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Prove that $sqrt[3]p+sqrt[3]p^5$ is irrational if $p$ is prime [on hold]

How to prove that $sqrt 3$ is an irrational number?Prove $x = sqrt[100]sqrt3 + sqrt2 + sqrt[100]sqrt3 - sqrt2$ is irrationalProve that $sqrt 2 +sqrt 3$ is irrational.Show that $sqrt[3]2 + sqrt[3]4$ is irrationalProve that $sqrt[5]672$ is irrationalProve that $sqrt3+ sqrt5+ sqrt7$ is irrationalProving $sqrt[4]4$ is irrationalLet $p,q$ be irrational numbers, such that, $p^2$ and $q^2$ are relatively prime. Show that $sqrtpq$ is also irrational.Square root of a prime is irrationalProving $sqrt3 + sqrt[3]2$ to be irrational

-1

$begingroup$

Prove that $sqrt[3]p+sqrt[3]p^5$ is irrational when $p$ is a prime.

First I suppose $x=sqrt[3]p+sqrt[3]p^5$. Cubing gives
$$x^3=p+p^5+p^2x$$
And then what properties of prime, and how to test its irrationallity?

number-theory irrational-numbers rationality-testing

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edited yesterday

Parcly Taxel

43.5k1375104

asked yesterday

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put on hold as off-topic by uniquesolution, Cesareo, Carl Mummert, B. Goddard, Parcly Taxel yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, Cesareo, Carl Mummert, B. Goddard, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure that is irrational?
  $endgroup$
  – uniquesolution
  yesterday
  

  
  
  
  

add a comment | 

-1

$begingroup$

Prove that $sqrt[3]p+sqrt[3]p^5$ is irrational when $p$ is a prime.

First I suppose $x=sqrt[3]p+sqrt[3]p^5$. Cubing gives
$$x^3=p+p^5+p^2x$$
And then what properties of prime, and how to test its irrationallity?

number-theory irrational-numbers rationality-testing

share|cite|improve this question

edited yesterday

Parcly Taxel

43.5k1375104

asked yesterday

HeartHeart

29318

$endgroup$

put on hold as off-topic by uniquesolution, Cesareo, Carl Mummert, B. Goddard, Parcly Taxel yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, Cesareo, Carl Mummert, B. Goddard, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure that is irrational?
  $endgroup$
  – uniquesolution
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Prove that $sqrt[3]p+sqrt[3]p^5$ is irrational when $p$ is a prime.

First I suppose $x=sqrt[3]p+sqrt[3]p^5$. Cubing gives
$$x^3=p+p^5+p^2x$$
And then what properties of prime, and how to test its irrationallity?

number-theory irrational-numbers rationality-testing

share|cite|improve this question

edited yesterday

Parcly Taxel

43.5k1375104

asked yesterday

HeartHeart

29318

$endgroup$

share|cite|improve this question

edited yesterday

Parcly Taxel

43.5k1375104

asked yesterday

HeartHeart

29318

share|cite|improve this question

edited yesterday

Parcly Taxel

43.5k1375104

asked yesterday

HeartHeart

29318

edited yesterday

Parcly Taxel

43.5k1375104

edited yesterday

Parcly Taxel

43.5k1375104

asked yesterday

HeartHeart

29318

asked yesterday

HeartHeart

29318

1 Answer
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2

$begingroup$

By Eisenstein's criterion, $X^3-p$ is irreducible over $Bbb Q$. Therefore $1$, $sqrt[3]p$ and $sqrt[3]p^2$ are linearly independent over $Bbb Q$.

Now observe that $x=sqrt[3]p+psqrt[3]p^2$.

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answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At this level one should really justify the inference "Therefore...."
  $endgroup$
  – Bill Dubuque
  yesterday
  
  

  
  
  
  

add a comment | 

2

$begingroup$

By Eisenstein's criterion, $X^3-p$ is irreducible over $Bbb Q$. Therefore $1$, $sqrt[3]p$ and $sqrt[3]p^2$ are linearly independent over $Bbb Q$.

Now observe that $x=sqrt[3]p+psqrt[3]p^2$.

share|cite|improve this answer

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At this level one should really justify the inference "Therefore...."
  $endgroup$
  – Bill Dubuque
  yesterday
  
  

  
  
  
  

add a comment | 

2

$begingroup$

By Eisenstein's criterion, $X^3-p$ is irreducible over $Bbb Q$. Therefore $1$, $sqrt[3]p$ and $sqrt[3]p^2$ are linearly independent over $Bbb Q$.

Now observe that $x=sqrt[3]p+psqrt[3]p^2$.

share|cite|improve this answer

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At this level one should really justify the inference "Therefore...."
  $endgroup$
  – Bill Dubuque
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

By Eisenstein's criterion, $X^3-p$ is irreducible over $Bbb Q$. Therefore $1$, $sqrt[3]p$ and $sqrt[3]p^2$ are linearly independent over $Bbb Q$.

Now observe that $x=sqrt[3]p+psqrt[3]p^2$.

share|cite|improve this answer

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

share|cite|improve this answer

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133