Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater.

Let *AB* and *C* be unequal magnitudes, and let *AB* be greater, and let *D* be another, arbitrary, magnitude.

I say that *AB* has to *D* a greater ratio than *C* has to *D,* and *D* has to *C* a greater ratio than it has to *AB.*

(V.Def.4)

Since *AB* is greater than *C,* make *EB* equal to *C.* Then the less of the magnitudes *AE* and *EB,* if multiplied, will eventually be greater than *D.*

First, let *AE* be less than *EB.* Let *AE* be multiplied, and let *FG* be a multiple of it which is greater than *D.* Make *GH* the same multiple of *EB* and *K* the same multiple of *C* that *FG* is of *AE.*

(V.Def.4)

Take *L* double of *D* and *M* triple of it, and successive multiples increasing by one, until what is taken is the first multiple of *D* that is greater than *K.* Let it be taken, and let it be *N* which is quadruple of *D* and the first multiple of it greater than *K.*

Since *K* is less than *N* first, therefore *K* is not less than *M.*

And, since *FG* is the same multiple of *AE* that *GH* is of *EB,* therefore *FG* is the same multiple of *AE* that *FH* is of *AB.*

But *FG* is the same multiple of *AE* that *K* is of *C,* therefore *FH* is the same multiple of *AB* that *K* is of *C.* Therefore *FH* and *K* are equimultiples of *AB* and *C.*

Again, since *GH* is the same multiple of *EB* that *K* is of *C,* and *EB* equals *C,* therefore *GH* equals *K.*

But *K* is not less than *M,* therefore neither is *GH* less than *M.*

And *FG* is greater than *D,* therefore the whole *FH* is greater than the sum of *D* and *M.*

But the sum of *D* and *M* equals *N,* inasmuch as *M* is triple *D,* and the sum of *M* and *D* is quadruple *D,* while *N* is also quadruple *D,* therefore the sum of *M* and *D* equals *N.*

But *FH* is greater than the sum of *M* and *D,* therefore *FH* is in excess of *N,* while *K* is not in excess of *N.*

And *FH* and *K* are equimultiples of *AB* and *C,* while *N* is another, arbitrary, multiple of *D,* therefore *AB* has to *D* a greater ratio than *C* has to *D.*

I say next, that *D* has to *C* a greater ratio than *D* has to *AB.*

With the same construction, we can prove similarly that *N* is in excess of *K,* while *N* is not in excess of *FH.*

And *N* is a multiple of *D,* while *FH* and *K* are other, arbitrary, equimultiples of *AB* and *C,* therefore *D* has to *C* a greater ratio than *D* has to *AB.*

Next, let *AE* be greater than *EB.*

(V.Def.4)

Then the less, *EB,* if multiplied, will eventually be greater than *D.*

Let it be multiplied, and let *GH* be a multiple of *EB* and greater than *D.* Make *FG* the same multiple of *AE,* and *K* the same multiple of *C* that *GH* is of *EB.*

(V.Def.4)

Then we can prove similarly that *FH* and *K* are equimultiples of *AB* and *C,* and, similarly, take *N* the first multiple of *D* that is greater than *FG,* so that *FG* is again not less than *M.*

But *GH* is greater than *D,* therefore the whole *FH* is in excess of the sum of *D* and *M,* that is, of *N.*

Now *K* is not in excess of *N,* inasmuch as *FG* also, which is greater than *GH,* that is, than *K,* is not in excess of *N.*

And in the same manner, by following the above argument, we complete the demonstration.

Therefore, *of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater.
*

Q.E.D.

At four points in the proof V.Def.4 is used as an axiom of comparability rather than a definition. The first instance:

- Then the less of the magnitudes

But algebra obscures much, too. Euclid carefully proved distributivity of multiplication by numbers over addition of magnitudes in V.1, which is used in this proof. We manipulate algebraic expressions almost automatically. In order to be as correct as Euclid, we should verify the rules of algebra and be aware when we use them.

With these preliminary qualifications, let’s look at a translation of the proof into symbolic algebra.

To prove: if a > c, then a : d > c : d, but d : c > d : a. |
a = AB
c = C = EB
d = D |

Let a > c. Either a – c < c, or a – c > c, [or a – c = c]. |
a – c = AE |

Case 1: Suppose a – c < c. Let m be a number such that m(a – c) > d. |
m(a – c) = FG
mc = GH = K |

Let n be the smallest number such that nd > mc. [What happens when n = 1 to Euclid’s proof?] |
nd = N
( n – 1)d = M |

Since mc is not less than (n – 1)d, and m(a – c) > d, therefore, by adding, ma > nd. But mc is not greater than nd. Therefore a : d > c : d. |
ma = FH |

Also nd > mc but nd is not greater than ma. Therefore d : c > d : a. | |

Case 2: Suppose c < a – c. Let Let m be a number such that mc > d. |
(Same as above) |

Let n be the smallest number such that nd > m(a – c).
Since Also
[Case 3 when Thus, the conclusion is reached in any case. Q.E.D. |