# Proposition 18

If magnitudes are proportional taken separately, then they are also proportional taken jointly.
V.Def.15

Let AE, EB, CF, and FD be magnitudes proportional taken separately, so that AE is to EB as CF is to FD.

V.Def.14

I say that they are also proportional taken jointly, that is, AB is to BE as CD is to FD.

For, if CD is not to DF as AB is to BE, then AB is to BE as CD is either to some magnitude less than DF or to a greater.

First, let it be in that ratio to a less magnitude DG.

V.17

Then, since AB is to BE as CD is to DG, they are magnitudes proportional taken jointly, so that they are also proportional taken separately. Therefore AE is to EB as CG is to GD.

V.11

But also, by hypothesis, AE is to EB as CF is to FD. Therefore CG is to GD as CF is to FD.

V.14

But the first CG is greater than the third CF, therefore the second GD is also greater than the fourth FD.

But it is also less, which is impossible. Therefore AB is to BE as CD is not to a less magnitude than FD.

Similarly we can prove that neither is it in that ratio to a greater, it is therefore in that ratio to FD itself.

Therefore, if magnitudes are proportional taken separately, then they are also proportional taken jointly.

Q.E.D.

## Guide

This proposition is the converse of the last one. It says that if w : x = y : z, then (w + x):x = (y + z):z. As in the last proposition, two of the magnitudes w and x can be of one kind while the other two y and z are of another.

#### On the existence of fourth proportionals

At the beginning of this proof we have, paraphrased:
If CD : DF does not equal AB : BE, then AB : BE = CD : DG where DG is some magnitude greater or less than DF.
Given the other three magnitudes, a fourth proportional DG is assumed. It is not asserted that the fourth proportional can be constructed; it is only hypothetical. This is the beginning of a proof by contradiction.

This technique of assuming the existence of a fourth proportional to derive a contradiction is also used in Book XII to prove various proportionalities of areas and volumes, for example, in proposition XII.2 which shows circles are proportional to the squares on their diameters. Eudoxus, who developed the techniques of both Books V and XII, or Euclid, or both of them, accepted this technique as valid.

The problem is: do fourth proportionals exist? They certainly can’t be constructed in all cases. The problems of doubling a cube, squaring a circle, and trisecting an angle cannot be solved by plane Euclidean methods, and they all involve inconstructable fourth proportionals. Take doubling a cube for example. If C is a cube with an edge A, then the inconstructable edge B of a cube with double the volume of C is the fourth proportional in C : (C+C) = A : B.

Is there a difference between existence and constructibility? Constructibility is a fairly clear concept since there are postulates for what can be constructed. There are no postulates for things that exist but aren’t constructed, but the existence of a fourth proportional is a good candidate for a such a postulate.

There is a similar situation in modern mathematics with the axiom of choice for set theory. That axiom says that in certain situations there is at least one set satisfying certain criteria. It does not construct anything in the usual sense of "construct," and it doesn’t even specify a particular set. Although it is useful in many situations, mathematicians prefer not to use it unless it’s necessary.

For this proposition, the assumption of the existence of fourth proportionals is unnecessary as the following alternate proof shows.

#### An alternate proof

Proposition: If w : x = y : z, then (w + x):x = (y + z):z.

Proof: Suppose w : x = y : z. Let n and m be any numbers. Either n < m or not.

Case 1: n < m.

Suppose n(w + x) >=< mx. Subtract nx to get nw >=< (m – n)x. But w : x = y : z, so ny >=< (m – n)z. Add nz to get n(y + z) >=< mz.

Case 2: n is not less than m.

Then both n(w + x) > mx, and n(y + z) > mz.

In any case n(w + x) >=< mx implies n(y + z) >=< mz. Therefore (w + x):x = (y + z):z. Q.E.D.

#### Use of this proposition

This proposition is used in proposition V.24 and a few in Books VI, X, XII, and XIII.