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GmIDytP af, (Y, 2) =f, (y — jt, z- ke), (4) . 
gr IPukD? f(y, z) =f, (yt ja, 2+ har). (5) 
Sir William Hamilton at once perceived the inaccuracy of 
these results, and referred it to its origin, which was the erro- 
neous supposition, that 
gidvikDs _ giDa gkDs, 
This last equivalence does noé subsist, because the symbols 7 
and # are not commutative. 
‘< Indeed, the consideration of a simple case might lead to 
the suspicion that the formula (4) was incorrect. Suppose that 
Si (y, z2)=yz: it becomes at once a question, what is the 
meaning of fi(y—jx, z-ha)? Is it (y—ju)(z-kax), or 
(z-khx)(y—jx)? for these expressions have different values. 
Thus, in the first instance, it is apparent that the assigned re- 
sult is ambiguous. - But from what follows it will appear that 
neither (y —jw) (z— zx), nor (z —kx)(y — jx), is equivalent to 
“EID ykDs) yz, 
‘* The question relative to the interpretation of the sym- 
bolic formule (2) and (3) being in this state, I have endea- 
voured to resolve it in the present Paper. 
‘* As a first step in our investigation, let us inquire what is 
the effect of the symbol, 
GME s, 
or 7, as it will be more convenient to denote it, upon the term 
y” 2", m and n being positive integers. 
‘«‘ Beginning with simple cases, we shall find by actual ex- 
pansion of the exponential symbol, 
Ll. x y™=(y4+ jr), and 7 2°=(2+ka)". 
2. 7 yz=yz+juz + kay. 
3. mt PZ=Y2 + Wayet hay? — 22 — tha’. 
4.07 PB=yr2z? + Ways + Skay?z? — x25 — 3a°y?z — kasz* 
— Qja®yz — katy? + avtz + thas. 
