September 11th, 2017, 02:34 PM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88  Fundamental Theorem of Algebra Proof
Fundamental Theorem of Algebra for Real or Complex Coefficients by Induction . $\displaystyle z^{n}+a^{n1}z^{n1}+....a^{1}z +a^{0}=0$ $\displaystyle z^{n}+a^{n1}[(zb_{1})(zb_{2})...(zb_{n1})]=0$ $\displaystyle z^{n}=a^{n1}[(zb_{1})(zb_{2})...(zb_{n1})]$ which has exactly n roots Ref https://en.wikipedia.org/wiki/Fundam...rem_of_algebra 
September 11th, 2017, 03:01 PM  #2  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,939 Thanks: 2266 Math Focus: Mainly analysis and algebra  Quote:
 
September 11th, 2017, 05:18 PM  #3 
Senior Member Joined: Aug 2012 Posts: 1,527 Thanks: 364  How do you know $a^{n1}$ isn't $0$? Like $z^{46} + z^2 + 1$, say. If $a^{n1} = 0$ then you can't pull it out of the last $n1$ terms like you did. And you can't fix this by trying $a^{n2}$ 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: 1,527 Thanks: 364 
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 highestorder 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: 1,527 Thanks: 364 
Now suppose we try to make your idea work for polynomials with all nonzero coefficients. Then you can indeed divide through by $a_{n2}$. [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. Last edited by Maschke; September 11th, 2017 at 08:02 PM. 
September 12th, 2017, 06:19 PM  #6  
Senior Member Joined: May 2016 From: USA Posts: 785 Thanks: 312  Quote:
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_(n1) 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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88 
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 $\displaystyle (xd)$ thinking to select $\displaystyle d$ to make remainder 0. This led to a standard result of algebra $\displaystyle P(x)=(xd)Q(x)+R(d)$. 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, $\displaystyle x+(3a)$ is the quotient and the remainder is set to 0 by setting coefficients of $\displaystyle x^{2}, x,$ and constant term to 0: 1) $\displaystyle (5b)(3a)a=0$ 2) $\displaystyle (7c)(3a)b=0$ 3) $\displaystyle 9 (3a)c=0$ From 3), $\displaystyle c=\frac{9}{3a}$. Sub into 2) to get $\displaystyle b(3a)^{2}+7(3a)+3=0$, which gives $\displaystyle (3a)=\frac{k}{b}, b=\frac{k}{3a}$. 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, $\displaystyle x+(3a)$ 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) $\displaystyle (5b)(3a)a=0$ 2) $\displaystyle (7c)(3a)b=0$ 3) $\displaystyle (9d) (3a)c=0$ 4) $\displaystyle 11(3a)d=0$ 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 xt for some t. 2) Then any polynomial $\displaystyle P_{n}$ is expressible by n factors $\displaystyle (xa_{i})$. 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  
Senior Member Joined: Aug 2012 Posts: 1,527 Thanks: 364  Quote:
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: 1,527 Thanks: 364 
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: 6,939 Thanks: 2266 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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