Saturday, May 20, 2006

Fundamental Theorem of Algebra: The Proof

In today's blog, I complete the proof for the Fundamental Theorem of Algebra. In my next blog, I will use this result to factor Fermat's Last Theorem into cyclotomic integers.

Today's proof is taken from David Antin's translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

Lemma 1: if an algebraic equation f(x) has a root α, then f(x) can be divided by x-α without a remainder and the degree of the result f'(x) is less than the degree of f(x).


(1) Let f(x) = xn + a1xn-1 + ... + an-1x + an

(2) Let α be a root such that f(α) = 0

(3) Now, if we divide the polynomial by (x-α), we get the following (see here if proof needed):

f(x)/(x - α) = f1(x) + R/(x-α)

where R is a constant and f1(x) is a polynomial with order n-1.

(4) Multiplying both sides with x-α gives us:

f(x) = (x - α)f1(x) + R

(5) Now, if we substitute α for x we get:

f(α) = 0 which means that the constant in the equation is 0 so R = 0.


Theorem: Fundamental Theorem of Algebra

For any polynomial equation of order n, there exist n roots ri such that:

xn + a1xn-1 + ... + an-1x + an = (x - r1)(x - r2)*...*(x - rn)


(1) Let f(x) = xn + a1xn-1 + ... + an-1x + an

(2) We know that f(x) has at least one solution α1. [See here for proof]

(3) Using Lemma 1 above, we know that:

f(x)/(x - α1) = f'(x) where deg f'(x) = n-1.

So that we have:

f(x) = (x - α1)f'(x)

(4) But we know f'(x) has at least one solution (from here) so we can repeat steps #2 and #3 to get:

f'(x)/(x - α2) = f''(x) where deg f''(x) = n-2.

which combined with step #3 gives us:

f(x) = (x - α1)(x - α2)f''(x)

(5) Eventually we get to the point where the degree of fn(x) = 1.

In this case, fn(x) = x - αn.

(6) This establishes that there are n roots for a given equation f(x) where the degree is n.

(7) Putting this all together gives us:

f(x) = (x - α1)(x - α2)*...*(x - αn)

(8) Now, since f(x)=0 only when one of the values αi=x, we see that the n roots αi are the only solutions.

(9) So, we have proven that each equation is equal to n roots.

One important point to remember is that the n roots are not necessarily distinct. That is, it is possible that αi = αj where i ≠ j.



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