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well as several of the others, has two of its parts suppressed, and are perfectly consistent with the theory previously explained.

4. Many Cruciferæ become tetrandrous by pelorization; others are normally so. In either case the four stamens are thus equal. This, I answer, is at least as easily explained in our theory as on that of the separation of stamens into two.

5. Finally, certain Cruciferæ instead of returning to the quaternary type recede from it. The single stamens undergo a change analogous or very similar to that of the double pair. One of us has observed flowers of Matthiola incana, in which the single stamens were cleft throughout their entire length, each portion being provided with half an anther and half a filament. M. Lestiboudois speaks of a Cheiranthus Cheiri in which these stamens were completely geminated, not laterally as the longer pair, but from without inwards, M. Lermye met with a flower of the same species, which had the lowerstamens doubled exactly as the upper. Now let these cases be fairly considered: the first appears to show that a stamen may 'oe occasionally slit vertically, but it is acknowledged that there is no increase in the real number of parts, each portion it is expressly stated consisting of half an anther (a single cell,) and half a filament. This may render more probable Dr. Lindley's explanation, of Fumariaceæ, destroying an analogy on which Dr. Gray greatly relies, but it supplies no argument in favour of a single Primitive organ having become two perfect ones with all their part. The case observed by Lestiboudcis is apparently not one Chorisis, but of development under the stimulus of cultivat on of the gland, which is often noticed within the short stamens that of M. Lermye requires to be more accurately described, 'out it must not be hastily assumed to have consisted in a division of the single stamen into two perfect ones, it may have been a case like that seen by one of the authors themselves, a mere fissure of the stamen into two parts; or it is perhaps just possible that the single stamen may have been suppressed, and the two glands which often appear at each side of it, developed into a pair of stamens. It is certainly not sufficient without more exact information, to support or overthrow a theory. Dr. Gray relies so completely on the arguments of Messrs. Moquin Tandon and Webb, that I need only farther observe that even if Chorisis furnishes the true explanation of the symmetry of Fumariaciæ, which I hold to be very doubtful, there is no such relation between that order and Brassicaceae as would oblige

us to extend the principle to this latter, and I cannot but conceive that a more probable explanation has been proposed.

My note on the structure of Primulaceæ relates to one point which I have not seen rightly explained. In this order the stamens are observed to be constantly opposite the petals, a circumstance which always seems to need some explanation. In the present case, I think it evident that it is due to the abortion of a circle of parts belonging to the intermediate position between the petals and stamens and alternating with both. A careful examination of almost any Primula, the Auricula affording an excellent example, shows that the coloured eye of the flower consists of a series of pieces like the petals, as it were fastened on to them, and in such an order that the middle of each arch of the eye is exactly placed between two of the petals. In the genus Aretia this is still more evident. In Samolus a set of abortive stamens occurs between the petals, and the same is the case with several species of Lysimachia; in Cyclamen this organ is also easily observed, and in Glaux the proper corolla as well as its double is suppressed. From these examples we are enabled ideally to restore the lost circle, where it is most completely suppressed, and thus to comprehend the true symmetry and the reason of a seeming departure from a general rule. In how many other cases of opposite circles a similar explanation may be justified, I will not presume to say. In respect to this order I think it entirely satisfactory, but it is not the only one conceivable, for any one who has carefully considered a Camelia, in which the numerous circles of petals, instead of alternating as is usual, are forced into regular lines radiating from the centre, will be ready to admit the possibility of parts which are normally alternate becoming opposite by a sort of twist, and what occurs occasionally as a variety, may occur uniformly or nearly so, from a like cause, more constantly operating on a particular tribe, so that we are by no means driven to imagine without evidence an intermediate circle, in every instance of opposite parts, nor is there any necessity for assuming the occurrence of Chorisis where it cannot be distinctly proved.

THE RELATION WHICH CAN BE PROVED TO SUBSIST BETWEEN THE AREA OF A PLANE TRIANGLE AND THE SUM OF THE ANGLES, ON THE HYPOTHESIS THAT EUCLID'S 12TH AXIOM IS FALSE.

BY THE REV. GEORGE PAXTON YOUNG, M.A.,
PROFESSOR OF LOGIC AND METAPHYSICS, KNOX COLLEGE, TORONTO.

Read before the Canadian Institute, 25th February, 1860.

I propose to prove in the present paper, that, if Euclid's 12th Axiom be supposed to fail in any case, a relation subsists between the area of a plane triangle and the sum of the angles. Call the area A; and the sum of the angles S; a right angle being taken as the unit of measure. Then

A = k (2-S);

k being a constant finite quantity, that is, a finite quantity which remains the same for all triangles. This formula may be considered as holding good even when Euclid's 12th Axiom is assumed to be true; only k is in that case infinite.

Before proceeding with the proof of the law referred to, I would observe, that, while on the one hand Euclid's 12th Axiom is assuredly not an Axiom in the proper sense of the term, that is, not a selfevident truth, on the other hand it has never been demonstrated to be true. I even feel satisfied, from metaphysical considerations, that a demonstration of its truth is impossible. Legendre's supposed demonstration, which Mathematicians appear to have accepted as valid, was shown by me, in the Canadian Journal for November, 1856, to be erroneous.' * For the sake of those who may not have the former

In an Essay on Mathematical Reasoning, appended to his Mathematical Euclid, Dr. Whewell refers to the attempts which have been made to dispense with Euclid's 12th Axiom. "No one," he writes, "has yet been able to construct a system of Mathematical truth by means of Definitions alone, to the exclusion of Axioms; though attempts having this tendency have been made constantly and earnestly. It is, for instance, well known to most readers, that many mathematicians have endeavoured to get rid of Euclid's Axioms respecting straight lines and parallel lines; but that none of these essays have been gener. ally considered satisfactory." The last clause in this statement calls for remark. Sir John Leslie objected to Legendre's reasoning; but on grounds which (as Professor Playfair showed in the Edinburgh Review) are altogether frivolous. Playfair maintained that Legendre's proof was satisfactory; and since then, till the publication in the Canadian Journal of the article above referred to, mathematicians have-by their silence at leastacquiesced in his verdict. If Legendre's proof has been generally considered unsatisfactory, why did none of those by whom such a view was taken show where the reasoning is defective

numbers of the Journal at hand, the substance of my refutation of Legendre is given in an Appendix to the present paper.

PROPOSITION I.

The sum of the angles of a triangle AHE (Fig. 1) is not greater than two right angles.

FIG I

A

For, produce HE to F. Bisect AE in M. Bisect AE in M. Draw HMB, making MB HM; and join BE. In like manner construct the triangle CHE; N being the middle point of BE; and CN being equal to HN. In like manner construct the triangle DHE; P being the middle point of CE; and DP being equal to PH. And so on indefinitely. Denote by S, S,, S., &c., the sum of the angles of the triangles AHE, BHE, CHE, &c., respectively; and by A,, A., A,. &c., the angles HBE, HCE, HDE, &c., respectively. Then it is plain that the quantities S, S,, S., &c., are all equal to one another. Also, as the number n becomes indefinitely great, the angle A, becomes indefinitely small. For, the sum of all the angles in the series, A, A1, A2, &c., is less than AEF; and, since the series, A, A,, &c., may be made to contain an indefinite number of terms, those terms which are ultimately obtained must be indefinitely small, in order that AEF may be a finite angle. But, the exterior angle DEF being greater than the interior and opposite angle DHE, S3 cannot exceed two right angles by D. And S, S. Therefore S cannot exceed two right angles by D or A,. In like manner it may be proved that S cannot exceed two right angles by A,, whatever a be. And A, is ultimately less than any assignable angle. Therefore S cannot exceed two right angles by any finite angle whatsoever.

3

COR. 1.—If a line AE (Fig. 2) be drawn from A, an angle of a triangle ADF, to a point in the opposite side; and if the sum of the

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a right angle being taken as the position, s is not greater than 2.

E

unit of measure. But, by the ProTherefore S is not greater than S1.

COR. 2.-From B, a point within the triangle DAF, draw BC to a point C in AF; and let S2 be the sum of the angles of the triangle ABC. Then S, is not less than S. For, produce AB to E; and join EC. Then, by Cor. 1, S, is not less than the sum of the angles of the triangle AEC; which sum, again, is not less than S1, or the sum of the angles of the triangle AEF; and S, is not less than S. Therefore S, is not less than S.

2

PROPOSITION II.

If any triangle CHE (Fig. 3) have S, the sum of its angles, equal to two right angles, every triangle has the sum of its angles equal to two right angles.

For, CE being a side which is not less than any other side of the triangle CHE, let fall HD perpendicular on CE. Then HD cannot fall without the base CE; else (supposing it to fall beyond E) the an

FIG 3

N

gles CEH would be greater than a right angle: hence, because CE is not less than CH, the angle CHE would be greater than a right angle so that S would be greater than two right angles: which (Prop. I.) is impossible. Produce CD to F; making DF = CD.

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