where the root of $f(x) $ belong? [closed]

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

let $f(x) in mathbbZ[x]$ be a monic polynomial.then root of $f $

choose the correct options

$1.$ can belong to $mathbbZ$

$2.$ always belong to ($mathbb R$ $mathbbQ )cup mathbbZ$

$3.$always belong to ($mathbb C$ $mathbbQ )cup mathbbZ$

$4.$ can belong to ($mathbbQ$ $mathbbZ)$

My attempts : i take $f(x) = x^2 -1$ then root will belong to $mathbbZ$,so option 1 is obiously true

here im confusing about the other options

Any Hints/solution will appreciated

thanks in advance

abstract-algebra

share|cite|improve this question

asked Jul 16 at 16:27

stupid

55819

closed as off-topic by Andres Mejia, Mostafa Ayaz, Xander Henderson, Leucippus, Parcly Taxel Jul 17 at 1:29

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andres Mejia, Mostafa Ayaz, Xander Henderson, Leucippus, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For option 4, you have the rational roots theorem.
  – Bernard
  Jul 16 at 16:29
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

let $f(x) in mathbbZ[x]$ be a monic polynomial.then root of $f $

choose the correct options

$1.$ can belong to $mathbbZ$

$2.$ always belong to ($mathbb R$ $mathbbQ )cup mathbbZ$

$3.$always belong to ($mathbb C$ $mathbbQ )cup mathbbZ$

$4.$ can belong to ($mathbbQ$ $mathbbZ)$

My attempts : i take $f(x) = x^2 -1$ then root will belong to $mathbbZ$,so option 1 is obiously true

here im confusing about the other options

Any Hints/solution will appreciated

thanks in advance

abstract-algebra

share|cite|improve this question

asked Jul 16 at 16:27

stupid

55819

closed as off-topic by Andres Mejia, Mostafa Ayaz, Xander Henderson, Leucippus, Parcly Taxel Jul 17 at 1:29

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andres Mejia, Mostafa Ayaz, Xander Henderson, Leucippus, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For option 4, you have the rational roots theorem.
  – Bernard
  Jul 16 at 16:29
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

let $f(x) in mathbbZ[x]$ be a monic polynomial.then root of $f $

choose the correct options

$1.$ can belong to $mathbbZ$

$2.$ always belong to ($mathbb R$ $mathbbQ )cup mathbbZ$

$3.$always belong to ($mathbb C$ $mathbbQ )cup mathbbZ$

$4.$ can belong to ($mathbbQ$ $mathbbZ)$

My attempts : i take $f(x) = x^2 -1$ then root will belong to $mathbbZ$,so option 1 is obiously true

here im confusing about the other options

Any Hints/solution will appreciated

thanks in advance

abstract-algebra

share|cite|improve this question

asked Jul 16 at 16:27

stupid

55819

let $f(x) in mathbbZ[x]$ be a monic polynomial.then root of $f $

choose the correct options

$1.$ can belong to $mathbbZ$

$2.$ always belong to ($mathbb R$ $mathbbQ )cup mathbbZ$

$3.$always belong to ($mathbb C$ $mathbbQ )cup mathbbZ$

$4.$ can belong to ($mathbbQ$ $mathbbZ)$

My attempts : i take $f(x) = x^2 -1$ then root will belong to $mathbbZ$,so option 1 is obiously true

here im confusing about the other options

Any Hints/solution will appreciated

thanks in advance

abstract-algebra

share|cite|improve this question

asked Jul 16 at 16:27

stupid

55819

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 16:27

stupid

55819

asked Jul 16 at 16:27

stupid

55819

asked Jul 16 at 16:27

stupid

55819

55819

closed as off-topic by Andres Mejia, Mostafa Ayaz, Xander Henderson, Leucippus, Parcly Taxel Jul 17 at 1:29

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andres Mejia, Mostafa Ayaz, Xander Henderson, Leucippus, Parcly Taxel

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

closed as off-topic by Andres Mejia, Mostafa Ayaz, Xander Henderson, Leucippus, Parcly Taxel Jul 17 at 1:29

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andres Mejia, Mostafa Ayaz, Xander Henderson, Leucippus, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For option 4, you have the rational roots theorem.
  – Bernard
  Jul 16 at 16:29
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For option 4, you have the rational roots theorem.
  – Bernard
  Jul 16 at 16:29
  

  
  
  
  

1

1

For option 4, you have the rational roots theorem.
– Bernard
Jul 16 at 16:29

For option 4, you have the rational roots theorem.
– Bernard
Jul 16 at 16:29

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

It follows from the rational root theorem that all rational roots of such a polynomial are in fact integers. So, the third option is true and the fourth option is false. And $x^2+1$ shows that the second option is false too.

share|cite|improve this answer

answered Jul 16 at 16:32

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ jose sir option 4 i have doubt that (QZ) = Q...now suppose if i take $ f(x) = x^2 - 1$ , then it will true ..i mean option 4 will true ..
  – stupid
  Jul 17 at 6:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @stupid Why do you say that $mathbbQsetminusmathbbZ=mathbb Q$? This is not true. The set $mathbbQsetminusmathbbZ$ is the set of non-integer rationals.
  – José Carlos Santos
  Jul 17 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ookks thanks u @jose sir
  – stupid
  Jul 17 at 6:38
  

  
  
  
  

add a comment | 

up vote
2
down vote

Hint:

A consequence of the rational roots theorem is that a rational root of a monic polynomial in $mathbfZ[X]$ is necessarily an integer.

share|cite|improve this answer

answered Jul 16 at 16:32

Bernard

110k635103

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

It follows from the rational root theorem that all rational roots of such a polynomial are in fact integers. So, the third option is true and the fourth option is false. And $x^2+1$ shows that the second option is false too.

share|cite|improve this answer

answered Jul 16 at 16:32

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ jose sir option 4 i have doubt that (QZ) = Q...now suppose if i take $ f(x) = x^2 - 1$ , then it will true ..i mean option 4 will true ..
  – stupid
  Jul 17 at 6:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @stupid Why do you say that $mathbbQsetminusmathbbZ=mathbb Q$? This is not true. The set $mathbbQsetminusmathbbZ$ is the set of non-integer rationals.
  – José Carlos Santos
  Jul 17 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ookks thanks u @jose sir
  – stupid
  Jul 17 at 6:38
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

It follows from the rational root theorem that all rational roots of such a polynomial are in fact integers. So, the third option is true and the fourth option is false. And $x^2+1$ shows that the second option is false too.

share|cite|improve this answer

answered Jul 16 at 16:32

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ jose sir option 4 i have doubt that (QZ) = Q...now suppose if i take $ f(x) = x^2 - 1$ , then it will true ..i mean option 4 will true ..
  – stupid
  Jul 17 at 6:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @stupid Why do you say that $mathbbQsetminusmathbbZ=mathbb Q$? This is not true. The set $mathbbQsetminusmathbbZ$ is the set of non-integer rationals.
  – José Carlos Santos
  Jul 17 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ookks thanks u @jose sir
  – stupid
  Jul 17 at 6:38
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

It follows from the rational root theorem that all rational roots of such a polynomial are in fact integers. So, the third option is true and the fourth option is false. And $x^2+1$ shows that the second option is false too.

share|cite|improve this answer

answered Jul 16 at 16:32

José Carlos Santos

114k1698177

It follows from the rational root theorem that all rational roots of such a polynomial are in fact integers. So, the third option is true and the fourth option is false. And $x^2+1$ shows that the second option is false too.

share|cite|improve this answer

answered Jul 16 at 16:32

José Carlos Santos

114k1698177

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 16:32

José Carlos Santos

114k1698177

answered Jul 16 at 16:32

José Carlos Santos

114k1698177

answered Jul 16 at 16:32

José Carlos Santos

114k1698177

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ jose sir option 4 i have doubt that (QZ) = Q...now suppose if i take $ f(x) = x^2 - 1$ , then it will true ..i mean option 4 will true ..
  – stupid
  Jul 17 at 6:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @stupid Why do you say that $mathbbQsetminusmathbbZ=mathbb Q$? This is not true. The set $mathbbQsetminusmathbbZ$ is the set of non-integer rationals.
  – José Carlos Santos
  Jul 17 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ookks thanks u @jose sir
  – stupid
  Jul 17 at 6:38
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ jose sir option 4 i have doubt that (QZ) = Q...now suppose if i take $ f(x) = x^2 - 1$ , then it will true ..i mean option 4 will true ..
  – stupid
  Jul 17 at 6:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @stupid Why do you say that $mathbbQsetminusmathbbZ=mathbb Q$? This is not true. The set $mathbbQsetminusmathbbZ$ is the set of non-integer rationals.
  – José Carlos Santos
  Jul 17 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ookks thanks u @jose sir
  – stupid
  Jul 17 at 6:38
  

  
  
  
  

@ jose sir option 4 i have doubt that (QZ) = Q...now suppose if i take $ f(x) = x^2 - 1$ , then it will true ..i mean option 4 will true ..
– stupid
Jul 17 at 6:34

@ jose sir option 4 i have doubt that (QZ) = Q...now suppose if i take $ f(x) = x^2 - 1$ , then it will true ..i mean option 4 will true ..
– stupid
Jul 17 at 6:34

1

1

@stupid Why do you say that $mathbbQsetminusmathbbZ=mathbb Q$? This is not true. The set $mathbbQsetminusmathbbZ$ is the set of non-integer rationals.
– José Carlos Santos
Jul 17 at 6:37

@stupid Why do you say that $mathbbQsetminusmathbbZ=mathbb Q$? This is not true. The set $mathbbQsetminusmathbbZ$ is the set of non-integer rationals.
– José Carlos Santos
Jul 17 at 6:37

ookks thanks u @jose sir
– stupid
Jul 17 at 6:38

ookks thanks u @jose sir
– stupid
Jul 17 at 6:38

add a comment | 

up vote
2
down vote

Hint:

A consequence of the rational roots theorem is that a rational root of a monic polynomial in $mathbfZ[X]$ is necessarily an integer.

share|cite|improve this answer

answered Jul 16 at 16:32

Bernard

110k635103

add a comment | 

up vote
2
down vote

Hint:

A consequence of the rational roots theorem is that a rational root of a monic polynomial in $mathbfZ[X]$ is necessarily an integer.

share|cite|improve this answer

answered Jul 16 at 16:32

Bernard

110k635103

add a comment | 

up vote
2
down vote

up vote
2
down vote

Hint:

A consequence of the rational roots theorem is that a rational root of a monic polynomial in $mathbfZ[X]$ is necessarily an integer.

share|cite|improve this answer

answered Jul 16 at 16:32

Bernard

110k635103

Hint:

A consequence of the rational roots theorem is that a rational root of a monic polynomial in $mathbfZ[X]$ is necessarily an integer.

share|cite|improve this answer

answered Jul 16 at 16:32

Bernard

110k635103

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 16:32

Bernard

110k635103

answered Jul 16 at 16:32

Bernard

110k635103

answered Jul 16 at 16:32

Bernard

110k635103

110k635103

add a comment | 

add a comment |