Commentary on Proposition VI.22¶
Back to Proposition VI.22.
The second assumption in the first step of the first part of the proof, viz. that, as CD is to O, so GH to P, should perhaps be explained. It is a deduction [by V. 11] from the facts that
AB is to CD as CD to O, EF is to GH as GH to P, and AB is to CD as EF to GH.
The defect in the proof of this proposition is well known, namely the assumption, without proof, that, because the figures NH, SR are equal, besides being similar and similarly situated, their corresponding sides GH, QR are equal. Hence the minimum addition necessary to make the proof complete is a proof of a lemma to the effect that, if two similar figures are also equal, any pair of corresponding sides are equal.
To supply this lemma is one alternative; another is to prove, as a preliminary proposition, a much more general theorem, viz. that, if the duplicate ratios of two ratios are equal, the two ratios are themselves equal. When this is proved, the second part of VI. 22 is an immediate inference from it, and the effect is, of course, to substitute a new proof instead of supplementing Euclid’s.
I. It is to be noticed that the lemma required as a minimum is very like what is needed to supplement VI. 28 and 29, in the proofs of which Euclid assumes that, if two similar parallelograms are unequal, any side in the greater is greater than the corresponding side in the smaller. Therefore, on the whole, it seems preferable to adopt the alternative of proving the simpler lemma which will serve to supplement all three proofs, viz. that, if of two similar rectilineal figures the first is greater than, equal to, or less than, the second, any side of the first is greater than, equal to, or less than, the corresponding side of the second respectively.
The case of equality of the figures is the case required for VI. 22; and the proof of it is given in the Greek text after the proposition. Since to give such a “lemma” after the proposition in which it is required is contrary to Euclid’s manner, Heiberg concludes that it is an interpolation, though it is earlier than Theon. The lemma runs thus:
“But that, if rectilineal figures be equal and similar, their corresponding sides are equal to one another we will prove thus.
Let NH, SR be equal and similar rectilineal figures, and suppose that,
as HG is to GN, so is RQ to QS;
I say that
RQ is equal to HG.
For, if they are unequal, one of them is greater; let RQ be greater than HG.
[The step marked * is easy to see if it is remembered that it is only necessary to prove its truth in the case of triangles (since similar polygons are divisible into the same number of similar and similarly situated triangles having the same ratio to each other respectively as the polygons have). If the triangles be applied to each other so that the two corresponding sides of each, which are used in the question, and the angles included by them coincide, the truth of the inference is obvious.]
The lemma might also be arrived at by proving that, if a ratio is greater than a ratio of equality, the ratio which is its duplicate is also greater than a ratio of equality; and if the ratio which is duplicate of another ratio is greater than a ratio of equality, the ratio of which it is the duplicate is also greater than a ratio of equality. It is not difficult to prove this from the particular case of V. 25 in which the second magnitude is equal to the third, i.e. from the fact that in this case the sum of the extreme terms is greater than double the middle term.
II. We now come to the alternative which substitutes a new proof for the second part of the proposition, making the whole proposition an immediate inference from one to which it is practically equivalent, viz. that
If two ratios be equal, their duplicate ratios are equal, and (2) conversely, if the duplicate ratios of two ratios be equal, the ratios are equal.
The proof of part (1) is after the manner of Euclid’s own proof of the first part of VI. 22.
Let $A$ be to $B$ as $C$ to $D$, and let $X$ be a third proportional to $A$, $B$, and $Y$ a third proportional to $C$, $D$, so that
$$\begin{align*} &A\text{ is to }B\text{ as }B\text{ to }X\text{,} \\ \text{and }&C\text{ is to }D\text{ as }D\text{ to }Y\text{;} \end{align*}$$
whence $A$ is to $X$ in the duplicate ratio of $A$ to $B$, and $C$ is to $Y$ in the duplicate ratio of $C$ to $D$. Since
$$A\text{ is to }B\text{ as }C\text{ to }D\text{,}$$
and
$$\begin{align*} B\text{ is to }X&\text{ as }A\text{ to }B\text{,} \\ &\text{i.e. as }C\text{ to }D\text{,} \\ &\text{i.e. as }D\text{ to }Y\text{,} &\text{[V. 11]} \end{align*}$$
therefore, ex aequali,
$$A\text{ is to }X\text{ as }C\text{ to }Y\text{.}$$
Part (2) is much more difficult and is the crux of the whole thing. Most of the proofs depend on the assumption that, $B$ being any magnitude and $P$ and $Q$ two magnitudes of the same kind, there does exist a magnitude $A$ which is to $B$ in the same ratio as $P$ to $Q$. It is this same assumption which makes Euclid’s proof of V. 18 illegitimate, since it is nowhere proved in Book V. Hence any proof of the proposition now in question which involves this assumption even in the case where $B$, $P$, $Q$ are all straight lines should not properly be given as an addition to Book V.; it should at least be postponed until we have learnt, by means of VI. 12, giving the actual construction of a fourth proportional, that such a fourth proportional exists.
Two proofs which are given of the proposition depend upon the following lemma.
If A, B, C be three magnitudes of one kind, and D, E, F three magnitudes of one kind, then, if
the ratio of A to B is greater than that of D to E, and the ratio of B to C greater than that of E to F, ex aequali, the ratio of A to C is greater than that of D to F.
One proof of this does not depend upon the assumption referred to, and therefore, if this proof is used, the theorem can be added to Book V. The proof is that of Hauber (Camerer’s Euclid, p. 358 of Vol. II.) and is reproduced by Mr H. M. Taylor. For brevity we will use symbols.
Take equimultiples $mA$, $mD$ of $A$, $D$ and $nB$, $nE$ of $B$, $E$ such that
$$mA > nB\text{, but } mD \ngtr nE\text{.}$$
Also let $pB$, $pE$ be equimultiples of $B$, $E$ and $qC$, $qF$ equimultiples of $C$, $F$ such that
$$pB > qC\text{, but } pE \ngtr qF\text{.}$$
Therefore, multiplying the first line by $p$ and the second by $n$, we have
$$\begin{align*} &pmA > pnB\text{, } pmD \ngtr pnE\text{,} \\ \text{and }&npB > nqC\text{, } npE \ngtr nqF\text{,} \\ \text{whence }&pmA > nqC\text{, } pmD \ngtr nqF\text{.} \end{align*}$$
Now $pmA$, $pmD$ are equimultiples of $mA$, $mD$, and $nqC$, $nqF$ equimultiples of $qC$, $qF$. Therefore [V. 3] they are respectively equimultiples of $A$, $D$ and of $C$, $F$. Hence [V. Def. 7]
$$A : C > D : F\text{.}$$
Another proof given by Clavius, though depending on the assumption referred to, is neat. Take G such that
G : C = E : F.
Therefore B : C > G : C, [V. 13] and B > G. [V. 10] Therefore A : G > A : B. [V. 8]
But A : B > D : E. Therefore, a fortiori, A : G > D : E.
Suppose H taken such that
H : G = D : E. Therefore A > H. [V. 13, 10] Hence A : C > H : C. [V. 8] But H : G = D : E, G : C = E : F. Therefore, ex aequali, H : C = D : F. [V. 22] Hence A : C > D : F. [V. 13]
Now we can prove that Ratios of which equal ratios are duplicate are equal.
$$\begin{align*} \text{Suppose that}\quad\quad\quad\quad A : B &= B : C\text{,} \\ \text{and}\quad\quad\quad\quad D : E &= E : F\text{,} \\ \text{and further that}\quad\quad\quad\quad A : C &= D : F\text{.} \end{align*}$$
it is required to prove that
$$A : B = D : E\text{.}$$
For, if not, one of the ratios must be greater than the other. Let $A : B$ be the greater.
$$\begin{align*} \text{Then, since}\quad\quad\quad\quad A : B &= B : C\text{,} \\ \text{and}\quad\quad\quad\quad D : E &= E : F\text{,} \\ \text{while}\quad\quad\quad\quad A : B &> D : E\text{,} \\ \text{it follows that}\quad\quad\quad\quad B : C &> E : F\text{.} &\text{[V. 13]} \end{align*}$$
Hence, by the lemma, ex aequali,
$$A : C > D : F\text{,}$$
which contradicts the hypothesis. Thus the ratios $A : B$ and $D : E$ cannot be unequal; that is, they are equal.
Another proof, given by Dr Lachlan, also assumes the existence of a fourth proportional, but depends upon a simpler lemma to the effect that It is impossible that two different ratios can have the same duplicate ratio. For, if possible, let the ratio $A : B$ be duplicate both of $A : X$ and $A : Y$, so that
$$\begin{align*} A : X &= X : B\text{,} \\ \text{and}\quad\quad\quad\quad A : Y &= Y : B\text{.} \end{align*}$$
Let $X$ be greater than $Y$.
$$\begin{align*} \text{Then}\quad\quad\quad\quad A : X &< A : Y\text{;} &\text{[V. 8]} \\ \text{that is,}\quad\quad\quad\quad X : B &< Y : B\text{,} &\text{[V. 11, 13]} \\ \text{or}\quad\quad\quad\quad X &< Y\text{.} &\text{[V. 10]} \end{align*}$$
But $X$ is greater than $Y$: which is absurd, etc.
$$\text{Hence}\quad\quad\quad\quad X = Y\text{.}$$
$$\begin{align*} \text{Now suppose that}\quad\quad\quad\quad A : B &= B : C\text{,} \\ D : E &= E : F\text{,} \\ \text{and}\quad\quad\quad\quad A : C &= D : F\text{.} \\ \text{To prove that}\quad\quad\quad\quad A : B &= D : E\text{.} \end{align*}$$
If this is not so, suppose that
$$A : B = D : Z\text{.}$$
$$\begin{align*} \text{Since}\quad\quad\quad\quad A : C &= D : F\text{,} \\ \text{therefore, inversely,}\quad\quad C : A &= F : D\text{.} \\ \text{Therefore, \textit{ex aequali},}& \\ C : B &= F : Z\text{,} &\text{[V. 22]} \\ \text{or, inversely,}\quad\quad\quad\quad B : C &= Z : F\text{.} \\ \text{Therefore}\quad\quad\quad\quad A : B &= Z : F\text{.} &\text{[V. 11]} \\ \text{But}\quad\quad\quad\quad A : B &= D : Z\text{, by hypothesis.} \\ \text{Therefore}\quad\quad\quad\quad D : Z &= Z : F\text{.} &\text{[V. 11]} \\ \text{Also, by hypothesis,}\quad\quad D : E &= E : F\text{;} \\ \text{whence, by the lemma,}\quad\quad E &= Z\text{.} \\ \text{Therefore}\quad\quad\quad\quad A : B &= D : E\text{.} \end{align*}$$
De Morgan remarks that the best way of remedying the defect in Euclid is to insert the proposition (the lemma to the last proof) that it is impossible that two different ratios can have the same duplicate ratio, “which,” he says, “immediately proves the second (or defective) case of the theorem.” But this seems to be either too much or too little: too much, if we choose to make the minimum addition to Euclid (for that addition is a lemma which shall prove that, if a duplicate ratio is a ratio of equality, the ratio of which it is duplicate is also one of equality), and too little if the proof is to be altered in the more fundamental manner explained above.
I think that, if Euclid’s attention had been drawn to the defect in his proof of VI. 22 and he had been asked to remedy it, he would have done so by supplying what I have called the minimum lemma and not by making the more fundamental alteration. This I infer from Prop. 24 of the Data, where he gives a theorem corresponding to the proposition that ratios of which equal ratios are duplicate are equal. The proposition in the Data is enunciated thus: If three straight lines be proportional, and the first have to the third a given ratio, it will also have to the second a given ratio.
$A$, $B$, $C$ being the three straight lines, so that
$$A : B = B : C\text{,}$$
and $A : C$ being a given ratio, it is required to prove that $A : B$ is also a given ratio.
Euclid takes any straight line $D$, and first finds another, $F$, such that
$$D : F = A : C\text{,}$$
whence $D : F$ must be a given ratio, and, as $D$ is given, $F$ is therefore given.
Then he takes $E$ a mean proportional between $D$, $F$, so that
$$D : E = E : F\text{.}$$
It follows [VI. 17] that > the rectangle $D$, $F$ is equal to the square on $E$.
But $D$, $F$ are both given; > therefore the square on $E$ is given, so that $E$ is also given.
[Observe that De Morgan’s lemma is here assumed without proof. It may be proved (1) as it is by De Morgan, whose proof is that given above, p. 245, (2) in the manner of the “minimum lemma,” pp. 242—3 above, or (3) as it is by Proclus on I. 46 (see note on that proposition).]
Hence the ratio $D$ : $E$ is given.
Now, since
$$A : C = D : F\text{,}$$
and
$$A : C = (\text{square on } A) : (\text{rect. } A, C)\text{,}$$
while
$$D : F = (\text{square on } D) : (\text{rect. } D, F)\text{,}\qquad \text{[VI. 1]}$$
therefore
$$(\text{square on } A) : (\text{rect. } A, C) = (\text{square on } D) : (\text{rect. } D, F)\text{.}\qquad \text{[v. 11]}$$
But, since $A : B = B : C$, $(\text{rect. } A, C) = (\text{sq. on } B)$; [VI. 17] and $(\text{rect. } D, F) = (\text{sq. on } E)$, from above; therefore
$$(\text{square on } A) : (\text{square on } B) = (\text{sq. on } D) : (\text{sq. on } E)\text{.}$$
$Therefore$, says Euclid,
$$A : B = D : E\text{,}$$
that is, he assumes the truth of VI. 22 for squares.
Thus he deduces his proposition from VI. 22, instead of proving VI. 22 by means of it (or the corresponding proposition used by Mr Taylor and Dr Lachlan).