If $q in mathbb Q setminus mathbb Z$ then $q(q+1) notin mathbbZ$? [on hold]

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If $q in mathbb Q setminus mathbb Z$ then how I can prove that $q(q+1)$ will never be an integer? I tried without luck.

elementary-number-theory arithmetic rational-numbers products

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put on hold as off-topic by amWhy, John Ma, user21820, Did, user91500 2 days 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 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." – amWhy, John Ma, user21820, Did, user91500

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

• 
  
  
  

  
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  start with writing $q=m/n$ such that $m$ and $n$ have no common factors and $n>1$.
  – Marco
  2 days ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

If $q in mathbb Q setminus mathbb Z$ then how I can prove that $q(q+1)$ will never be an integer? I tried without luck.

elementary-number-theory arithmetic rational-numbers products

share|cite|improve this question

edited 2 days ago

user21820

36.2k440140

asked 2 days ago

user591656

182

New contributor

user591656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

put on hold as off-topic by amWhy, John Ma, user21820, Did, user91500 2 days 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 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." – amWhy, John Ma, user21820, Did, user91500

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  start with writing $q=m/n$ such that $m$ and $n$ have no common factors and $n>1$.
  – Marco
  2 days ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

If $q in mathbb Q setminus mathbb Z$ then how I can prove that $q(q+1)$ will never be an integer? I tried without luck.

elementary-number-theory arithmetic rational-numbers products

share|cite|improve this question

edited 2 days ago

user21820

36.2k440140

asked 2 days ago

user591656

182

New contributor

user591656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

If $q in mathbb Q setminus mathbb Z$ then how I can prove that $q(q+1)$ will never be an integer? I tried without luck.

elementary-number-theory arithmetic rational-numbers products

elementary-number-theory arithmetic rational-numbers products

share|cite|improve this question

edited 2 days ago

user21820

36.2k440140

asked 2 days ago

user591656

182

New contributor

user591656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 days ago

user21820

36.2k440140

asked 2 days ago

user591656

182

New contributor

user591656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

share|cite|improve this question

edited 2 days ago

user21820

36.2k440140

edited 2 days ago

user21820

36.2k440140

edited 2 days ago

user21820

36.2k440140

36.2k440140

asked 2 days ago

user591656

182

New contributor

user591656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

user591656

182

asked 2 days ago

user591656

182

182

New contributor

user591656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

user591656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

user591656 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

put on hold as off-topic by amWhy, John Ma, user21820, Did, user91500 2 days 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 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." – amWhy, John Ma, user21820, Did, user91500

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

put on hold as off-topic by amWhy, John Ma, user21820, Did, user91500 2 days 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 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." – amWhy, John Ma, user21820, Did, user91500

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  start with writing $q=m/n$ such that $m$ and $n$ have no common factors and $n>1$.
  – Marco
  2 days ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  start with writing $q=m/n$ such that $m$ and $n$ have no common factors and $n>1$.
  – Marco
  2 days ago
  

  
  
  
  

2

2

start with writing $q=m/n$ such that $m$ and $n$ have no common factors and $n>1$.
– Marco
2 days ago

start with writing $q=m/n$ such that $m$ and $n$ have no common factors and $n>1$.
– Marco
2 days ago

add a comment | 

2 Answers
2

active

oldest

votes

up vote
6
down vote

Assume the product is an integer. Call it $z$.

Therefore

$$qtimes (q+1)=z$$
$$iff q^2-q-z=0$$

Applying the Rational root theorem, and there is a contradiction.

_{(Remark: short proof. Not so simple if you don't know the theorem in the first place)}

share|cite|improve this answer

answered 2 days ago

user202729

1,235412

add a comment | 

up vote
6
down vote

Since $qinmathbbQsetminusmathbbZ$ it is $q=fracab$ and $bnmid a$.

[Edit: Where without loss of generality $fracab$ is a reduced fraction.]

Then $q+1=fracab+1=fraca+bb$ and $qcdot (q+1)=fraca(a+b)b^2$.

We have to show, that $b^2nmid a(a+b)$.

Suppose $b^2mid a(a+b)$. Then exists some $kinmathbbZ$ with $b^2k=a(a+b)Leftrightarrow a^2=b(bk-a)$

Which means $bmid a^2$. But then $bmid a$, which is a contradiction.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Cornman

2,84221228

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  The last line - $b=9$ and $a=6$, easy fix once use reduced terms.
  – AHusain
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for pointing that out.
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $a^2=b(bk-a)$ then why are you being presumptuous that $b$ divides $a^2$ ? You haven't proved that $(bk-a)$ can't be a divisor of $a^2$ .
  – user591656
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user591656 We can write $a^2=b(bk-a)$ that means $bmid a^2$ and $bk-amid a^2$.
  – Cornman
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @user3856370 $frac159$ is not a reduced fraction, that's what the edit was about. The edit could alternatively be stated as "Assume wlog that $frac ab$ is a reduced fraction, i.e. $gcd(a,b)=1$. Then $bnmid aimplies bnmid a^2$", allowing $bmid a^2$ to be the contradiction, if that is easier to understand.
  – hvd
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
6
down vote

Assume the product is an integer. Call it $z$.

Therefore

$$qtimes (q+1)=z$$
$$iff q^2-q-z=0$$

Applying the Rational root theorem, and there is a contradiction.

_{(Remark: short proof. Not so simple if you don't know the theorem in the first place)}

share|cite|improve this answer

answered 2 days ago

user202729

1,235412

add a comment | 

up vote
6
down vote

Assume the product is an integer. Call it $z$.

Therefore

$$qtimes (q+1)=z$$
$$iff q^2-q-z=0$$

Applying the Rational root theorem, and there is a contradiction.

_{(Remark: short proof. Not so simple if you don't know the theorem in the first place)}

share|cite|improve this answer

answered 2 days ago

user202729

1,235412

add a comment | 

up vote
6
down vote

up vote
6
down vote

Assume the product is an integer. Call it $z$.

Therefore

$$qtimes (q+1)=z$$
$$iff q^2-q-z=0$$

Applying the Rational root theorem, and there is a contradiction.

_{(Remark: short proof. Not so simple if you don't know the theorem in the first place)}

share|cite|improve this answer

answered 2 days ago

user202729

1,235412

Assume the product is an integer. Call it $z$.

Therefore

$$qtimes (q+1)=z$$
$$iff q^2-q-z=0$$

Applying the Rational root theorem, and there is a contradiction.

_{(Remark: short proof. Not so simple if you don't know the theorem in the first place)}

share|cite|improve this answer

answered 2 days ago

user202729

1,235412

share|cite|improve this answer

share|cite|improve this answer

answered 2 days ago

user202729

1,235412

answered 2 days ago

user202729

1,235412

answered 2 days ago

user202729

1,235412

1,235412

add a comment | 

add a comment | 

up vote
6
down vote

Since $qinmathbbQsetminusmathbbZ$ it is $q=fracab$ and $bnmid a$.

[Edit: Where without loss of generality $fracab$ is a reduced fraction.]

Then $q+1=fracab+1=fraca+bb$ and $qcdot (q+1)=fraca(a+b)b^2$.

We have to show, that $b^2nmid a(a+b)$.

Suppose $b^2mid a(a+b)$. Then exists some $kinmathbbZ$ with $b^2k=a(a+b)Leftrightarrow a^2=b(bk-a)$

Which means $bmid a^2$. But then $bmid a$, which is a contradiction.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Cornman

2,84221228

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  The last line - $b=9$ and $a=6$, easy fix once use reduced terms.
  – AHusain
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for pointing that out.
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $a^2=b(bk-a)$ then why are you being presumptuous that $b$ divides $a^2$ ? You haven't proved that $(bk-a)$ can't be a divisor of $a^2$ .
  – user591656
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user591656 We can write $a^2=b(bk-a)$ that means $bmid a^2$ and $bk-amid a^2$.
  – Cornman
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @user3856370 $frac159$ is not a reduced fraction, that's what the edit was about. The edit could alternatively be stated as "Assume wlog that $frac ab$ is a reduced fraction, i.e. $gcd(a,b)=1$. Then $bnmid aimplies bnmid a^2$", allowing $bmid a^2$ to be the contradiction, if that is easier to understand.
  – hvd
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

up vote
6
down vote

Since $qinmathbbQsetminusmathbbZ$ it is $q=fracab$ and $bnmid a$.

[Edit: Where without loss of generality $fracab$ is a reduced fraction.]

Then $q+1=fracab+1=fraca+bb$ and $qcdot (q+1)=fraca(a+b)b^2$.

We have to show, that $b^2nmid a(a+b)$.

Suppose $b^2mid a(a+b)$. Then exists some $kinmathbbZ$ with $b^2k=a(a+b)Leftrightarrow a^2=b(bk-a)$

Which means $bmid a^2$. But then $bmid a$, which is a contradiction.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Cornman

2,84221228

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  The last line - $b=9$ and $a=6$, easy fix once use reduced terms.
  – AHusain
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for pointing that out.
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $a^2=b(bk-a)$ then why are you being presumptuous that $b$ divides $a^2$ ? You haven't proved that $(bk-a)$ can't be a divisor of $a^2$ .
  – user591656
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user591656 We can write $a^2=b(bk-a)$ that means $bmid a^2$ and $bk-amid a^2$.
  – Cornman
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @user3856370 $frac159$ is not a reduced fraction, that's what the edit was about. The edit could alternatively be stated as "Assume wlog that $frac ab$ is a reduced fraction, i.e. $gcd(a,b)=1$. Then $bnmid aimplies bnmid a^2$", allowing $bmid a^2$ to be the contradiction, if that is easier to understand.
  – hvd
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

up vote
6
down vote

up vote
6
down vote

Since $qinmathbbQsetminusmathbbZ$ it is $q=fracab$ and $bnmid a$.

[Edit: Where without loss of generality $fracab$ is a reduced fraction.]

Then $q+1=fracab+1=fraca+bb$ and $qcdot (q+1)=fraca(a+b)b^2$.

We have to show, that $b^2nmid a(a+b)$.

Suppose $b^2mid a(a+b)$. Then exists some $kinmathbbZ$ with $b^2k=a(a+b)Leftrightarrow a^2=b(bk-a)$

Which means $bmid a^2$. But then $bmid a$, which is a contradiction.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Cornman

2,84221228

Since $qinmathbbQsetminusmathbbZ$ it is $q=fracab$ and $bnmid a$.

[Edit: Where without loss of generality $fracab$ is a reduced fraction.]

Then $q+1=fracab+1=fraca+bb$ and $qcdot (q+1)=fraca(a+b)b^2$.

We have to show, that $b^2nmid a(a+b)$.

Suppose $b^2mid a(a+b)$. Then exists some $kinmathbbZ$ with $b^2k=a(a+b)Leftrightarrow a^2=b(bk-a)$

Which means $bmid a^2$. But then $bmid a$, which is a contradiction.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Cornman

2,84221228

share|cite|improve this answer

share|cite|improve this answer

edited 2 days ago

edited 2 days ago

edited 2 days ago

answered 2 days ago

Cornman

2,84221228

answered 2 days ago

Cornman

2,84221228

answered 2 days ago

Cornman

2,84221228

2,84221228

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  The last line - $b=9$ and $a=6$, easy fix once use reduced terms.
  – AHusain
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for pointing that out.
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $a^2=b(bk-a)$ then why are you being presumptuous that $b$ divides $a^2$ ? You haven't proved that $(bk-a)$ can't be a divisor of $a^2$ .
  – user591656
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user591656 We can write $a^2=b(bk-a)$ that means $bmid a^2$ and $bk-amid a^2$.
  – Cornman
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @user3856370 $frac159$ is not a reduced fraction, that's what the edit was about. The edit could alternatively be stated as "Assume wlog that $frac ab$ is a reduced fraction, i.e. $gcd(a,b)=1$. Then $bnmid aimplies bnmid a^2$", allowing $bmid a^2$ to be the contradiction, if that is easier to understand.
  – hvd
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  The last line - $b=9$ and $a=6$, easy fix once use reduced terms.
  – AHusain
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for pointing that out.
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $a^2=b(bk-a)$ then why are you being presumptuous that $b$ divides $a^2$ ? You haven't proved that $(bk-a)$ can't be a divisor of $a^2$ .
  – user591656
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user591656 We can write $a^2=b(bk-a)$ that means $bmid a^2$ and $bk-amid a^2$.
  – Cornman
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @user3856370 $frac159$ is not a reduced fraction, that's what the edit was about. The edit could alternatively be stated as "Assume wlog that $frac ab$ is a reduced fraction, i.e. $gcd(a,b)=1$. Then $bnmid aimplies bnmid a^2$", allowing $bmid a^2$ to be the contradiction, if that is easier to understand.
  – hvd
  2 days ago
  

  
  
  
  

4

4

The last line - $b=9$ and $a=6$, easy fix once use reduced terms.
– AHusain
2 days ago

The last line - $b=9$ and $a=6$, easy fix once use reduced terms.
– AHusain
2 days ago

Thanks for pointing that out.
– Cornman
2 days ago

Thanks for pointing that out.
– Cornman
2 days ago

If $a^2=b(bk-a)$ then why are you being presumptuous that $b$ divides $a^2$ ? You haven't proved that $(bk-a)$ can't be a divisor of $a^2$ .
– user591656
2 days ago

If $a^2=b(bk-a)$ then why are you being presumptuous that $b$ divides $a^2$ ? You haven't proved that $(bk-a)$ can't be a divisor of $a^2$ .
– user591656
2 days ago

1

1

@user591656 We can write $a^2=b(bk-a)$ that means $bmid a^2$ and $bk-amid a^2$.
– Cornman
2 days ago

@user591656 We can write $a^2=b(bk-a)$ that means $bmid a^2$ and $bk-amid a^2$.
– Cornman
2 days ago

2

2

@user3856370 $frac159$ is not a reduced fraction, that's what the edit was about. The edit could alternatively be stated as "Assume wlog that $frac ab$ is a reduced fraction, i.e. $gcd(a,b)=1$. Then $bnmid aimplies bnmid a^2$", allowing $bmid a^2$ to be the contradiction, if that is easier to understand.
– hvd
2 days ago

@user3856370 $frac159$ is not a reduced fraction, that's what the edit was about. The edit could alternatively be stated as "Assume wlog that $frac ab$ is a reduced fraction, i.e. $gcd(a,b)=1$. Then $bnmid aimplies bnmid a^2$", allowing $bmid a^2$ to be the contradiction, if that is easier to understand.
– hvd
2 days ago

 | 
show 1 more comment