567 
1907-8.] Algebra after Hamilton, or Multenions, § 11. 
this 2 n -tic in q involves even positive powers of the arbitrary scalar x. 
The coefficient (a rational integral function of q — Sq) of every such power 
is separately zero. From any such equation satisfied by q, q~ x may be 
derived, and this is a solution of the problem, given q, required q -1 . But 
that it is practically a very clumsy one is obvious from taking n — 10, in 
which case the 2 n -tic is of degree higher than a thousand, and it involves 
a determinant the number of whose constituents is over a million. 
We saw in § 4 that to find q~ l we need only find ( Kq.q )~\ The 
problem of finding p~ x when p is self -conjugate is slightly but not much 
simpler than the general problem. Thus let p be a given self -conjugate 
multenion, r an arbitrary self -conjugate multenion, and let 
i f,r = l(pr + rp) (3) 
Then \fr is a self -conjugate multilinity and \fsr a self -conjugate multenion. 
Such a multilinity clearly satisfies an equation of degree equal to the 
number of self -conjugate multits, that is 
2 n— 1 4- 2 i(n_1) cos \{n - i 2 )7r. 
p satisfies the same identity. The roots are all real. When they recur, if 
a v a 2 , .... a k are the different roots, then 
(P ~ a i )(P ~ « 2 ) (p-a k ) = 0 . . . . (4) 
with no repetitions of roots. If we put 
Pi = (p- a 2)(P ~ a s) (P~ «n)/(«i “ « 2 )( a i ~ « 8 ) K “ a n) • (5) 
and similarly for p 2 , p s , . , then the following are true, 
PPl = a i Pl> PP ‘2 = a 2 Pv • • • • ( 6 ) 
Pl=PvP2=Pv ,lhP2=P2Pl = ° ( 7 ) 
l=Pi+P2+ +Pk ( 8 ) 
P = a iPi + a 2P2 +••••+ <hPk • • • (9) 
more generally ; always for positive integral values of m ; also for negative 
integral values of m, when no a is zero ; 
If a x = 0 
or generally 
p m = a l m p 1 + +a%p k . 
Lt(l +xp )~ 1 = p 1 
X=<x> 
Lt[l + x(p - a,)] -1 . 
3=00 
( 10 ) 
(ii) 
Up to ^ = 4, (p — Sp) 2 is a scalar, so that the inversion of p is simple. 
This suggests that given n, the maximum value of k is decidedly smaller 
generally than the number 2 n_1 + 2 l(n_1) cos \{n — i 2 ) 7 r mentioned above. 
Thus with n = 4 this last is 6, whereas k is 2. What the maximum value 
