Commentary on Proposition X.2¶
Back to Proposition X.2.
This proposition states the test for incommensurable magnitudes, founded on the usual operation for finding the greatest common measure. The sign of the incommensurability of two magnitudes is that this operation never comes to an end, while the successive remainders become smaller and smaller until they are less than any assigned magnitude.
Observe that Euclid says “let this process be repeated continually until there is left some magnitude which is less than E.” Here he evidently assumes that the process will some time produce a remainder less than any assigned magnitude E. Now this is by no means self-evident, and yet Heiberg (though so careful to supply references) and Lorenz do not refer to the basis of the assumption, which is in reality X. 1, as Billingsley and Williamson were shrewd enough to see. The fact is that, if we set off a smaller magnitude once or oftener along a greater which it does not exactly measure, until the remainder is less than the smaller magnitude, we take away from the greater more than its half. Thus, in the figure, FD is more than the half of CD, and BG more than the half of AB. If we continued the process, AG marked off along CF as many times as possible would cut off more than its half; next, more than half AG would be cut off, and so on. Hence along CD, AB alternately the process would cut off more than half, then more than half the remainder and so on, so that on both lines we should ultimately arrive at a remainder less than any assigned length.
The method of finding the greatest common measure exhibited in this proposition and the next is of course again the same as that which we use and which may be shown thus :
$$ \begin{array}{r@{}l} b) & \ a \ (p \\ & \underline{pb} \\ & c) \ b \ (q \\ & \quad \underline{qc} \\ & \quad d) \ c \ (r \\ & \qquad \underline{rd} \\ & \qquad e \end{array} $$
The proof too is the same as ours, taking just the same form, as shown in the notes to the similar propositions VII. 1, 2 above. In the present case the hypothesis is that the process never stops, and it is required to prove that a, b cannot in that case have any common measure, as f. For suppose that f is a common measure, and suppose the process to be continued until the remainder e, say, is less than f.
Then, since f measures a, b, it measures $a - pb$, or c. Since f measures b, c, it measures $b - qc$, or d; and, since f measures c, d, it measures $c - rd$, or e: which is impossible, since $e < f$.
Euclid assumes as axiomatic that, if f measures a, b, it measures $ma \pm nb$.
In practice, of course, it is often unnecessary to carry the process far in order to see that it will never stop, and consequently that the magnitudes are incommensurable. A good instance is pointed out by Allman (Greek Geometry from Thales to Euclid, pp. 42, 137—8). Euclid proves in XIII. 5 that, if AB be cut in extreme and mean ratio at C, and if DA equal to AC be added, then DB is also cut in extreme and mean ratio at A. This is indeed obvious from the proof of II. 11. It follows conversely that, if BD is cut into extreme and mean ratio at A, and AC, equal to the lesser segment AD, be subtracted from the greater AB, AB is similarly divided at C. We can then
mark off from AC a portion equal to CB, and AC will then be similarly divided, and so on. Now the greater segment in a line thus divided is greater than half the line, but it follows from XIII. 3 that it is less than twice the lesser segment, i.e. the lesser segment can never be marked off more than once from the greater. Our process of marking off the lesser segment from the greater continually is thus exactly that of finding the greatest common measure. If, therefore, the segments were commensurable, the process would stop. But it clearly does not; therefore the segments are incommensurable.
Allman expresses the opinion that it was rather in connexion with the line cut in extreme and mean ratio than with reference to the diagonal and side of a square that the Pythagoreans discovered the incommensurable. But the evidence seems to put it beyond doubt that the Pythagoreans did discover the incommensurability of $\sqrt{2}$ and devoted much attention to this particular case. The view of Allman does not therefore commend itself to me, though it is likely enough that the Pythagoreans were aware of the incommensurability of the segments of a line cut in extreme and mean ratio. At all events the Pythagoreans could hardly have carried their investigations into the incommensurability of the segments of this line very far, since Theaetetus is said to have made the first classification of irrationals, and to him is also, with reasonable probability, attributed the substance of the first part of Eucl. XIII., in the sixth proposition of which occurs the proof that the segments of a rational straight line cut in extreme and mean ratio are apotomes.
Again, the incommensurability of $\sqrt{2}$ can be proved by a method practically equivalent to that of X. 2, and without carrying the process very far. This method is given in Chrystal’s Text-book of Algebra (I. p. 270). Let d, a be the diagonal and side respectively of a square ABCD. Mark off AF along AC equal to a. Draw FE at right angles to AC meeting BC in E.
It is easily proved that
$$BE = EF = FC,$$
$$CF = AC - AB = d - a \dots\dots\dots\dots (1),$$
$$CE = CB - CF = a - (d - a)$$
$$\phantom{CE} = 2a - d \dots\dots\dots\dots (2).$$
Suppose, if possible, that d, a are commensurable. If d, a are both commensurably expressible in terms of any finite unit, each must be an integral multiple of a certain finite unit.
But from (1) it follows that CF, and from (2) it follows that CE, is an integral multiple of the same unit.
And CF, CE are the side and diagonal of a square CFEG, the side of which is less than half the side of the original square. If a1, d1 are the side and diagonal of this square,
$$ \left. \begin{aligned} a_1 &= d - a \\ d_1 &= 2a - d \end{aligned} \right\}. $$
Similarly we can form a square with side $a_2$ and diagonal $d_2$ which are less than half $a_1$, $d_1$ respectively, and $a_2$, $d_2$ must be integral multiples of the same unit, where
$$ \begin{aligned} a_2 &= d_1 - a_1, \\ d_2 &= 2a_1 - d_1; \end{aligned} $$
and this process may be continued indefinitely until (x. 1) we have a square as small as we please, the side and diagonal of which are integral multiples of a finite unit : which is absurd.
It will be observed that this method is the opposite of that shown in the Pythagorean series of side- and diagonal-numbers, the squares being successively smaller instead of larger.