Does the square root of negative one lead to a contradiction?

[Tom Mattson]

Why don't you look it up yourself? It's not that hard to find. Is your objective here to learn something or is it to test everyone's patience?

 

What you are being asked to do is to learn the statement of the theorem so that you stop with these ridiculous strawman arguments (and if what you're doing isn't childish, then I don't know what is).

 

The theorem does not say that every nth degree polynomial will have n points of intersection with the x-axis. That is the result of you forcing the word "distinct" in between "n" and "roots" in the following statement:

 

Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers.

 

Simply put: You are arguing against a statement that is not equivalent to the statement of the Fundamental Theorem of Algebra. Since the thing you are arguing against is a non-theorem, it's no surprise that you are able to topple it so easily. No one is impressed.

[Algebracus]

The fundamental theorem of algebra:

Every polynomial p(x) with complex coefficients of degree >= 1 has a complex root.

 

The statement that a complex polynomial with degree n >= 1 has n roots is in fact a statement where multiplicity is encounted. The "real" theorem is the one stated above (if you allow multiple roots, then the second statement can easily be shown by induction).

 

Other formulations of the theorem is:

- Every complex polynomial have a factorization where no polynomial factor is of degree greater than 1.

- C splits over C.

- The field C of complex numbers is algebraically closed.

I suggest you should allow yourself to look into some books introducing you to the field of abstract algebra. Then you hopefully would understand the concept of i; it is more to it than lower calculus is telling you.

[Johnny5]

The fundamental theorem of algebra:

Every polynomial p(x) with complex coefficients of degree >= 1 has a complex root.

 

The statement that a complex polynomial with degree n >= 1 has n roots is in fact a statement where multiplicity is encounted. The "real" theorem is the one stated above (if you allow multiple roots' date=' then the second statement can easily be shown by induction).

 

Other formulations of the theorem is:

- Every complex polynomial have a factorization where no polynomial factor is of degree greater than 1.

- C splits over C.

- The field C of complex numbers is algebraically closed.

I suggest you should allow yourself to look into some books introducing you to the field of abstract algebra. Then you hopefully would understand the concept of [i']i[/i]; it is more to it than lower calculus is telling you.

 

Thank you, now we are getting to the bottom of things.

[Johnny5]

Why don't you look it up yourself? It's not that hard to find. Is your objective here to learn something or is it to test everyone's patience?

 

What you are being asked to do is to learn the statement of the theorem so that you stop with these ridiculous strawman arguments (and if what you're doing isn't childish' date=' then I don't know what is).

 

The theorem does not say that every nth degree polynomial will have n points of intersection with the x-axis. That is the result of you forcing the word "distinct" in between "n" and "roots" in the following statement:

 

[i']Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers.[/i]

 

Simply put: You are arguing against a statement that is not equivalent to the statement of the Fundamental Theorem of Algebra. Since the thing you are arguing against is a non-theorem, it's no surprise that you are able to topple it so easily. No one is impressed.

 

Ok let me think about what you are saying. By the way you should know me by now, I don't play games.

 

I just don't like contradictions. So let me slowly sift through what you have said up there.

 

Ok, firstly you say look up the theorem. Well i have done that. I will do it again, then quote it below.

 

Fundamental theorem of algebra

 

Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers.

 

 

Gauss' proof of the FTA

 

Here is another link:

 

Fundamental theorem of algebra

 

Here is how that site explains it:

 

The fundamental theorem of algebra (now considered something of a Quick Facts about: misnomer

An incorrect or unsuitable namemisnomer by many mathematicians) states that every complex Quick Facts about: polynomial

A mathematical expression that is the sum of a number of termspolynomial of degree n has exactly n zeroes, counted with multiplicity. More formally, if

(where the coefficients a0, ..., an−1 can be Quick Facts about: real

An old small silver Spanish coinreal or Quick Facts about: complex

(psychoanalysis) a combination of emotions and impulses that have been rejected from awareness but still influence a person's behaviorcomplex numbers),

then there exist (not necessarily distinct) complex numbers z1, ..., zn such that

 

This shows

that the Quick Facts about: field

A piece of land cleared of trees and usually enclosedfield of Quick Facts about: complex numbers

A number of the form a+bi where a and b are real numbers and i is the square root of -1complex numbers, unlike the field of Quick Facts about: real numbers

Any rational or irrational numberreal numbers, is Quick Facts about: algebraically closed

Quick Summary not found for this subjectalgebraically closed. An easy consequence is that the product of all the roots equals (−1)n a0 and the sum of all the roots equals -an−1.

 

Notice how the above site says, "NOT NECESSARILY DISTINCT'

 

Ahem...

[Dave]

Quite frankly, the quality of posts in this thread is quite appauling. It's being closed because I don't want this forum turned into a place where people try to disprove the FTA for no good reason. There's a reason it's called "fundamental", you know.