will be 8 𐅵, and that of the 2nd 11 𐅵. The ⬜’s themselves will be, one 72 4, the other 132 4, and things of the proposition are obvious.