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Question about a degree 5 polynomial with no rational roots

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Question about a degree 5 polynomial with no rational roots

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2

$begingroup$

Is it possible to find a 5th degree polynomial with no rational roots, and at least one irrational root?

Edit: the polynomial must only have rational coefficients

functions polynomials

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edited 9 hours ago

Diba

asked 9 hours ago

DibaDiba

1355 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  how about $(x-pi)^5$ or $(x-e)^5$ or $(x-sqrt2)^5$?
  $endgroup$
  – J. W. Tanner
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Or even just $x^5-2$?
  $endgroup$
  – Lord Shark the Unknown
  9 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

Is it possible to find a 5th degree polynomial with no rational roots, and at least one irrational root?

Edit: the polynomial must only have rational coefficients

functions polynomials

share|cite|improve this question

edited 9 hours ago

Diba

asked 9 hours ago

DibaDiba

1355 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  how about $(x-pi)^5$ or $(x-e)^5$ or $(x-sqrt2)^5$?
  $endgroup$
  – J. W. Tanner
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Or even just $x^5-2$?
  $endgroup$
  – Lord Shark the Unknown
  9 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Is it possible to find a 5th degree polynomial with no rational roots, and at least one irrational root?

Edit: the polynomial must only have rational coefficients

functions polynomials

share|cite|improve this question

edited 9 hours ago

Diba

asked 9 hours ago

DibaDiba

1355 bronze badges

$endgroup$

share|cite|improve this question

edited 9 hours ago

Diba

asked 9 hours ago

DibaDiba

1355 bronze badges

share|cite|improve this question

edited 9 hours ago

Diba

asked 9 hours ago

DibaDiba

1355 bronze badges

asked 9 hours ago

DibaDiba

1355 bronze badges

asked 9 hours ago

DibaDiba

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5 Answers
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4

$begingroup$

In fact we can even achieve this with a polynomial with rational coefficients: [Edit: After I wrote this answer, OP added the rationality of the coefficients as a condition.]

The Rational Root Theorem implies that $$x^5 - 2$$ has no rational roots, but since its degree is odd, it has at least one real---and hence irrational---root. (In fact, can replace $2$ here with any rational number that is not a perfect $5$th power of a rational number.)

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

TravisTravis

69.9k88 gold badges7777 silver badges162162 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  in fact with integer coefficients
  $endgroup$
  – J. W. Tanner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed! That said, we can always clear the denominators of any rational polynomial by multiplying by their $operatorname{lcm}$, and this operation preserves the roots. (But it's true that this example is a monic polynomial with integer coefficients.)
  $endgroup$
  – Travis
  9 hours ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

That is pretty easy, if there are not other propertys required. Just consider a polynomial like:

$(x-sqrt{2})^5$

share|cite|improve this answer

answered 9 hours ago

CornmanCornman

6,16122 gold badges1313 silver badges3333 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That doesn't have rational coefficients.
  $endgroup$
  – fleablood
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood OP added the requirement that the coefficients be rational a few minutes after this answer was posted.
  $endgroup$
  – Travis
  1 hour ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

If $alpha$ is irrational, then the polynomial $(x-alpha)^5$ has degree $5$, and the only (repeated) root is irrational.

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

16.9k11 gold badge1212 silver badges3535 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  More generally, $(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)$ for any irrational $a_1, dots, a_5$.
  $endgroup$
  – 79037662
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I realize now that I might have phrased the question badly- it is only considering polynomials with rational coefficients.
  $endgroup$
  – Diba
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DibaHeydary: In that case, see the answer by Travis
  $endgroup$
  – J. W. Tanner
  9 hours ago
  
  
  

  
  
  
  

add a comment
 | 

1

$begingroup$

$$ x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1 $$

Five real irrational roots,
$$ 2 cos left( frac{2k
pi}{11} right) $$

If you prefer the roots to be $cos( 2k pi /11),$ adjust the quintic to

$$ 32x^5 + 16x^4 - 32 x^3 - 12 x^2 + 6 x + 1 $$

gp-pari:

? x = cos( 2 * Pi / 11)

%1 = 0.8412535328311811688618116489

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%2 = -1.009741959 E-28

? 

? x = cos( 4 * Pi / 11)

%3 = 0.4154150130018864255292741493

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%4 = -2.019483917 E-28

? 

? x = cos( 6 * Pi / 11)

%5 = -0.1423148382732851404437926686

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%6 = -7.57306469 E-29

? 

? x = cos( 8 * Pi / 11)

%7 = -0.6548607339452850640569250725

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%8 = 0.E-28

? 

=============================

share|cite|improve this answer

edited 4 hours ago

answered 8 hours ago

Will JagyWill Jagy

109k55 gold badges107107 silver badges209209 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused, how did you connect polynomials to triginometry? Through something like taylor series?
  $endgroup$
  – vikarjramun
  38 mins ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

Well, $sqrt[5]{m}$ is irrational if $m$ is an integer that isn't a perfect power of $5$.

And $sqrt[5]{m}$ is a solution to $x^5 -m =0$.

And as $x^5 -m =0implies x^5 = mimplies x =sqrt [5]{m}$ so that is the only solution.

share|cite|improve this answer

answered 3 hours ago

fleabloodfleablood

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4

$begingroup$

In fact we can even achieve this with a polynomial with rational coefficients: [Edit: After I wrote this answer, OP added the rationality of the coefficients as a condition.]

The Rational Root Theorem implies that $$x^5 - 2$$ has no rational roots, but since its degree is odd, it has at least one real---and hence irrational---root. (In fact, can replace $2$ here with any rational number that is not a perfect $5$th power of a rational number.)

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

TravisTravis

69.9k88 gold badges7777 silver badges162162 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  in fact with integer coefficients
  $endgroup$
  – J. W. Tanner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed! That said, we can always clear the denominators of any rational polynomial by multiplying by their $operatorname{lcm}$, and this operation preserves the roots. (But it's true that this example is a monic polynomial with integer coefficients.)
  $endgroup$
  – Travis
  9 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

In fact we can even achieve this with a polynomial with rational coefficients: [Edit: After I wrote this answer, OP added the rationality of the coefficients as a condition.]

The Rational Root Theorem implies that $$x^5 - 2$$ has no rational roots, but since its degree is odd, it has at least one real---and hence irrational---root. (In fact, can replace $2$ here with any rational number that is not a perfect $5$th power of a rational number.)

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

TravisTravis

69.9k88 gold badges7777 silver badges162162 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  in fact with integer coefficients
  $endgroup$
  – J. W. Tanner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed! That said, we can always clear the denominators of any rational polynomial by multiplying by their $operatorname{lcm}$, and this operation preserves the roots. (But it's true that this example is a monic polynomial with integer coefficients.)
  $endgroup$
  – Travis
  9 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

In fact we can even achieve this with a polynomial with rational coefficients: [Edit: After I wrote this answer, OP added the rationality of the coefficients as a condition.]

The Rational Root Theorem implies that $$x^5 - 2$$ has no rational roots, but since its degree is odd, it has at least one real---and hence irrational---root. (In fact, can replace $2$ here with any rational number that is not a perfect $5$th power of a rational number.)

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

TravisTravis

69.9k88 gold badges7777 silver badges162162 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

TravisTravis

69.9k88 gold badges7777 silver badges162162 bronze badges

answered 9 hours ago

TravisTravis

69.9k88 gold badges7777 silver badges162162 bronze badges

answered 9 hours ago

TravisTravis

69.9k88 gold badges7777 silver badges162162 bronze badges

1

$begingroup$

That is pretty easy, if there are not other propertys required. Just consider a polynomial like:

$(x-sqrt{2})^5$

share|cite|improve this answer

answered 9 hours ago

CornmanCornman

6,16122 gold badges1313 silver badges3333 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That doesn't have rational coefficients.
  $endgroup$
  – fleablood
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood OP added the requirement that the coefficients be rational a few minutes after this answer was posted.
  $endgroup$
  – Travis
  1 hour ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

That is pretty easy, if there are not other propertys required. Just consider a polynomial like:

$(x-sqrt{2})^5$

share|cite|improve this answer

answered 9 hours ago

CornmanCornman

6,16122 gold badges1313 silver badges3333 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That doesn't have rational coefficients.
  $endgroup$
  – fleablood
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood OP added the requirement that the coefficients be rational a few minutes after this answer was posted.
  $endgroup$
  – Travis
  1 hour ago
  

  
  
  
  

add a comment
 | 

$begingroup$

That is pretty easy, if there are not other propertys required. Just consider a polynomial like:

$(x-sqrt{2})^5$

share|cite|improve this answer

answered 9 hours ago

CornmanCornman

6,16122 gold badges1313 silver badges3333 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

CornmanCornman

6,16122 gold badges1313 silver badges3333 bronze badges

answered 9 hours ago

CornmanCornman

6,16122 gold badges1313 silver badges3333 bronze badges

answered 9 hours ago

CornmanCornman

6,16122 gold badges1313 silver badges3333 bronze badges

1

$begingroup$

If $alpha$ is irrational, then the polynomial $(x-alpha)^5$ has degree $5$, and the only (repeated) root is irrational.

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

16.9k11 gold badge1212 silver badges3535 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  More generally, $(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)$ for any irrational $a_1, dots, a_5$.
  $endgroup$
  – 79037662
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I realize now that I might have phrased the question badly- it is only considering polynomials with rational coefficients.
  $endgroup$
  – Diba
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DibaHeydary: In that case, see the answer by Travis
  $endgroup$
  – J. W. Tanner
  9 hours ago
  
  
  

  
  
  
  

add a comment
 | 

1

$begingroup$

If $alpha$ is irrational, then the polynomial $(x-alpha)^5$ has degree $5$, and the only (repeated) root is irrational.

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

16.9k11 gold badge1212 silver badges3535 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  More generally, $(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)$ for any irrational $a_1, dots, a_5$.
  $endgroup$
  – 79037662
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I realize now that I might have phrased the question badly- it is only considering polynomials with rational coefficients.
  $endgroup$
  – Diba
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DibaHeydary: In that case, see the answer by Travis
  $endgroup$
  – J. W. Tanner
  9 hours ago
  
  
  

  
  
  
  

add a comment
 | 

$begingroup$

If $alpha$ is irrational, then the polynomial $(x-alpha)^5$ has degree $5$, and the only (repeated) root is irrational.

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

16.9k11 gold badge1212 silver badges3535 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

J. W. TannerJ. W. Tanner

16.9k11 gold badge1212 silver badges3535 bronze badges

answered 9 hours ago

J. W. TannerJ. W. Tanner

16.9k11 gold badge1212 silver badges3535 bronze badges

answered 9 hours ago

J. W. TannerJ. W. Tanner

16.9k11 gold badge1212 silver badges3535 bronze badges

1

$begingroup$

$$ x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1 $$

Five real irrational roots,
$$ 2 cos left( frac{2k
pi}{11} right) $$

If you prefer the roots to be $cos( 2k pi /11),$ adjust the quintic to

$$ 32x^5 + 16x^4 - 32 x^3 - 12 x^2 + 6 x + 1 $$

gp-pari:

? x = cos( 2 * Pi / 11)

%1 = 0.8412535328311811688618116489

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%2 = -1.009741959 E-28

? 

? x = cos( 4 * Pi / 11)

%3 = 0.4154150130018864255292741493

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%4 = -2.019483917 E-28

? 

? x = cos( 6 * Pi / 11)

%5 = -0.1423148382732851404437926686

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%6 = -7.57306469 E-29

? 

? x = cos( 8 * Pi / 11)

%7 = -0.6548607339452850640569250725

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%8 = 0.E-28

? 

=============================

share|cite|improve this answer

edited 4 hours ago

answered 8 hours ago

Will JagyWill Jagy

109k55 gold badges107107 silver badges209209 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused, how did you connect polynomials to triginometry? Through something like taylor series?
  $endgroup$
  – vikarjramun
  38 mins ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

$$ x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1 $$

Five real irrational roots,
$$ 2 cos left( frac{2k
pi}{11} right) $$

If you prefer the roots to be $cos( 2k pi /11),$ adjust the quintic to

$$ 32x^5 + 16x^4 - 32 x^3 - 12 x^2 + 6 x + 1 $$

gp-pari:

? x = cos( 2 * Pi / 11)

%1 = 0.8412535328311811688618116489

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%2 = -1.009741959 E-28

? 

? x = cos( 4 * Pi / 11)

%3 = 0.4154150130018864255292741493

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%4 = -2.019483917 E-28

? 

? x = cos( 6 * Pi / 11)

%5 = -0.1423148382732851404437926686

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%6 = -7.57306469 E-29

? 

? x = cos( 8 * Pi / 11)

%7 = -0.6548607339452850640569250725

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%8 = 0.E-28

? 

=============================

share|cite|improve this answer

edited 4 hours ago

answered 8 hours ago

Will JagyWill Jagy

109k55 gold badges107107 silver badges209209 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused, how did you connect polynomials to triginometry? Through something like taylor series?
  $endgroup$
  – vikarjramun
  38 mins ago
  

  
  
  
  

add a comment
 | 

$begingroup$

$$ x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1 $$

Five real irrational roots,
$$ 2 cos left( frac{2k
pi}{11} right) $$

If you prefer the roots to be $cos( 2k pi /11),$ adjust the quintic to

$$ 32x^5 + 16x^4 - 32 x^3 - 12 x^2 + 6 x + 1 $$

gp-pari:

? x = cos( 2 * Pi / 11)

%1 = 0.8412535328311811688618116489

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%2 = -1.009741959 E-28

? 

? x = cos( 4 * Pi / 11)

%3 = 0.4154150130018864255292741493

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%4 = -2.019483917 E-28

? 

? x = cos( 6 * Pi / 11)

%5 = -0.1423148382732851404437926686

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%6 = -7.57306469 E-29

? 

? x = cos( 8 * Pi / 11)

%7 = -0.6548607339452850640569250725

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%8 = 0.E-28

? 

=============================

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edited 4 hours ago

answered 8 hours ago

Will JagyWill Jagy

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$endgroup$

? x = cos( 2 * Pi / 11)

%1 = 0.8412535328311811688618116489

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%2 = -1.009741959 E-28

? 

? x = cos( 4 * Pi / 11)

%3 = 0.4154150130018864255292741493

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%4 = -2.019483917 E-28

? 

? x = cos( 6 * Pi / 11)

%5 = -0.1423148382732851404437926686

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%6 = -7.57306469 E-29

? 

? x = cos( 8 * Pi / 11)

%7 = -0.6548607339452850640569250725

? f = 32 * x^5 + 16 * x^4 - 32 * x^3 - 12 * x^2 + 6 * x + 1

%8 = 0.E-28

? 

share|cite|improve this answer

edited 4 hours ago

answered 8 hours ago

Will JagyWill Jagy

109k55 gold badges107107 silver badges209209 bronze badges

answered 8 hours ago

Will JagyWill Jagy

109k55 gold badges107107 silver badges209209 bronze badges

answered 8 hours ago

Will JagyWill Jagy

109k55 gold badges107107 silver badges209209 bronze badges

1

$begingroup$

Well, $sqrt[5]{m}$ is irrational if $m$ is an integer that isn't a perfect power of $5$.

And $sqrt[5]{m}$ is a solution to $x^5 -m =0$.

And as $x^5 -m =0implies x^5 = mimplies x =sqrt [5]{m}$ so that is the only solution.

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answered 3 hours ago

fleabloodfleablood

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$endgroup$

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1

$begingroup$

Well, $sqrt[5]{m}$ is irrational if $m$ is an integer that isn't a perfect power of $5$.

And $sqrt[5]{m}$ is a solution to $x^5 -m =0$.

And as $x^5 -m =0implies x^5 = mimplies x =sqrt [5]{m}$ so that is the only solution.

share|cite|improve this answer

answered 3 hours ago

fleabloodfleablood

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$endgroup$

add a comment
 | 

$begingroup$

Well, $sqrt[5]{m}$ is irrational if $m$ is an integer that isn't a perfect power of $5$.

And $sqrt[5]{m}$ is a solution to $x^5 -m =0$.

And as $x^5 -m =0implies x^5 = mimplies x =sqrt [5]{m}$ so that is the only solution.

share|cite|improve this answer

answered 3 hours ago

fleabloodfleablood

81.3k22 gold badges3232 silver badges9898 bronze badges

$endgroup$

share|cite|improve this answer

answered 3 hours ago

fleabloodfleablood

81.3k22 gold badges3232 silver badges9898 bronze badges

answered 3 hours ago

fleabloodfleablood

81.3k22 gold badges3232 silver badges9898 bronze badges

answered 3 hours ago

fleabloodfleablood

81.3k22 gold badges3232 silver badges9898 bronze badges