Fundamental Theorem of Algebra - Numberphile
video description
But the facts remain very clear that human definitions are never any true discoveries in mathematics, especially that most human definitions in mathematics strictly are easily refutble, since it contradict strictly the Pythagoras discovered theorem,
Which is why the immaginary number was designed in order to illegally solve the cubic equation beside designing the very false fundamental theorem of algebra
Date: 2022-04-08
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Comments and reviews: 9
bassam
Please, note that I don't have time to reexplain & repeat many elementary proofs about the flowed so-called Fundamental theorem of algebra, since every thing is already & publically published freely under my name, and to provide a fast hint about the complete falsehood of FTG, please, consider this arbitrary odd degree polynomial that must have at least one real root and 48 other complete roots that are strictly dependent on that existing alleged real root, the polynomial (x--49- + x-7 =1, where I claimed, it hasn't any root at all since every alleged Real root must necessarily be a counter examples to Fermat's last theorem, which is of course impossible, hence no real root exists nor all other 48 dependent roots exist, where this is only one public published proof of mine under my name, out of other dozen of proofs about the flawed Fundamental theorem of algebra and many more of course
So, kindly, don't remove my comment unless you prove me wrong
Thanking your attention, & spreading the truth that innocent school students deserve to know
Bassam Karzeddin
reply
Please, note that I don't have time to reexplain & repeat many elementary proofs about the flowed so-called Fundamental theorem of algebra, since every thing is already & publically published freely under my name, and to provide a fast hint about the complete falsehood of FTG, please, consider this arbitrary odd degree polynomial that must have at least one real root and 48 other complete roots that are strictly dependent on that existing alleged real root, the polynomial (x--49- + x-7 =1, where I claimed, it hasn't any root at all since every alleged Real root must necessarily be a counter examples to Fermat's last theorem, which is of course impossible, hence no real root exists nor all other 48 dependent roots exist, where this is only one public published proof of mine under my name, out of other dozen of proofs about the flawed Fundamental theorem of algebra and many more of course
So, kindly, don't remove my comment unless you prove me wrong
Thanking your attention, & spreading the truth that innocent school students deserve to know
Bassam Karzeddin
reply
Michael
I disagree with the creation of concepts like imaginary numbers. They open doors for people to make huge mistakes that travel for years without being corrected. In the case of the imaginary number. Rather than making essentially an entire new field of math, which it isn't, we should have basically left a blank by saying the square root of -1 is -1 after it is squared or the square root of -1 is unsquared. I am tired of seeing the misled destruction causing problems for hundreds of years before anyone corrects them. Physics is riddled with these kind of concepts. Example: motion is relative. garbage. The observer is not important. All physics and math is only true from the perspective of the universe not the creatures who float through it.
reply
I disagree with the creation of concepts like imaginary numbers. They open doors for people to make huge mistakes that travel for years without being corrected. In the case of the imaginary number. Rather than making essentially an entire new field of math, which it isn't, we should have basically left a blank by saying the square root of -1 is -1 after it is squared or the square root of -1 is unsquared. I am tired of seeing the misled destruction causing problems for hundreds of years before anyone corrects them. Physics is riddled with these kind of concepts. Example: motion is relative. garbage. The observer is not important. All physics and math is only true from the perspective of the universe not the creatures who float through it.
reply
Topi
15: 20 Like he says complex numbers follow all algebraic rules. So why are they called complex and not simplex. I hate the name as it tells students that complex numbers are complex, which they are not. The only thing you lose when you move from real line to complex plane is that on real line if x1-x2 is non-zero the sign tells you if x1 < x2 or x2 < x1 but on complex plane z1-z2 is non zero tells you just that z1 is not z2. There no order on complex plane same way that there is on real line. The triangle inequality still holds though. So -z1- + -z2- >= -z1+z2-.
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15: 20 Like he says complex numbers follow all algebraic rules. So why are they called complex and not simplex. I hate the name as it tells students that complex numbers are complex, which they are not. The only thing you lose when you move from real line to complex plane is that on real line if x1-x2 is non-zero the sign tells you if x1 < x2 or x2 < x1 but on complex plane z1-z2 is non zero tells you just that z1 is not z2. There no order on complex plane same way that there is on real line. The triangle inequality still holds though. So -z1- + -z2- >= -z1+z2-.
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Keldor314
I thought of a completely different approach during the video.
Observe that when you raise z to any power, you map the complex plane to the complex plane, with no holes. Observe that the same thing happens for addition and multiplication by a constant. This means that each of these three operations will map some point on the plane to the origin. Since every polynomial can be written as a tree-like composite function of these three elementry operations, it follows by induction that the entire polynomial must map some point to the origin.
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I thought of a completely different approach during the video.
Observe that when you raise z to any power, you map the complex plane to the complex plane, with no holes. Observe that the same thing happens for addition and multiplication by a constant. This means that each of these three operations will map some point on the plane to the origin. Since every polynomial can be written as a tree-like composite function of these three elementry operations, it follows by induction that the entire polynomial must map some point to the origin.
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dingus49ovi
My problem is visualization of the two different coordinate system. f(x) = x--2 +1 usually gets plotted on a coordinate system with and x axis and a y axis, both real. Imaginary numbers also get plotted on a coordinate system with an x axis and a y axis, but this case the y axis represents the imaginary part. Where do I visualize the imaginary part when both x and y axis represent reals?
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My problem is visualization of the two different coordinate system. f(x) = x--2 +1 usually gets plotted on a coordinate system with and x axis and a y axis, both real. Imaginary numbers also get plotted on a coordinate system with an x axis and a y axis, but this case the y axis represents the imaginary part. Where do I visualize the imaginary part when both x and y axis represent reals?
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Ken
Okay, so: I was a math major (in fact, I went to college with David Eisenbud's son, who was also a math & music person like me) and I'm still a big math geek, and SOMEHOW despite knowing the Fundamental Theorem of Algebra since I was pretty young, it never occurred to me that I'd never seen a proof of it. What a fantastic proof! Now I need -3blue1brown to make an awesome animation of it. =)
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Okay, so: I was a math major (in fact, I went to college with David Eisenbud's son, who was also a math & music person like me) and I'm still a big math geek, and SOMEHOW despite knowing the Fundamental Theorem of Algebra since I was pretty young, it never occurred to me that I'd never seen a proof of it. What a fantastic proof! Now I need -3blue1brown to make an awesome animation of it. =)
reply
Patrick
I guess the concept is sound, but you would have to show that the circle in the domain will map to a continuous path, and also show that as you shrink the circle, that this path cannot somehow -skip- zero. I guess its kind of obvious but the actual proof requires some theories from analysis.
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I guess the concept is sound, but you would have to show that the circle in the domain will map to a continuous path, and also show that as you shrink the circle, that this path cannot somehow -skip- zero. I guess its kind of obvious but the actual proof requires some theories from analysis.
reply
Anit
Hello, I am very familiar with this theorem. If I were to try to express the real numbers as squares of i with factors and minus signs where needed, put them on the number line, is that forbidden or could it be a way to show the imaginary roots as part of the new imaginary part of the graph?
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Hello, I am very familiar with this theorem. If I were to try to express the real numbers as squares of i with factors and minus signs where needed, put them on the number line, is that forbidden or could it be a way to show the imaginary roots as part of the new imaginary part of the graph?
reply
Cameron
A more rigorous proof is that if f is non zero then abs(f) has a non zero lower bound for the whole of C: hence 1/f(z) is holomorphic everywhere in C and bounded above, so 1/f(z) must be a constant, hence f(z) is a constant which is a contradiction.
Hence f must have a root
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A more rigorous proof is that if f is non zero then abs(f) has a non zero lower bound for the whole of C: hence 1/f(z) is holomorphic everywhere in C and bounded above, so 1/f(z) must be a constant, hence f(z) is a constant which is a contradiction.
Hence f must have a root
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