The (almost really) Complete Works of Lewis Carroll

So Far as it Relates to Commensurable Magnitudes, with Notes

Preface

The theory of Incommensurable Magnitudes, without which the whole subject of Geometrical Proportion is so incomplete as to be, from a logical point of view, utterly valueless, is nevertheless omitted, as far as possible, from the following treatise.

My reasons for this omission are two: first, that I believe it to be much too abstruse a subject for the ordinary Pass Examination; secondly, that it is not required in it. The exemption is a most necessary one, though the effect of it is to reduce the V^th and VI^th Books, in the form in which they are now learned and accepted in the Schools, to a logical absurdity.

Whether it would not be preferable to substitute for these Books an equivalent quantity of Algebra, perhaps as far as Permutations and Combinations, is a question I do not here enter on. To supply, in its shortest form, that knowledge of the subject which is at present required and accepted in the Schools, is my object in putting forth this treatise. I hope that it may not be long wanted.

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Hints on Proving Propositions

Most of the Propositions of the V^th Book may be proved by observing the following three rules:—

1. Express the ‘data’ in algebraical language and reduce it to its simplest form.

2. Do the same with the ‘conclusion’.

3. Seek for a connecting link between these two results. In many cases this is obvious, but, where it is not so, obtain, from the data, values of some of the mangitudes in terms of others, then take the terms of the conclusion one by one, and substitute in them the values obtained from the data.

These rules will be easily understood by applying them to an example. Let us take Prop. II. “If the first magnitude be the same multiple of the second that the third is of the fourth, and the fifth the same multiple of the second that the sixth is of the fourth; then the first and fifth together shall be the same multiple of the second that the third and sixth together are of the fourth.”

1. We represent the mangitudes by a, b, c, d, e, f, making $a=mb$, so that $c=md$, and making $e=nb$, so that $f=nd$. This cannot be reduced to a simpler form.

2. The ‘conclusion’ is that $\left(a+e\right)$ is the same multiple of b as $\left(c+f\right)$ is of d. And this also cannot be reduced to a simpler form.

3. No connecting link is obvious: we have therefore to obtain, from the data, values of some of the mangitudes in terms of others; but these we already have, viz. $a=mb$, $c=md$, $e=nb$, $f=nd$. Nothing remains then but to take the terms of the conclusion one by one, and substitute in them these values. Thus $\left(a+e\right)=\left(mb+nb\right)=\left(m+n\right).b$, and $\left(c+f\right)=\left(md+nd\right)=\left(m+n\right).d$. Hence we see that $\left(a+e\right)$ is $\left(m+n\right)$ times b, and that $\left(c+f\right)$ is $\left(m+n\right)$ times d, which proves the ‘conclusion’: we then arrange the proof this:—

$$\begin{array}{rl}\therefore & a+e=mb+nb=\left(m+n\right).b;\\ \text{and }& c+f=md+nd=\left(m+n\right).d\text{.}\end{array}$$ Q.E.D.

Again, let us take Prop. IV. “If the first of four magnitudes has the same ratio to the second which the third has to the fourth; then any equimultiples whatever of the first and third shall have the same ratio to any equimultiples of the second and fourth, viz. ‘the equimultiple of the first shall have the same ratio to that of the second, which the equimultiple of the third has to that of the fourth’.”

1. We represent the magnitudes by a, b, c, d, so that $a:b::c:d$. We represent the equimultiples of a, c, by A, C, making $A=ma$, so that $C=mc$; and we represent the equimultiples of b, d, by B, D, making $B=nb$, so that $D=nd$. We then reduce the proportion, $a:b::c:d$, to its simplest form, viz. $\frac{a}{b}=\frac{c}{d}$.

2. The ‘conclusion’ is $A:B::C:D$; and this, reduced to its simplest form, becomes $\frac{A}{B}=\frac{C}{D}$.

We have now to seek for a link to connect the equation $\frac{a}{b}=\frac{c}{d}$ with the equation $\frac{A}{B}=\frac{C}{D}$, and it is tolerably obvious that this may be done by multiplying both sides of the first by $\frac{m}{n}$; we thus obtain $\frac{ma}{nb}=\frac{mc}{nd}$, and thence $\frac{A}{B}=\frac{C}{D}$: we then arrange the proof this:— $$\begin{array}{rl}\because & a:b::c:d;\\ \therefore & \frac{a}{b}=\frac{c}{d};\\ \therefore & \text{(×^ng by <math><mfrac><mi>m</mi><mi>n</mi></mfrac></math>),}\frac{ma}{nb}=\frac{mc}{nd};\\ \therefore & \frac{A}{B}=\frac{C}{D};\\ \therefore & A:B::C:D\text{.}\end{array}$$ Q.E.D.

But suppose this connecting link were not obvious. We then obtain, from the data, values of some of the mangitudes in terms of others, thus: $\frac{a}{b}=\frac{c}{d}=k$; ∴ $a=kb$, $c=kd$. Now the terms of the conclusion are $\frac{A}{B}$, $\frac{C}{D}$; taking these one by one, and substituting in them the values obtained from the data, we get $\frac{A}{B}=\frac{ma}{nb}=\frac{mkb}{nb}=\frac{mk}{n}$, and $\frac{C}{D}=\frac{mc}{nd}=\frac{mkd}{nd}=\frac{mk}{n}$. Having thus proved that $\frac{A}{B}$ and $\frac{C}{D}$ are each equal to the same thing, viz. $\frac{mk}{n}$, we see that they are equal to each other, and the conclusion follows as before: we then arrange the proof thus:— $$\begin{array}{rl}\because & a:b::c:d;\\ \therefore & \frac{a}{b}=\frac{c}{d}=k\text{ (say)};\\ \therefore & a=kb,c=kd;\\ \text{now }& \frac{A}{B}=\frac{ma}{nb}=\frac{mkb}{nb}=\frac{mk}{n};\\ \text{and }& \frac{C}{D}=\frac{mc}{nd}=\frac{mkd}{nd}=\frac{mk}{n};\\ \therefore & \frac{A}{B}=\frac{C}{D};\\ \therefore & A:B::C:D\text{.}\end{array}$$ Q.E.D.

The method of proving the proposition by introducing ‘k’, and thus reducing the equation ‘$\frac{a}{b}=\frac{c}{d}$’ to the form ‘$a=kb$, $c=kd$’, is generally rather longer than the other, and should not be adopted without first seeking for a shorter connecting link.

The student is recommended to attempt to prove the propositions for himself before consulting the solutions here given.

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