(229 ) 
latter cannot of course be realized. Solving the equation 
(a, —Ab,) (a, —Ab,) — (a,,—Ab,,)? = 0 
we find: 
pn boch) HM Nb 
2(b, b, ay 2) 1 
This equation can be satisfied by a mal value of À if 
bb. (are de th, one ONE te Mt 
ia Bee tds % 12 0. 
4b, a ie b, 1 b, Ore b, Big > 
a 
a 
: . ° : ye e sl as 
This equation is certainly satisfied if — is both smaller than — 
12 Ji 
a : bo ; 
and than —, but it may also be satisfied in other cases. Let us 
b, 
a, at a, . 
assume that — > and ee pe If we then have: 
BIE bb 
12 12 2 
My, ty a Ee bb, (a, GoNs 
Wig b, b, Ds Ab,” b, b, 
a minimum value of 2 occurs indeed, but in this ease it corresponds, 
} ; ; 2 
according to our previous observations, to a negative value of jae 
5 SUH 
We arrive at the same result starting from the equation of Cont. II p. 20. 
For a ternary system we have, putting Ss 
xy 
(a, —ab, )(1—a«—y)? + (a, —2b,)x? + (a, —1b,)y? + 2(a,,—Ab,,)a(1-a-y) + 
+ 2(a,,—4b,,)e(1 — «—y)+2(a,,—4b,,)ay = 0. 
We may represent this by the sum of three squares: 
(a,b) Gra )+(4,.—46, ,)a+ (do Mra 
(a,—4b,) 
—Ab »— Ab, .)(a,3;—40,5)) |? 
v aren aa)" +y [ett Oreste he) 
a, Jb a, —db, 
ue a + 
(a,,—4,,)* 
(a,—Ab,) k — 
a,—db, 
| | Zeh ae 
as) | “Va 
—db 
SN Sa ae ‘ Ser 
: . a,—db, ( al ) (4,.— Ab)’ 
_ & ). —_—s———- _ SS 
= ; a, — db, 
In the case that a,—ab,>0 and (a,—ab,) (a,—Ad,) ) > (aa aD.) 
this equation cannot be satisfied if the coefficient of y* is positive. 
If this coefficient decreases to zero, then the equation is satisfied by 
only one set of values for ez and y, namely by those values for 
