548 
Proceedings of Royal Society of Edinburgh. [sess. 
In the course of some recent investigations I noticed that if <f> 
have real roots, so also has 
whatever real strain if/ may be. This is, of course, obvious, for 
they are if/a, if/fi, i fry, if a, /3, y he the roots oi <f>. At first sight 
this appeared to me to he a generalisation of the theorem above, 
of a nature inconsistent with some of the steps of the proof. 
But it is easy to see that it is not so. For all expressions of 
the form 
if/onfr' 
correspond to pure strains if oo is pure. Hence 
if/cfnf/ -1 — ifrwwif/ -1 — if/anf /' . xf/'^unf/" 1 
and is thus, as required by the theorem, the product of two pure 
strains. 
Of course we might have decomposed it into other pairs of 
factors, thus 
\f/(jL>ifr ~ l . if/unf/' 1 , if/oy^g 1 . ■^mif/~ 1 J etc. 
In the former case the factors have each three real roots, in the 
latter they have not generally more than one. 
A great number of curious developments at once suggest them- 
selves, of which I mention one or two. 
Thus, let there be three successive pure strains (which may 
obviously represent any strain). We may alter them individually, 
as below, in an infinite number of ways without altering the 
whole. 
(Dtojtog = Wf 1 . 0)2000)! . 00 2 = 00 . 0)20)20)! . OOf 1 
= OOf 1 0) -1 0)f 1 . 0)20)0)20)2" 1 0) 1 0)0)2 . 00 2 
= OOf 1 00~ 1 00f 1 . 0)2000)20)0)2 . 0) 2 = etc. 
The expression oouj itself, when its three roots are given, i.e., 
a, /?, y with g v g 2 , g 3i gives oo and m separately, with three 
independent scalars. For we may take 
0 op = ^aSa p + (SS/Sp + . . . 
rap = ?/iY/3yS/Iyp + y. 2 VyaSyap + .... 
and then obviously there are three conditions only, viz. 
