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 September 11th, 2017, 02:34 PM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Fundamental Theorem of Algebra Proof Fundamental Theorem of Algebra for Real or Complex Coefficients by Induction . $\displaystyle z^{n}+a^{n-1}z^{n-1}+....a^{1}z +a^{0}=0$ $\displaystyle z^{n}+a^{n-1}[(z-b_{1})(z-b_{2})...(z-b_{n-1})]=0$ $\displaystyle z^{n}=-a^{n-1}[(z-b_{1})(z-b_{2})...(z-b_{n-1})]$ which has exactly n roots Ref https://en.wikipedia.org/wiki/Fundam...rem_of_algebra
September 11th, 2017, 03:01 PM   #2
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 Originally Posted by Your reference In spite of its name, there is no purely algebraic proof of the theorem, since any proof must use the completeness of the reals (or some other equivalent formulation of completeness), which is not an algebraic concept.
I think your attempt has other problems too.

September 11th, 2017, 05:18 PM   #3
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 Originally Posted by zylo Fundamental Theorem of Algebra for Real or Complex Coefficients by Induction . $\displaystyle z^{n}+a^{n-1}z^{n-1}+....a^{1}z +a^{0}=0$ $\displaystyle z^{n}+a^{n-1}[(z-b_{1})(z-b_{2})...(z-b_{n-1})]=0$
How do you know $a^{n-1}$ isn't $0$? Like $z^{46} + z^2 + 1$, say. If $a^{n-1} = 0$ then you can't pull it out of the last $n-1$ terms like you did. And you can't fix this by trying $a^{n-2}$ because that might be $0$ as well. So at the very least you have to take the first nonzero coefficient you find and divide through by that. Your proof will get more complicated.

Last edited by Maschke; September 11th, 2017 at 05:21 PM.

 September 11th, 2017, 05:55 PM #4 Senior Member   Joined: Aug 2012 Posts: 2,342 Thanks: 731 ps -- The FTA has been on my mind a bit lately ever since someone posted this thread about using the Intermediate Value theorem (IVT) to prove that some particular polynomial has a zero. In that particular case it was easy to find a couple of input values whose corresponding output had opposite signs. And it was an odd degree polynomial so you can just look at the limiting behavior as the input goes to infinity in both directions, which someone noted. I noted that for very small (in absolute value) input values, a polynomial is dominated by its constant term; and for large values, it's dominated by its highest-order term. So that if the degree is odd, we have our proof. That observation happens to be at the heart of one of the standard proofs of FTA. If $\sum_{i=0}^n a_n z^n = 0$ where $z \in \mathbb C$, then a tiny region around the origin is mapped into a tiny little region about $a_0$. And if $z$ is very large, a tiny region about the origin is mapped to a huge region of the plane that must include the origin, because the output is unbounded. You just take $z$ as large as you need it to be. As $z$ goes from very small (in absolute value) to very large, there must be some particular $z$ such that the boundary of the output region necessarily crosses the origin. That's our zero! As you can see, this argument depends crucially on the topological properties of continuous functions on the plane. If there's a hole at the origin where there should be a point, the proof fails. Every known proof of FTA depends on concepts from analysis. It's interesting (as Wiki notes) that Gauss gets credit for the first rigorous proof (by the standards of the time) in 1799; but the last little detail wasn't addressed completely till 1920. Last edited by Maschke; September 11th, 2017 at 06:18 PM.
 September 11th, 2017, 07:55 PM #5 Senior Member   Joined: Aug 2012 Posts: 2,342 Thanks: 731 Now suppose we try to make your idea work for polynomials with all nonzero coefficients. Then you can indeed divide through by $a_{n-2}$. [Sorry I wrote those as superscripts above by mistake, I hope that's clear. Oh wait YOU made that mistake and I just copied it!] In your third line you have some other polynomial and you claim by induction that the right hand side has $n - 1$ roots. I agree. But what's the $n$-th root? Are you saying it's $0$? But $0$ is not a root of the original poly you started with. What do you claim is the $n$-th root? How have you demonstrated that the ORIGINAL poly you started with has $n$ roots? I don't see how you are establishing your inductive step even for the special case of all nonzero coefficients. Thanks from JeffM1 Last edited by Maschke; September 11th, 2017 at 08:02 PM.
September 12th, 2017, 06:19 PM   #6
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 Originally Posted by Maschke I don't see how you are establishing your inductive step even for the special case of all nonzero coefficients.
I think his reasoning is that there is one root of every polynomial of degree one when all coefficients (except possibly the constant term) are non-zero. That justifies assuming that there exists at least one integer k such that there are k roots of every polynomial of degree k when all coefficients (except possibly the constant term) are not zero. Then his n is k + 1 and k = n - 1. Consequently, he can indeed factor the sum of all of the terms of degree less than n into a product of n factors. He moves that product to the other side of the equation. So far so good.

He sees n factors on one side of the equation and concludes n roots by relying on (1) the zero product property, which does not apply except in the trivial case where z = 0, plus (2) the possibility that the coefficient a_(n-1) may be 0, which reduces to the trivial case and is excluded by hypothesis.

If my understanding is correct, his reasoning is completely bogus.

Last edited by JeffM1; September 12th, 2017 at 06:23 PM.

 September 14th, 2017, 10:56 AM #7 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 OP attempt led nowhere. But sometimes you have to plunge in. I wasn't fishing, I thought I could do the geometry. I next tried dividing $\displaystyle P_{n}(x)$ by $\displaystyle (x-d)$ thinking to select $\displaystyle d$ to make remainder 0. This led to a standard result of algebra $\displaystyle P(x)=(x-d)Q(x)+R(d)$. BuT $\displaystyle R(d)=P(d)=0$ and you are right back where you started from. But that suggested the next step which gave OP proof, illustrated by example. Example 1 Divide $\displaystyle x^{4}+3x^{3}+5x^{2}+7x+9$ by $\displaystyle x^{3}+ax^{2}+bx+c$. Think of numerical coefficients as symbols to make algebra (latex) easier, and illustrate patterns. After 2 steps of standard long division process, $\displaystyle x+(3-a)$ is the quotient and the remainder is set to 0 by setting coefficients of $\displaystyle x^{2}, x,$ and constant term to 0: 1) $\displaystyle (5-b)-(3-a)a=0$ 2) $\displaystyle (7-c)-(3-a)b=0$ 3) $\displaystyle 9- (3-a)c=0$ From 3), $\displaystyle c=\frac{9}{3-a}$. Sub into 2) to get $\displaystyle b(3-a)^{2}+7(3-a)+3=0$, which gives $\displaystyle (3-a)=\frac{k}{b}, b=\frac{k}{3-a}$. Sub into 1) to get a cubic in $\displaystyle a$ which is solvable. Example 2 Divide $\displaystyle x^{5}+3x^{4}+5x^{3}+7x^{2}+9x+11$ by $\displaystyle x^{4}+ax^{3}+bx^{2}+cx+d$. Think of numerical coefficients as symbols to make algebra (latex) easier, and illustrate patterns. After 2 steps of standard long division process, $\displaystyle x+(3-a)$ is the quotient and the remainder is set to 0 by setting coefficients of $\displaystyle x^{3}, x^{2}, x$, and constant term to 0: 1) $\displaystyle (5-b)-(3-a)a=0$ 2) $\displaystyle (7-c)-(3-a)b=0$ 3) $\displaystyle (9-d)- (3-a)c=0$ 4) $\displaystyle 11-(3-a)d=0$ Work backwards from 4) following the same procedure as in Example 1 to get a cubic in "a," which is solvable. The procedure and pattern is obvious, always leading to a cubic in "a," and holds for any complex polynomial because only the simple arithmetic of a field was used. Conclusion: 1) Any polynomial is divisible by x-t for some t. 2) Then any polynomial $\displaystyle P_{n}$ is expressible by n factors $\displaystyle (x-a_{i})$. Standard proof by induction. 3) Any (not just to quartic) polynomial is regressively solvable for its n roots by the elementary (but complicated) procedure above. Last edited by skipjack; September 19th, 2017 at 07:07 AM.
September 14th, 2017, 11:14 AM   #8
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Quote:
 Originally Posted by zylo 3) Any (not just to quartic) polynomial is regressively solvable for it's n roots by the elementary (but complicated) procedure above.
That contradicts a standard result known since Niels Henrik Abel's 1824 proof of the unsolvabiity of the quintic.

It's one thing to provide a false proof of something known to be true. The next level downward is to provide a false proof of something known to be false.

Or as Lloyd Bentsen might have said: I knew Niels Henrik Abel. Niels Henrik Abel was a friend of mine. And you Sir are no Niels Henrik Abel.

 September 14th, 2017, 12:54 PM #9 Senior Member   Joined: Aug 2012 Posts: 2,342 Thanks: 731 Can you please walk through your method to find a zero of $f(x) = x^5 â€“ x â€“ 1$ This example of an unsolvable quintic was given by Van der Waerden. https://en.wikipedia.org/wiki/Galois_theory Last edited by Maschke; September 14th, 2017 at 01:28 PM.
 September 14th, 2017, 03:15 PM #10 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra The method may work, but it does rely on one being able to find $t$.

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