511 
1907-8.] Algebra after Hamilton, or Multenions, § 3. 
For some purposes it is more convenient to suppose quaternions included 
in multenions, thus: Do not assume q 2 = q 2 =— 1, but assume (1) below. 
Put i — qq _1 , j — q* 3 -1 , h = qq _1 , or 
w + xi + yj + zk = w + £t, 3 i 2 _1 + 2/qi 3 -1 + zqq -1 * 
Unless the contrary is stated, we shall always suppose that 
(Law A) l 1 2 = t 2 2= B±l. ..... (1) 
[We leave ourselves the liberty, very rarely availed of, to abandon 
Law A. For a motor calculus it would be convenient to put 
qq> • • ■ • — 1 = q 2 — q 2 = q J == q 2 — • • . . 
but we nowhere, except in the Supplement, adopt this below.] 
Is there any good reason for choosing the upper or the lower sign in 
(1) ? What is the reason in quaternions why the lower sign is adopted ? 
It is because if we assume / 1 2 = / 2 2 = -4-1 we obtain without any ambiguity 
that (q* 2 ) 2 = — 1 , so that if we would have 
q 2 = t 2 2 = ( t i t 2 ) 2 
we must put each equal to — 1. 
If we suppose N to be given, we are similarly compelled to accept 
definitely the upper or the lower sign in (1), as will be shown immediately. 
Meanwhile we may say that for most purposes N may be taken as large 
as we please, and in that case there is no reason on present grounds for 
either sign in preference to the other. I have decided, however (at a little 
inconvenience in some applications), to leave the sign ambiguous. When 
the ambiguity affects the form of an expression below, it will be rendered 
evident by the presence therein of i 2 which will be put for any one of the 
equals q 2 , l 2 , .... Moreover, it is almost invariably true (though 
generally not obvious) that a formula in which l 2 does not explicitly occur 
is true even when Law A is not assumed, but only Laws (1) to (4) of § 2. 
To show how the magnitude of N governs the sign of (1) put 
— qq • * • • L rv> ^N = t l l 2 ••••%• . • • (2) 
By transpositions it is evident that even when Law A is not assumed 
n(n— 1 ) 
CJ2 = (-) * vV l n .... (3) 
or when Law A is assumed 
n(n— 1 ) 
£T 2 = (-) 2 (±l) n (4) 
Hence 
N(N-l) 
®n 2 = (-) 2 (±1) N 
(0 
