Prove that the following equation has no constructible solutions.
Prove that the following equation has no constructible solution:
$ x^3 - 6x + 2sqrt{pi} = 0$
The way I am trying to approach is that: I want to transform the equation into some integer coefficient equation and use the Rational Root Theorem to prove that the corresponding equation has no rational root, then by Theorem: if a cubic polynomial with rational coefficients has a constructible root, then it must also have a rational root. (using contrapositive) to conclude that the original equation has no constructible solution.
However, I get stuck since $sqrt{pi}$ is not a constructible number. Therefore I can't come up with a corresponding integer coefficients equation and use the Rational Root Theorem to proceed.
Can you please point me in the right direction. Thanks in advance!
abstract-algebra geometric-construction
asked Dec 9 at 18:13
Edward
132
add a comment |
Prove that the following equation has no constructible solution:
$ x^3 - 6x + 2sqrt{pi} = 0$
The way I am trying to approach is that: I want to transform the equation into some integer coefficient equation and use the Rational Root Theorem to prove that the corresponding equation has no rational root, then by Theorem: if a cubic polynomial with rational coefficients has a constructible root, then it must also have a rational root. (using contrapositive) to conclude that the original equation has no constructible solution.
However, I get stuck since $sqrt{pi}$ is not a constructible number. Therefore I can't come up with a corresponding integer coefficients equation and use the Rational Root Theorem to proceed.
Can you please point me in the right direction. Thanks in advance!
abstract-algebra geometric-construction
asked Dec 9 at 18:13
Edward
132
add a comment |
Prove that the following equation has no constructible solution:
$ x^3 - 6x + 2sqrt{pi} = 0$
The way I am trying to approach is that: I want to transform the equation into some integer coefficient equation and use the Rational Root Theorem to prove that the corresponding equation has no rational root, then by Theorem: if a cubic polynomial with rational coefficients has a constructible root, then it must also have a rational root. (using contrapositive) to conclude that the original equation has no constructible solution.
However, I get stuck since $sqrt{pi}$ is not a constructible number. Therefore I can't come up with a corresponding integer coefficients equation and use the Rational Root Theorem to proceed.
Can you please point me in the right direction. Thanks in advance!
abstract-algebra geometric-construction
asked Dec 9 at 18:13
Edward
132
Prove that the following equation has no constructible solution:
$ x^3 - 6x + 2sqrt{pi} = 0$
The way I am trying to approach is that: I want to transform the equation into some integer coefficient equation and use the Rational Root Theorem to prove that the corresponding equation has no rational root, then by Theorem: if a cubic polynomial with rational coefficients has a constructible root, then it must also have a rational root. (using contrapositive) to conclude that the original equation has no constructible solution.
However, I get stuck since $sqrt{pi}$ is not a constructible number. Therefore I can't come up with a corresponding integer coefficients equation and use the Rational Root Theorem to proceed.
Can you please point me in the right direction. Thanks in advance!
abstract-algebra geometric-construction
abstract-algebra geometric-construction
asked Dec 9 at 18:13
Edward
132
asked Dec 9 at 18:13
Edward
132
asked Dec 9 at 18:13
Edward
132
asked Dec 9 at 18:13
Edward
132
asked Dec 9 at 18:13
Edward
132
132
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
If a solution of $p(x) = x^3 -6x+2 sqrt{pi}$ was constructible, then
$$pi= frac{1}{2}left(6x -x^3right)^2$$
would be constructible as the square, the cube and the division by an integer of a constructible number is constructible.
But that can’t be as $pi$ is transcendental while a constructible number is algebraic.
answered Dec 9 at 18:34
mathcounterexamples.net
24.3k21753
add a comment |
Suppose that that equation has a constructible root $r$.
Then $r^3-6r+2sqrt{pi}=0$, or
$$sqrt{pi}=frac{6r-r^3}{2},$$
which is in $Bbb{Q}(r)$, contradicting that $sqrt{pi}$ isn't constructible.
answered Dec 9 at 18:31
jgon
12.7k21940
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
If a solution of $p(x) = x^3 -6x+2 sqrt{pi}$ was constructible, then
$$pi= frac{1}{2}left(6x -x^3right)^2$$
would be constructible as the square, the cube and the division by an integer of a constructible number is constructible.
But that can’t be as $pi$ is transcendental while a constructible number is algebraic.
answered Dec 9 at 18:34
mathcounterexamples.net
24.3k21753
add a comment |
If a solution of $p(x) = x^3 -6x+2 sqrt{pi}$ was constructible, then
$$pi= frac{1}{2}left(6x -x^3right)^2$$
would be constructible as the square, the cube and the division by an integer of a constructible number is constructible.
But that can’t be as $pi$ is transcendental while a constructible number is algebraic.
answered Dec 9 at 18:34
mathcounterexamples.net
24.3k21753
add a comment |
If a solution of $p(x) = x^3 -6x+2 sqrt{pi}$ was constructible, then
$$pi= frac{1}{2}left(6x -x^3right)^2$$
would be constructible as the square, the cube and the division by an integer of a constructible number is constructible.
But that can’t be as $pi$ is transcendental while a constructible number is algebraic.
answered Dec 9 at 18:34
mathcounterexamples.net
24.3k21753
If a solution of $p(x) = x^3 -6x+2 sqrt{pi}$ was constructible, then
$$pi= frac{1}{2}left(6x -x^3right)^2$$
would be constructible as the square, the cube and the division by an integer of a constructible number is constructible.
But that can’t be as $pi$ is transcendental while a constructible number is algebraic.
answered Dec 9 at 18:34
mathcounterexamples.net
24.3k21753
answered Dec 9 at 18:34
mathcounterexamples.net
24.3k21753
answered Dec 9 at 18:34
mathcounterexamples.net
24.3k21753
answered Dec 9 at 18:34
mathcounterexamples.net
24.3k21753
24.3k21753
add a comment |
add a comment |
Suppose that that equation has a constructible root $r$.
Then $r^3-6r+2sqrt{pi}=0$, or
$$sqrt{pi}=frac{6r-r^3}{2},$$
which is in $Bbb{Q}(r)$, contradicting that $sqrt{pi}$ isn't constructible.
answered Dec 9 at 18:31
jgon
12.7k21940
add a comment |
Suppose that that equation has a constructible root $r$.
Then $r^3-6r+2sqrt{pi}=0$, or
$$sqrt{pi}=frac{6r-r^3}{2},$$
which is in $Bbb{Q}(r)$, contradicting that $sqrt{pi}$ isn't constructible.
answered Dec 9 at 18:31
jgon
12.7k21940
add a comment |
Suppose that that equation has a constructible root $r$.
Then $r^3-6r+2sqrt{pi}=0$, or
$$sqrt{pi}=frac{6r-r^3}{2},$$
which is in $Bbb{Q}(r)$, contradicting that $sqrt{pi}$ isn't constructible.
answered Dec 9 at 18:31
jgon
12.7k21940
Suppose that that equation has a constructible root $r$.
Then $r^3-6r+2sqrt{pi}=0$, or
$$sqrt{pi}=frac{6r-r^3}{2},$$
which is in $Bbb{Q}(r)$, contradicting that $sqrt{pi}$ isn't constructible.
answered Dec 9 at 18:31
jgon
12.7k21940
answered Dec 9 at 18:31
jgon
12.7k21940
answered Dec 9 at 18:31
jgon
12.7k21940
answered Dec 9 at 18:31
jgon
12.7k21940
12.7k21940
add a comment |
add a comment |
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