Malefs Proof that every Equation has Boots. 
223 
§ 1. Malet's proof is (in substance) : — 
luet f{x) =aoX'^ + a^x''~' + ... +a,, be an algebraic function of x of degree 
n, with the coefficients a^, a^,... a„ real. We proceed to prove that f(x) = 0 
always has a root, real or imaginary. 
We know by the principle of continuity that if n is odd, f{x) = 0 has a 
real root. If n is even, let x = a + [3. 
Then 
f{x) EE + A.ft"-' + A^/S"-^ + . . . + A„, 
where 
A»./(a), 
and 
f^(a) 
A„_i = / • (^^6 derived functions) ; Ao = 6to- 
Now try to determine a and /3 so that the even and odd parts of f{x) 
are separately 0. 
That is 
Ao/3'^ + A,/3'^-^+...+A„_,/3^ + A„ = 0 ] 
and h (i) 
/3(A,/3«-^ + A3/3'^-^ + ... + A„_3/3^ + A„_,) = 0) 
The possible solution /3 = 0, A^=/(a) = 0 we need not discuss. Other 
solutions are obtained by eliminating (3^ by the " dialytic " method, which 
gives an equation for a of degree n{n — i)/2. Moreover, the process also 
gives /3^ as a unique function of any root a of the final equation. There- 
fore, if a certain equation of degree n{n — i)l'2 has a root (real or imaginary), 
f{x) = 0 has two roots, x = a + fD. 
Now, if n is odd/(.9^) =0 has a real root. Assume n to have a factor 
2", then n{n — i)/2 has 2 only to the degree 
Thus f(x)-^0, when n contains 2^, has a root if a certain equation 
whose degree contains 2''"' has a root. It follows that this second 
equation has a root if one of degree containing 2"~' has a root, and as this 
process ultimately leads us to an equation of odd degree which has a (real) 
root, the proposition is proved. 
§ 2. This (Malet's) proof of the " Fundamental Theorem of Algebra " 
leaves nothing to be desired ; but the following detailed discussion of the 
work of eliminating [3^ by the dialytic method, including a more direct 
proof that f{x) = 0 has n roots real or imaginary, offers some points of 
interest. 
We shall give the work in detail for n = 8 for the sake of brevity ; but 
we shall show that the metbod is quite general. 
