Bibliotheca Polyglotta

If there be any number of magnitudes whatever which are, respectively, equimultiples of any magnitudes equal in multitude, then, whatever multiple one of the magnitudes is of one, that multiple also will all be of all. ma+mb+mc...=m(a+b+c...).

此數幾何。彼數幾何。此之各率。同幾倍於彼之各率。則此之幷率。亦幾倍於彼之幷率。

Let any number of magnitudes whatever AB, CD be respectively equimultiples of any magnitudes E, F equal in multitude; I say that, whatever multiple AB is of E, that multiple will AB, CD also be of E, F.

For, since AB is the same multiple of E that CD is of F, as many magnitudes as there are in AB equal to E, so many also are there in CD equal to F. Let AB be divided into the magnitudes AG, GB equal to E, and CD into CH, HD equal to F; then the multitude of the magnitudes AG, GB will be equal to the multitude of the magnitudes CH, HD. Now, since AG is equal to E, and CH to F, therefore AG is equal to E, and AG, CH to E, F. For the same reason GB is equal to E, and GB, HD to E, F; therefore, as many magnitudes as there are in AB equal to E, so many also are there in AB, CD equal to E, F; therefore, whatever multiple AB is of E, that multiple will AB, CD also be of E, F.

Therefore etc. Q. E. D.

Permanent link

http://www2.hf.uio.no/common/apps/permlink/permlink.php?app=polyglotta&context=ctext&uid=6aec53be-3ed4-11e1-aecb-00215aecadea

Choose languages

Choose images, etc.

Choose languages

Choose display