It is of much importance, in connection with Legendre's view, to remember the point mentioned at the end of ANGLE, namely, that angle, of all magnitudes, is the only one of which number is a function, and the converse.
The proof of M. Bertrand is as follows :—Let it be granted that two spaces, whether finite or infinite, are when one can be placed upon the other so that any point whatsoever of either lies upon a point of the other : that is, let it be legitimate to say, of the word equal as thus used, that spaces which are equal to the same space are equal to one another. Granting this, it is easily shown-1, that the infinite spaces of equal angles are equal; 2, that of two angles, the infinite space contained in the greater is greater than the infinite space con tained in the less.
Let there be two lines, 0 F, A o, making with o A internal angles F 0 A, a A 0, equal to two right angles. Then 0 F and A G are parallel.
Take A B, B C, Co, &C., each equal to 0 A, and make the angles H B 0, KOD,LDE, itc., all equal to r o A or GAD. Let all lines with arrow heads be produced without limit in the direction to which the arrow head points. Then if o A be placed on A B, the lines o r and A a will in their new positions coincide with A G and B n, or the infinite space FOAG is equal to the infinite space OABH: and similarly H B C K, K C D L, itc., are all equal to one another and to F o A a. But it is obvious that no number of these spaces, however great, will fill up the infinite space of the angle F a E. Now let a line o 1 be drawn in such a manner that 1 o A and a A o are together less than two right angles ; whence o I falla nearer to o E than does 0 F. Take F o 2 double of 01, F 0 3, treble of F 0 1 and so on : whence, since of two quantities which bear a ratio, the less may be multiplied so as to exceed the greater, some multiple of F 0 1 must be greater than F 0 E, whence some multiple of the infinite space F o I is greater than the infinite space FOE. But no multiple of roAo will he so great as FOE; whence the infinite space F 0 1 must exceed the infinite space F 0 A O. Therefore 0 1 produced must cut A a ; for if not, the space F o 1 would be entirely contained in F 0 A a, and the former could not exceed the latter.
We have not noticed the numerous attempts at the solution of the difficulty which proceed by tacit assumption or illogical process, under the pretence of aruiding the axiom of Euclid, without substitution of any other. W. shall conclude this article with an account of a result
tantalised In a paper, by Legendre, in the 12th volume of the • Memoirs the . Institute,' being is latest attempt at the volution of the problem. It is %bet he properly styles a geometrical translation of the analytical ',reef already noticed, and its chain of reasoning proceeds through three prepositions.
1. It is imp oodhle that the sum of the angles of a triangle can in any osae exceed two right angles.
2. If there be any one triangle which has angles together equal to two right angles, the same must be true of all triangles.
3. If the angles of a triangle be not equal to two right angles, then angles alone may be made to determine a straight lino.
1. If it be possible, let there be • triangle B A o having its angles together greater than two right angles. In A o produced take c E, E a, n t, &a each equal to a c, and make the angles D o B, tt o, &o. each equal to B • C, and let (.1), E r, &c. be each equal to A B. Join B D, D F, &C.
with r v the angles of P V x are together equal to two right angles, and because rvx has an angle in common with r Q n, the angles of r Q B are also equal to two right angles. Hence, if any one triangle have its angles equal to two right angles, the same must be true of all triangles whatsoever.
3. If then we deny the preceding truth in the case of any one triangle, we must deny it in the case of all. Let it then be which are not known to be in the same straight lino, though they are so. Then the triangles D c E, F H 0, &e. are severally equal in all respects to B A c. Aud because the angles at A, B, c, are together (by hyp.) greater than two right angles, and two of them are equal to two of the angles at c, it follows that the angle A is o is greater than n C D, whence A B and B C being severally equal to D C and C B, it follows that A e is greater than 11 D. Similarly c E is greater than D r, whence • l exceeds B D 11. 11 K by as many times the excess of A C over IS 13 as A c is contained in A r. Now however small this excess may be, it may be multiplied until the multiple exceeds twice A B, or A B and K I together. That is, A I, one side of a figure, may be made greater than A n, a D, D r . K I, the sum of all the other sides ; which is absurd. Hence, it is not true that the angles of any triangle are together greater than two right angles.