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2

1

$begingroup$

I was working on a problem, and I narrowed it down to the following:

$$c=sqrtfracp^2+2a^2q^2-a^2a^2q^2$$

such that $p$ and $q$ are integers, and $a$ and $c$ are rationals.

Does $c$ exist? So obviously, the numerator is not a perfect square, so it cannot be trivially simplified. Basically my question is to remove the radical in any way to solve for $c.$

I have accepted an answer, however, it is not quite what I was looking for.

This question is very intricately related to another question of mine, that I have not been able to solve: (I don't know if promoting questions through others is legal)

Rational Solution Set to these equations and inequalities?

Please help me!

algebra-precalculus

share|cite|improve this question

edited 8 hours ago

Atharv Sampath

asked 8 hours ago

Atharv SampathAtharv Sampath

4255 bronze badges

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What was the original question?
  $endgroup$
  – kingW3
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it is rather complicated to explain, should i make a new post about it or just edit this post?
  $endgroup$
  – Atharv Sampath
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did you accept the answer, if it's not what you are looking for?
  $endgroup$
  – TonyK
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, it correctly answers this question, but doesn't help much on the other
  $endgroup$
  – Atharv Sampath
  8 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

I was working on a problem, and I narrowed it down to the following:

$$c=sqrtfracp^2+2a^2q^2-a^2a^2q^2$$

such that $p$ and $q$ are integers, and $a$ and $c$ are rationals.

Does $c$ exist? So obviously, the numerator is not a perfect square, so it cannot be trivially simplified. Basically my question is to remove the radical in any way to solve for $c.$

I have accepted an answer, however, it is not quite what I was looking for.

This question is very intricately related to another question of mine, that I have not been able to solve: (I don't know if promoting questions through others is legal)

Rational Solution Set to these equations and inequalities?

Please help me!

algebra-precalculus

share|cite|improve this question

edited 8 hours ago

Atharv Sampath

asked 8 hours ago

Atharv SampathAtharv Sampath

4255 bronze badges

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What was the original question?
  $endgroup$
  – kingW3
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it is rather complicated to explain, should i make a new post about it or just edit this post?
  $endgroup$
  – Atharv Sampath
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did you accept the answer, if it's not what you are looking for?
  $endgroup$
  – TonyK
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, it correctly answers this question, but doesn't help much on the other
  $endgroup$
  – Atharv Sampath
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I was working on a problem, and I narrowed it down to the following:

$$c=sqrtfracp^2+2a^2q^2-a^2a^2q^2$$

such that $p$ and $q$ are integers, and $a$ and $c$ are rationals.

Does $c$ exist? So obviously, the numerator is not a perfect square, so it cannot be trivially simplified. Basically my question is to remove the radical in any way to solve for $c.$

I have accepted an answer, however, it is not quite what I was looking for.

This question is very intricately related to another question of mine, that I have not been able to solve: (I don't know if promoting questions through others is legal)

Rational Solution Set to these equations and inequalities?

Please help me!

algebra-precalculus

share|cite|improve this question

edited 8 hours ago

Atharv Sampath

asked 8 hours ago

Atharv SampathAtharv Sampath

4255 bronze badges

New contributor

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

$endgroup$

share|cite|improve this question

edited 8 hours ago

Atharv Sampath

asked 8 hours ago

Atharv SampathAtharv Sampath

4255 bronze badges

New contributor

Atharv Sampath 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 8 hours ago

Atharv Sampath

asked 8 hours ago

Atharv SampathAtharv Sampath

4255 bronze badges

New contributor

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

asked 8 hours ago

Atharv SampathAtharv Sampath

4255 bronze badges

New contributor

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

asked 8 hours ago

Atharv SampathAtharv Sampath

4255 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

First of all, we need to require that $a$ and $q$ both be nonzero (look at the denominator of the right hand side). On the plus side, once we make this stipulation we have $2a^2q^2>a^2$ and so $sqrtp^2+2a^2q^2-a^2$ - and hence $c$ itself - is at least a real number!

---

Now with that stipulation, note that the right hand side simplifies to $$sqrtp^2+2a^2q^2-a^2over vert aqvert.$$ This is rational iff $p^2+2a^2q^2-a^2$ is the square of a rational number; sometimes it will be and sometimes it won't be. For example, taking $q=1$ we get $p^2+2a^2q^2-a^2=p^2+a^2$. Sometimes this will be the square of a rational (e.g. $p=3, a=4$) and other times it won't be (e.g. $p=1, a=1$). So the answer to your question is a resounding maybe.

share|cite|improve this answer

answered 8 hours ago

Noah SchweberNoah Schweber

133k1010 gold badges159159 silver badges302302 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

By setting $r = fracpaq$ and $s=frac1q$, you can simplify this to
$$c = sqrt2 + r^2 - s^2
$$
This leads to a counterexample. If $r=s=1$ then $c=sqrt2$, which is not rational. And to get $r=s=1$ we simply set $a=p=q=1$.

share|cite|improve this answer

answered 8 hours ago

Lee MosherLee Mosher

55.9k44 gold badges3838 silver badges9696 bronze badges

$endgroup$

add a comment | 

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5

$begingroup$

First of all, we need to require that $a$ and $q$ both be nonzero (look at the denominator of the right hand side). On the plus side, once we make this stipulation we have $2a^2q^2>a^2$ and so $sqrtp^2+2a^2q^2-a^2$ - and hence $c$ itself - is at least a real number!

---

Now with that stipulation, note that the right hand side simplifies to $$sqrtp^2+2a^2q^2-a^2over vert aqvert.$$ This is rational iff $p^2+2a^2q^2-a^2$ is the square of a rational number; sometimes it will be and sometimes it won't be. For example, taking $q=1$ we get $p^2+2a^2q^2-a^2=p^2+a^2$. Sometimes this will be the square of a rational (e.g. $p=3, a=4$) and other times it won't be (e.g. $p=1, a=1$). So the answer to your question is a resounding maybe.

share|cite|improve this answer

answered 8 hours ago

Noah SchweberNoah Schweber

133k1010 gold badges159159 silver badges302302 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

First of all, we need to require that $a$ and $q$ both be nonzero (look at the denominator of the right hand side). On the plus side, once we make this stipulation we have $2a^2q^2>a^2$ and so $sqrtp^2+2a^2q^2-a^2$ - and hence $c$ itself - is at least a real number!

---

Now with that stipulation, note that the right hand side simplifies to $$sqrtp^2+2a^2q^2-a^2over vert aqvert.$$ This is rational iff $p^2+2a^2q^2-a^2$ is the square of a rational number; sometimes it will be and sometimes it won't be. For example, taking $q=1$ we get $p^2+2a^2q^2-a^2=p^2+a^2$. Sometimes this will be the square of a rational (e.g. $p=3, a=4$) and other times it won't be (e.g. $p=1, a=1$). So the answer to your question is a resounding maybe.

share|cite|improve this answer

answered 8 hours ago

Noah SchweberNoah Schweber

133k1010 gold badges159159 silver badges302302 bronze badges

$endgroup$

add a comment | 

$begingroup$

First of all, we need to require that $a$ and $q$ both be nonzero (look at the denominator of the right hand side). On the plus side, once we make this stipulation we have $2a^2q^2>a^2$ and so $sqrtp^2+2a^2q^2-a^2$ - and hence $c$ itself - is at least a real number!

---

Now with that stipulation, note that the right hand side simplifies to $$sqrtp^2+2a^2q^2-a^2over vert aqvert.$$ This is rational iff $p^2+2a^2q^2-a^2$ is the square of a rational number; sometimes it will be and sometimes it won't be. For example, taking $q=1$ we get $p^2+2a^2q^2-a^2=p^2+a^2$. Sometimes this will be the square of a rational (e.g. $p=3, a=4$) and other times it won't be (e.g. $p=1, a=1$). So the answer to your question is a resounding maybe.

share|cite|improve this answer

answered 8 hours ago

Noah SchweberNoah Schweber

133k1010 gold badges159159 silver badges302302 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Noah SchweberNoah Schweber

133k1010 gold badges159159 silver badges302302 bronze badges

answered 8 hours ago

Noah SchweberNoah Schweber

133k1010 gold badges159159 silver badges302302 bronze badges

answered 8 hours ago

Noah SchweberNoah Schweber

133k1010 gold badges159159 silver badges302302 bronze badges

1

$begingroup$

By setting $r = fracpaq$ and $s=frac1q$, you can simplify this to
$$c = sqrt2 + r^2 - s^2
$$
This leads to a counterexample. If $r=s=1$ then $c=sqrt2$, which is not rational. And to get $r=s=1$ we simply set $a=p=q=1$.

share|cite|improve this answer

answered 8 hours ago

Lee MosherLee Mosher

55.9k44 gold badges3838 silver badges9696 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

By setting $r = fracpaq$ and $s=frac1q$, you can simplify this to
$$c = sqrt2 + r^2 - s^2
$$
This leads to a counterexample. If $r=s=1$ then $c=sqrt2$, which is not rational. And to get $r=s=1$ we simply set $a=p=q=1$.

share|cite|improve this answer

answered 8 hours ago

Lee MosherLee Mosher

55.9k44 gold badges3838 silver badges9696 bronze badges

$endgroup$

add a comment | 

$begingroup$

By setting $r = fracpaq$ and $s=frac1q$, you can simplify this to
$$c = sqrt2 + r^2 - s^2
$$
This leads to a counterexample. If $r=s=1$ then $c=sqrt2$, which is not rational. And to get $r=s=1$ we simply set $a=p=q=1$.

share|cite|improve this answer

answered 8 hours ago

Lee MosherLee Mosher

55.9k44 gold badges3838 silver badges9696 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Lee MosherLee Mosher

55.9k44 gold badges3838 silver badges9696 bronze badges

answered 8 hours ago

Lee MosherLee Mosher

55.9k44 gold badges3838 silver badges9696 bronze badges

answered 8 hours ago

Lee MosherLee Mosher

55.9k44 gold badges3838 silver badges9696 bronze badges