514 
Proceedings of the Koyal Society of Edinburgh. [Sess. 
When q is given, then, q _1 may be infinite, but even in this case (4) and 
(5) hold; in general q~ 1 is finite; and when it is finite it has a unique 
value such that qq ~ 1 = 1 = q~ x q. 
The reader may have a suspicion that h {m) is in no real case zero. That 
it is sometimes zero can be seen thus : According as vv = ± vv we have 
or 
(xv + yv')(xv + yv) = x 2 v 2 + y 2 v' 2 
(xv + yv) = ————— . 
x l v l + y L v 1 
Choosing v and v so that v 2 = d= v 2 , we have 
(v + if) -1 = 00 . 
Indeed, choosing v , v , so that v 2 = — v 2 and vv = — vv , we have 
(v + v) 2 = 0. We can clearly choose v, v in a number of different ways to 
satisfy these conditions whether i 2 = +1 or — 1 (n being >2 for (h + i/) -1 = oo 
and > 3 for ( v + v) 2 = 0 when l 2 = — 1). 
In § 11 below I give an account of my somewhat unsuccessful attempts 
to simplify the problem of finding q~ x when q is given. This much seemed 
necessary for our immediate purpose. 
[One remark seems desirable. When the reciprocal of any self -conjugate 
or of any self-reversate is known, then also is the reciprocal of any 
multenion, for 
qr 1 = Kq(qKq)-' = Q 2 ( 2 Qff)^ = (K q.q)-'Kq = (Q q^Qq]. 
Before considering the rigid replacement, note that q( )q 1 is not in 
general commutative with S OJ that is, we have not in general 
q(S a r)q~l = S aiqrq- 1 ), 
as we are tempted to assume from quaternion analogy. We can prove 
this by the following particular case. Put 
q = cos \ 0 + q sin J 6 , = cos J 0 - sin when t 2 = — 1, 
q = cosh \ 6 + q sinh q~ l = cosh — q sinh J 6 when l 2 — +1. 
It will be found that we have 
respectively. 
qi 2 q~ ' x = t. 2 cos 6 + qq sin 0 , 
qc 2 q~ l = q cosh 6 + t x t 2 sinh 0, 
If, however, q is a jictor product, that is, q = cqct 2 . . . . a a where a v a 2 
.... are fictors, q( )q~ x is commutative with S & . To prove this we have 
to show that if v b is a multit of order b,qv b q _1 is of order b, that is, 
cqcq .... 0>a v b a a~ 1(l a-l~ 1 • • • • a i _1 
