518 
Proceedings of the Royal Society of Edinburgh. [Sess. 
In order that qp may be equal to p'q, q must satisfy the following n 
linear equations (and that it need satisfy no more conditions will be 
proved subsequently). 
qii=X 1 q,qi 2 = \ 2 q, (11) 
or (^-1)^0, (4> 2 -% = 0, (12) 
where <f> L , 0 2 . . . . are the “ multilinities ” (that is, linear multenion 
functions of a multenion) given by 
^> 1 7’ = X 1 n 1 -1 , cf) 2 r = X 2 Tt 2 ~ 1 , .... , . . . (13) 
Here if 1 , q _1 , .... have finite meanings for q -1 = i 2 .i h etc. 
(f> 2 , . . . . satisfy quadratic equations and are commutative, that is, 
<£l 2 =l> 02 2 =1> • ■ • •> = = 'Mr ( 14 ) 
Hence if we put 
g = ^r=2 -M (l + <£i)(l +<f> 2 ) .... (1 + <f* n ) 7 ’ • • • (15) 
q will satisfy all the equations (12). Here r is any multenion whatever. 
The doubt remains, of course, whether q~ x is finite. We will return to 
this immediately. Meanwhile note that by (14) and (15) \Js satisfies the 
quadratic equation 
= ( 16 ) 
since from (14) (1 + ^ 1 ) 2 = 2(1 + ^). Also notice that by expanding \fs in 
full in terms of <p v (p 2 , ... . 
2Y=2<M> 2 . . . • 
or from (13) we have q explicitly in terms of q, i 2 . . . . , \ v \ 2 , ... . thus 
q = if/r = 2~ n ^,vrv~ 1 (17) 
where v is any one of the 2 ?l multits and v is what v becomes when in it q 
is replaced by \ v i 2 by X 2 , etc. 
It is easy to show from (11) that qp=p'q. First suppose p is a multit 
such as qqqq. 
qp = qi i<'2 t 3 t 4 = ^■l ( I l 2 L 3 L 4: = = = ~P ( i’ 
Next suppose p — Hxu, 
qp - 'Zxqv = %xv'q=pq. 
It is also easy to show that by changing the sign of \ c if it should 
happen that 
{<t>c + 1 )(<A-i + 1 ) . • • • {4>i + l)r = 0 
q will not be zero. Thus put 
2i = (<£i + 1 )r, 2 2 = W> 2 +%i> , % n q = (<l>n+1.)q n -i 
and suppose that q c _ x is not zero but q c is. Thus 
q c - 1 =t 0, q c = Xpi^L- 1 + q c _ x = 0. 
