| Any number is either a part or parts of any number, the less of the greater. | ||
| Let A and BC be two numbers, and let BC be the less.
I say that BC is either a part, or parts, of A. | ||
| Either A and BC are relatively prime or they are not.
First, let A and BC be relatively prime. | ||
| Then, if BC is divided into the units in it, then each unit of those in BC is some part of A, so that BC is parts of A. | VII.Def.4 | |
| Next let A and BC not be relatively prime, then BC either measures, or does not measure, A. | ||
| Now if BC measures A, then BC is a part of A. But, if not, take the greatest common measure D of A and BC, and divide BC into the numbers equal to D, namely BE, EF, and FC. | VII.Def.3
VII.2 | |
| Now, since D measures A, therefore D is a part of A. But D equals each of the numbers BE, EF, and FC, therefore each of the numbers BE, EF, and FC is also a part of A, so that BC is parts of A. | ||
| Therefore, any number is either a part or parts of any number, the less of the greater. | ||
| Q.E.D. | ||
This proposition says that if b is a smaller number than a, then b is either a part of a, that is, b is a unit fraction of a, or b is parts of a, that is, a proper fraction, but not a unit fraction, of a. For instance, 2 is one part of 6, namely, one third part; but 4 is parts of 6, namely, 2 third parts of 6.
It seems obvious that when one number b is less than another a, then in all cases b would be parts of a, namely b consists of b of the ath parts of a. For instance, 4 consists of 4 sixth parts of 6. Yet, the proof of this proposition ignores that possibility, except in the special case when b and a are relatively prime. In the case of 4 and 6, the proof will find that 4 is 2 third parts of 6. Thus, it appears that a satisfactory answer to the question "How mary parts of a is b?" requires finding the least number of parts.
The proof has three cases.
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