Contradiction of Galios theory

That is a simple example of why approximate computation does not necessarily give exact values.In summary, Galois theory tells us that the equation x^5 - 6x + 3 is not solvable by radicals. However, using Newton's method, we can find one root of the equation and factor it to be (x-1.4)(ax^4+px^3+qx^2+sx+t). This shows that the equation is solvable. The confusion lies in the difference between approximate roots and exact roots, as shown by the example of x^3+3x+1=0.
  • #1
footmath
26
0
Hello.
Galois theory tell us $x^5-6x+3$ is not solvable by radical but every equation lower than fifth can solve by radical.
If $G$ is solvable and $H$ is solvable too $G*H$ are solvable. For $x^5-6x+3$ we can use Newton’s method and find one root of this equation.
We obtain $x=1.4$ and factor this equation.
$x^5-6x+3=(x-1.4)(ax^4+px^3+qx^2+sx+t)$
$a=1$
$p=1.4$
$q=(1.4)^2$
$s=(1.4)^3$
$t=-6+(1.4)^4$
$x^5-6x+3=(x-1.4)(x^4+1.4x^3+(1.4)x^2+(1.4)^3x-6+(1.4)^4)$
$(x-1.4)$ is solvable and $(x^4+1.4x^3+(1.4)x^2+(1.4)^3x-6+(1.4)^4)$ is solvable too so $(x-1.4)*(x^4+1.4x^3+(1.4)x^2+(1.4)3x-6+(1.4)^4)$ are solvable.
Why did Galois theory say that $x^5-6x+3$ is not solvable by radical?
 
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  • #2
[itex](1.4)^5- 6(1.4)+ 3= -0.02176[/itex], not 0.

1.4 is NOT a root of the equation so x- 1.4 is NOT a factor.

Don't confuse approximate roots with roots.
 
  • #3
When we want to calculate the root of a equation for example x^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546 $ but if you substitute $-0.3221853546 $ in the equation x^3+3x+1=0 you will see (-0.3221853546)^3+3(-0.3221853546)+1=0.00000000000846 . we cannot obtain a rational root but we say this equation is solvable by radical . $ \sqrt[5]{5.40985+0i} $ is the root of x^5- 6x+ 3
 
  • #4
footmath said:
When we want to calculate the root of a equation for example x^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546 $ but if you substitute $-0.3221853546 $ in the equation x^3+3x+1=0 you will see (-0.3221853546)^3+3(-0.3221853546)+1=0.00000000000846 . we cannot obtain a rational root but we say this equation is solvable by radical . $ \sqrt[5]{5.40985+0i} $ is the root of x^5- 6x+ 3
You are not understanding approximate roots, such as x = -0.3221853546, with exact roots. BTW, why are you putting $ symbols around your numbers?

Iterative approximation techniques such as Newton's method, give approximate values for the roots of equations, even equations of degree five and higher. Galois theory has to do with finding exact roots of equations.

As a simple example of the distinction between approximate roots and exact roots, you can use Newton's method to find a root of the equation x2 = 2. One approximation is 1.414. The exact value is [itex]\sqrt{2}[/itex], which is not the same as 1.414.
 
  • #5
Mark44 said:
You are not understanding approximate roots, such as x = -0.3221853546, with exact roots. BTW, why are you putting $ symbols around your numbers?

Iterative approximation techniques such as Newton's method, give approximate values for the roots of equations, even equations of degree five and higher. Galois theory has to do with finding exact roots of equations.

As a simple example of the distinction between approximate roots and exact roots, you can use Newton's method to find a root of the equation x2 = 2. One approximation is 1.414. The exact value is [itex]\sqrt{2}[/itex], which is not the same as 1.414.

What is your opinion about$ x^5 + 20x^3 + 20x^2 + 30x + 10 =0 $
Are you accept that the one root of this equation is $\sqrt[5]{2}-\sqrt[5]{2^2}+\sqrt[5]{2^3}-\sqrt[5]{2^4} $ ? if you substitute this answer in the equation you will see very, very close to 0 but NOT 0.
$ \sqrt[5]{5.40985+0i}$ is close to 0 for the equation $ x^{5}-6x+3 $
 
  • #6
footmath said:
What is your opinion about [itex]x^5 + 20x^3 + 20x^2 + 30x + 10 =0[/itex]
Are you accept that the one root of this equation is [itex]\sqrt[5]{2}-\sqrt[5]{2^2}+\sqrt[5]{2^3}-\sqrt[5]{2^4} [/itex] ? if you substitute this answer in the equation you will see very, very close to 0 but NOT 0.
[itex]\sqrt[5]{5.40985+0i}[/itex] is close to 0 for the equation [itex]x^{5}-6x+3 [/itex]

I see that the $ signs can be replaced by [ tex ] tags.

WolframAlpha shows that [itex]\sqrt[5]{2}-\sqrt[5]{2^2}+\sqrt[5]{2^3}-\sqrt[5]{2^4} [/itex] is a exact root of [itex]x^5 + 20x^3 + 20x^2 + 30x + 10 = 0\,.[/itex]

See the LINK.
 
  • #7
What SammyS is saying is that your statement that putting that into the equation gives a number that "is very, very close to 0 but NOT 0" is incorrect. I suspect what you did was use a calculator to find an approximate value for that solution. If so, again, you are confusing approximate computation with exact values.

Notice that, by the quadratic formula, [itex]1+\sqrt{6}[/itex] is a root of [itex]x^2- 2x- 5= 0[/itex] and it is easy to see that this is true by exact computation:
If [itex]x= 1+ \sqrt{6}[/itex] then [itex]x^2= 1+ 2\sqrt{6}+ 6= 7+ 2\sqrt{6}[/itex] so that [itex]x^2- 2x+ 5= 7+ 2\sqrt{6}- 2- 2\sqrt{6}- 5= 0[/itex]

But if we use a calculator to find that [itex]\sqrt{6}= 2.4494897427831780981972840747059[/itex], then x= 3.4494897427831780981972840747059 and [itex]x^2- 2x- 6[/itex] becomes
2.5265767520093026238086822867067e-37 which is "0" to 36 decimal places but NOT 0.
 

FAQ: Contradiction of Galios theory

What is the contradiction of Galios theory?

The contradiction of Galios theory refers to the fact that there are certain mathematical equations, known as quintic equations, that cannot be solved using the methods proposed by Galios. This goes against the main premise of Galios theory, which states that any algebraic equation can be solved using a combination of addition, subtraction, multiplication, division, and taking roots.

How did Galios theory come about?

Galios theory, also known as Galois theory, was developed by French mathematician Évariste Galois in the 19th century. He was interested in finding solutions to polynomial equations and was able to prove that some equations could not be solved using the traditional methods of algebra. This led to the development of the theory that bears his name.

What impact did the contradiction of Galios theory have on mathematics?

The contradiction of Galios theory had a significant impact on mathematics. It showed that there are limitations to the methods of algebra and that some equations cannot be solved using these methods. This led to the development of new branches of mathematics, such as abstract algebra and group theory, which have been used to solve previously unsolvable equations.

Can the contradiction of Galios theory be resolved?

While the contradiction of Galios theory cannot be resolved in the sense that all quintic equations can be solved using the traditional methods of algebra, mathematicians have found workarounds and alternative methods for solving these equations. Additionally, the development of new branches of mathematics has allowed for a deeper understanding of the limitations of traditional algebraic methods.

How does the contradiction of Galios theory impact other fields of science?

The contradiction of Galios theory has had a significant impact on other fields of science, particularly physics and engineering. These fields heavily rely on mathematical equations to describe and solve problems, and the fact that there are unsolvable equations challenges our understanding of the physical world. It also highlights the importance of constantly questioning and refining our theories and methods in order to better understand and solve complex problems.

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