286 Mr. Graves’s Reply to Prof. De Morgan’s Remarks 
(13.), assume v2 = ca, ¢ being some algebraic multiplier, 
since it is impossible to arrive at any simpler proof than the 
mere form of the proposition. 
For those to whom this assumption may not seem satisfac- 
tory, we may make the proof that f 2 = c¢ w rest upon dif- 
ferent data. Whatever be the form of & and value of 4, 
Wie ty +h)—V(e+y) 
h 
whether / be regarded as the increment of 2 or y. We have 
therefore in general 
dV (x+y) _ dv (r+y) 
PAS GT dy (14) 
or, in this particular case, performing the respective partial 
differentiations on the equivalent expression in (13.), we ob- 
tain 
the expression remains unaltered, 
d(Vatvy) dvr _dWartVy) dwy 
ani Pans or = ee eet (15.) 
dVa dy 
a? it must be independent 
but since is equal to 
da 
daz 
be equal to some quantity fed. We assume that the ge- 
neral form of fcdwz is ¢ + ¢.x, but 0 is the only value of cin 
the equation P2 = ¢ + cz, which will be found to satisfy 
equation (13.). Hence Pa = cz. 
Having satisfied ourselves in whatever way, that Va = cx 
is the general solution of (13.), we have, by (12.), Ya = 6+ex, 
Hence fbx or fx = f(§t+ex); but f(i+cx) = fb. f(c x) 
=1.f(er) =f(cr). Hence fr =f (cz). 
= ¢, then ‘x must 
of x, and therefore constant*. Let 
Q. E. D. 
Hence, if any function fx could be found to satisfy (9.), 
Ff (cx), ¢ being wholly arbitrary, would be the general solution 
of (9.). Equation (10.), which defines the base of the system, 
limits the otherwise arbitrary c to such values that fc may be 
equal to 1. Hence a* = f(cx), ¢ assuming in succession all 
the values of f—1a, and none other. 
The next step in the investigation is to find some function 
(no matter what) that fulfils condition (9.), and to determine 
the general form of its inverse. Such a function I find in 
cos 6 + s/f —] sin 6. 
* By similar considerations we might find at once from equation (9.), 
P : é dQu 
without having recourse to Taylor's theorem, that oa = cOxn, 
