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Proposition 14

If a first magnitude has to a second the same ratio as a third has to a fourth, and the first is greater than the third, then the second is also greater than the fourth; if equal, equal; and if less, less.
Let a first magnitude A have the same ratio to a second B as a third C has to a fourth D, and let A be greater than C. I say that B is also greater than D.
| Since A is greater than C, and B is another, arbitrary, magnitude, therefore A has to B a greater ratio than C has to B. | V.8
But A is to B as C is to D, therefore C has to D a greater ratio than C has to B. | V.13
But that to which the same has a greater ratio is less, therefore D is less than B, so that B is greater than D. | V.10
Similarly we can prove that, if A equals C, then B equals D, and, if A is less than C, then B is less than D.
Therefore, if a first magnitude has to a second the same ratio as a third has to a fourth, and the first is greater than the third, then the second is also greater than the fourth; if equal, equal; and if less, less.
Q.E.D.

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The statement is that

if a:b = c:d and a >=< c, then b >=< d.

In this form all four magnitudes need to be of the same kind.

#### The alternate form of the proposition

Curiously, sometimes the alternate form

if a:b = c:d and a >=< b, then c >=< d

is used. This other form is more general since a and b may be of one kind while c and d can be of a different kind. (See definition V.Def.12 and proposition V.16 for alternate proportions.) For example, in proposition VI.25 there are the statements:

...the triangle ABC is to the triangle KGH as the parallelogram BE is to the parallelogram EF. Therefore, alternately, the triangle ABC is to the parallelogram BE as the triangle KGH is to the parallelogram EF. But the triangle ABC equals the parallelogram BE, therefore the triangle KGH also equals the parallelogram EF.

First, the proportion is converted to its alternate form by V.16. Then, it is claimed that since the first equals the second, therefore the third equals the fourth. Clearly, V.14 is not being invoked otherwise the alternate form of the proportion would not be mentioned.

Another example comes from X.112.

... the rectangle BC by EF equals the rectangle BD by G, therefore CB is to BD as G is to EF. But CB is greater than BD, therefore G is also greater than EF.

Here the first is greater than the second, so the third is greater than the fourth. Proposition VI.16 (if the rectangle contained by the extremes equals the rectangle contained by the means, then the four straight lines are proportional) was used to derive the proportion CB:BD = G:EF from the equality of the rectangles, but it would have been just as easy to conclude CB:G = BD:EF, and then V.14 could be used.

Clearly, this proposition V.14 is not being invoked in either of these propositions, but the alternate form is used instead. That suggests that the proofs of VI.25 and X.112 were written when V.14 wasn't available.

The proof of the statement

if a:b = c:d and a >=< b, then c >=< d

is not difficult using the definition V.Def.5. Since a >=< b, therefore 2a >=< 2b. From the proportion a:b = c:d it follows that 2c >=< 2d. Therefore c >=< d. Q.E.D. The proof is even easier when 1 is considered to be a number.

Although this alternate form does not rely on using V.Def.4 as an axiom of comparability, the original form does. The statement of the proposition is false when infinitesimals are allowed. For a particular example, take the proportion x:(x + y) = x: (x + 2y) or its inverse.

#### Use of this proposition

This proposition is used in V.16 and a few other propositions in Books V, VI, X, XII, and XIII.

Book V Introduction - Proposition V.13 - Proposition V.15.

© 1996, 1997
D.E.Joyce
Clark University