Double Products and Strains in Hyperspace. 9 
(n—3)rd class and the elements a; aj ax of the third class. And 
so on, until finally 
(6”) (ar; Og... Gn) (@1 Gg .+. @n) = 1 
These equations are entirely analogous to those obtained in (6) for 
the reciprocals themselves. The extension of the idea of reciprocal 
sets to other than the primary elements is therefore suggested; and 
if equations analogous to (5) be formed for such elements, the theorem 
may be stated that: The reciprocals of the combinatory products 
of the set of ~ primary elements are the products of the reciprocals 
of those elements. Or it may be preferable to regard this state- 
ment taken with the equations (6’) and analogous equations as the 
definition of the reciprocals without appealing to equations analogous 
to (5). The sum of the class-number of any one of a set of ele- 
ments and of the classnumber of any one of the reciprocal set is 7. 
Equation (6“) shows that (a’‘; ag... @’n) == 0, and hence the reci- 
procals a‘;, a‘s,..., @’n are themselves independent. From this 

1 The proofs are very simple. For instance to show that 
(a’ p' y’) («# By) = 1, it is merely necessary to analyze as follows. 
(a By’) (@By) = a B[y! (@ By)] = & B[(y7) (« B) + (7 &) (By) 
+ (7 B) (vy &)] = (a B’) (8); 
for y‘y is 1 and y/a and 78 are both 0. <A repetition of the process 
shows the desired relation. In a similar manner the other relations may 
be proved. All the relations are, however, but special cases of an im- 
portant formula. Let @, 8, 7, ... be elements of class »—1, in number 
less than or equal to ~, and let «, 8, y, ... be an equal number of ele- 
ments of the first class. The product («@fBy...) (@By...) is evidently a 
scalar and is given by the formula 
(«@By...) @By...) =| Ba Bp By... 
DOK Seg MTS al G Ys uc 
Consider («8y...) as a single element and apply the associative law and 
the rule for expanding : 
(a@By...) @py...) = (@By...a) By...) = [(«a) (By...) — (Ba) (Ginza) 
+ (ya) (@B...)—...] By...) 
By a repetition of this process on each of the terms of the form 
Cpe (ys): * (pee e y one (Baan) (BY oo.)e 
a further reduction is accomplished, and so on. The final result will 
clearly be equal to the determinant—in fact the step already taken 
appears as the expansion of the determinant according to elements of 
the first column. If this formula be used, the relations between the 
reciprocals are immediate. 
