22 
Proceedings of Royal Society of Edinburgh. [sess. 
by reason of tbe old identity (used also by Cayley in this paper), 
356. 416 - 346.516 + 316.546 = 0. 
Consequently, as a first step, the equation in question is shortened to 
145.256 . 423 . MI + 245.123. 346.516 - 245.156.423.361=0. 
The process of condensation, however, can be continued, and 
that in two different ways. If we combine in the first instance the 
last two terms, which are seen to have the common factor 245 . 516, 
we read the result 
145 . 256 . 423 . 361 + 245. 516.143. 263 = 0 , 
or, say, 
145.136.234.256) 
+ 143 . 156 . 24 5 . 23Q j 
(A) 
whereas if we combine the first and last terms, which have the 
common factor 423. 361, we obtain 
or, say, 
423 '361- 125. 456 + 245 • 123 • 346 • 516 = 0 , 
125. 136.423. 456 ) 
-123* 156.425. 4Mj 
(B) 
Of these equally simple forms (A) and (B), the latter is the one 
previously obtained, so that the requisite reconcilement is effected. 
Since, however, the left-hand side of Cayley’s equation has been 
shown to be reducible either to the left-hand side of (A) or to the 
left-hand side of (B), a new question arises, viz., How it comes that 
the two reduced forms are the same, or how it is that 
145. 136.234. 256 + 143 . 156 . 245 . 236 
- 125. 136.423. 456 + 123 • 156 • 425 . 436 
vanishes identically? One answer is, that the aggregate of the first 
and third terms being 
= 136.234 (145.256 - 125.456) 
= 136. 234. 165. 425, 
and the aggregate of the second and fourth being 
= 156.245 (143. 236-123.436) 
= 156.245 . 163.423, 
the two aggregates differ only in sign, and consequently their sum 
is zero. 
