170 
paper. In fact, I at first imagined that the proof of them was 
necessary to my purpose. They are obtained as follows :— 
_* In virtue of the laws of combination of the imaginaries 
t, j, k, we have 
(ta ae jb a1 ke pee — (- Pye (a? a3 b+ ey rs (8) 
Now, the coefficient of a?4#c” in the left-hand member of this 
equation is = (2A, 2u, 2v), in conformity with the notation 
explained in p. 165: and the same coefficient in the right-hand 
member is plainly 
ies (r tpt v) ! 
Tillie Tp Ge 
Consequently, we have the theorem I. 
sie A+ ut v)! 
= (2A, 2u, 2v) = (- 1)* e ae 
‘‘ Multiplying both sides of the equation (8) by 2a +jb+ ke, 
we get 
(ia +jb + he)? = (— 1)" (ta +b + ke) (a? + B+ ye, 
The coefficient of a*10*c” in the left-hand member is 
= (2A +1, 2u, 2v); and the same coefficient in the right-hand 
member is 
(- 1pm CD 
We have, therefore, I1., 
S(2A+1, Qu, 2v) =(- ae ee i, 
and similar expressions for 7 and £. 
« Again, multiplying (8) by (éa+jb + ke)? =— (a? + B* +c). 
we get 
(ia + jb + ke) — (— 1) sl (9? + B+ 2) se 
The coefficient of a6" c” in the left-hand member is 
=(2A+1, 2u+1, 2v); whilst in the development of the 
right-hand member no such term appears, as all the exponents 
of a, b, c, must be even numbers. We have, therefore, II. 
— ee 
