Fdtxjr

Why does cron require MTA for logging?

What's the 'present simple' form of the word "нашла́" in 3rd person singular female?

What ability score modifier does a javelin's damage use?

How can I find out information about a service?

Which classes are needed to have access to every spell in the PHB?

Shifting between bemols and diesis in the key signature

I reported the illegal activity of my boss to his boss. My boss found out. Now I am being punished. What should I do?

Doesn't allowing a user mode program to access kernel space memory and execute the IN and OUT instructions defeat the purpose of having CPU modes?

How to resolve: Reviewer #1 says remove section X vs. Reviewer #2 says expand section X

What do you call someone who likes to pick fights?

How many characters using PHB rules does it take to be able to have access to any PHB spell at the start of an adventuring day?

Outlet with 3 sets of wires

Power Strip for Europe

PTIJ: Why does only a Shor Tam ask at the Seder, and not a Shor Mu'ad?

Why is a very small peak with larger m/z not considered to be the molecular ion?

When Schnorr signatures are part of Bitcoin will it be possible validate each block with only one signature validation?

Professor forcing me to attend a conference, I can't afford even with 50% funding

MySQL importing CSV files really slow

Why is gluten-free baking possible?

Signed and unsigned numbers

Was it really inappropriate to write a pull request for the company I interviewed with?

Would an aboleth's Phantasmal Force lair action be affected by Counterspell, Dispel Magic, and/or Slow?

Specifying a starting column with colortbl package and xcolor

What is the generally accepted pronunciation of “topoi”?

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

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 23 hours ago

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 23 hours ago

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
1

active

oldest

votes

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 | 

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