Prove that $a+b sqrt2$ is not a field [closed]

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I have a problem with understanding the top answer from here:
prove that $(a + b sqrt2,+,*)$ is not a field.

I'm probably making a trivial mistake along the way, but here's my reasoning: We want to show that there exists $a+bsqrt2$ such that there's no inverse $frac1a+bsqrt2$ in the field that suffices the equation $a+bsqrt2 * frac1a+bsqrt2 = 1$.

lhf starts with $frac1a+bsqrt2 = 2$ and draws a contradiction. What I don't understand is - aren't we working the other way around, i.e. showing that for a particular inverse (in this case equal to $2$) there's no corresponding number in the field? How does it prove that it's not a field? I mean, $1/pi$ is not an inverse of any number in $mathbbQ$, but it doesn't prove that $mathbbQ$ is not a field.

abstract-algebra

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edited Jul 22 at 14:14

Vinícius Lopes Simões

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asked Jul 22 at 13:56

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closed as off-topic by amWhy, Dietrich Burde, José Carlos Santos, max_zorn, Leucippus Jul 23 at 1:42

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, Dietrich Burde, José Carlos Santos, Leucippus

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

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  You have not quoted the question fully: the extra stipulation is that $a$, $binBbb Z$. $sqrt2$ has no reciprocal $a+bsqrt2$ with $a$, $b$ integers.
  – Lord Shark the Unknown
  Jul 22 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you don't say what are $;a,b;$ your question hardly makes sense...
  – DonAntonio
  Jul 22 at 14:32
  

  
  
  
  

add a comment | 

up vote
-2
down vote

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I have a problem with understanding the top answer from here:
prove that $(a + b sqrt2,+,*)$ is not a field.

I'm probably making a trivial mistake along the way, but here's my reasoning: We want to show that there exists $a+bsqrt2$ such that there's no inverse $frac1a+bsqrt2$ in the field that suffices the equation $a+bsqrt2 * frac1a+bsqrt2 = 1$.

lhf starts with $frac1a+bsqrt2 = 2$ and draws a contradiction. What I don't understand is - aren't we working the other way around, i.e. showing that for a particular inverse (in this case equal to $2$) there's no corresponding number in the field? How does it prove that it's not a field? I mean, $1/pi$ is not an inverse of any number in $mathbbQ$, but it doesn't prove that $mathbbQ$ is not a field.

abstract-algebra

share|cite|improve this question

edited Jul 22 at 14:14

Vinícius Lopes Simões

409211

asked Jul 22 at 13:56

Dumb Dumb

293

closed as off-topic by amWhy, Dietrich Burde, José Carlos Santos, max_zorn, Leucippus Jul 23 at 1:42

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, Dietrich Burde, José Carlos Santos, Leucippus

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You have not quoted the question fully: the extra stipulation is that $a$, $binBbb Z$. $sqrt2$ has no reciprocal $a+bsqrt2$ with $a$, $b$ integers.
  – Lord Shark the Unknown
  Jul 22 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you don't say what are $;a,b;$ your question hardly makes sense...
  – DonAntonio
  Jul 22 at 14:32
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

I have a problem with understanding the top answer from here:
prove that $(a + b sqrt2,+,*)$ is not a field.

I'm probably making a trivial mistake along the way, but here's my reasoning: We want to show that there exists $a+bsqrt2$ such that there's no inverse $frac1a+bsqrt2$ in the field that suffices the equation $a+bsqrt2 * frac1a+bsqrt2 = 1$.

lhf starts with $frac1a+bsqrt2 = 2$ and draws a contradiction. What I don't understand is - aren't we working the other way around, i.e. showing that for a particular inverse (in this case equal to $2$) there's no corresponding number in the field? How does it prove that it's not a field? I mean, $1/pi$ is not an inverse of any number in $mathbbQ$, but it doesn't prove that $mathbbQ$ is not a field.

abstract-algebra

share|cite|improve this question

edited Jul 22 at 14:14

Vinícius Lopes Simões

409211

asked Jul 22 at 13:56

Dumb Dumb

293

I have a problem with understanding the top answer from here:
prove that $(a + b sqrt2,+,*)$ is not a field.

I'm probably making a trivial mistake along the way, but here's my reasoning: We want to show that there exists $a+bsqrt2$ such that there's no inverse $frac1a+bsqrt2$ in the field that suffices the equation $a+bsqrt2 * frac1a+bsqrt2 = 1$.

lhf starts with $frac1a+bsqrt2 = 2$ and draws a contradiction. What I don't understand is - aren't we working the other way around, i.e. showing that for a particular inverse (in this case equal to $2$) there's no corresponding number in the field? How does it prove that it's not a field? I mean, $1/pi$ is not an inverse of any number in $mathbbQ$, but it doesn't prove that $mathbbQ$ is not a field.

abstract-algebra

share|cite|improve this question

edited Jul 22 at 14:14

Vinícius Lopes Simões

409211

asked Jul 22 at 13:56

Dumb Dumb

293

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 14:14

Vinícius Lopes Simões

409211

edited Jul 22 at 14:14

Vinícius Lopes Simões

409211

edited Jul 22 at 14:14

Vinícius Lopes Simões

409211

409211

asked Jul 22 at 13:56

Dumb Dumb

293

asked Jul 22 at 13:56

Dumb Dumb

293

asked Jul 22 at 13:56

Dumb Dumb

293

293

closed as off-topic by amWhy, Dietrich Burde, José Carlos Santos, max_zorn, Leucippus Jul 23 at 1:42

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, Dietrich Burde, José Carlos Santos, Leucippus

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

closed as off-topic by amWhy, Dietrich Burde, José Carlos Santos, max_zorn, Leucippus Jul 23 at 1:42

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, Dietrich Burde, José Carlos Santos, Leucippus

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You have not quoted the question fully: the extra stipulation is that $a$, $binBbb Z$. $sqrt2$ has no reciprocal $a+bsqrt2$ with $a$, $b$ integers.
  – Lord Shark the Unknown
  Jul 22 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you don't say what are $;a,b;$ your question hardly makes sense...
  – DonAntonio
  Jul 22 at 14:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You have not quoted the question fully: the extra stipulation is that $a$, $binBbb Z$. $sqrt2$ has no reciprocal $a+bsqrt2$ with $a$, $b$ integers.
  – Lord Shark the Unknown
  Jul 22 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you don't say what are $;a,b;$ your question hardly makes sense...
  – DonAntonio
  Jul 22 at 14:32
  

  
  
  
  

2

2

You have not quoted the question fully: the extra stipulation is that $a$, $binBbb Z$. $sqrt2$ has no reciprocal $a+bsqrt2$ with $a$, $b$ integers.
– Lord Shark the Unknown
Jul 22 at 13:59

You have not quoted the question fully: the extra stipulation is that $a$, $binBbb Z$. $sqrt2$ has no reciprocal $a+bsqrt2$ with $a$, $b$ integers.
– Lord Shark the Unknown
Jul 22 at 13:59

If you don't say what are $;a,b;$ your question hardly makes sense...
– DonAntonio
Jul 22 at 14:32

If you don't say what are $;a,b;$ your question hardly makes sense...
– DonAntonio
Jul 22 at 14:32

add a comment | 

4 Answers
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To show that $(a,b in mathbbZ, + , times)$ is not a field, it suffices to find a particular non-zero element from the set that does not have an inverse.

$2 in a,b in mathbbZ$, hence, if it doesn't have a multiplicative inverse, then $(a,b in mathbbZ, + , times)$ is not a field.

$pi notin mathbbQ$, it can't be used to show that $mathbbQ$ is not a field. (and actually $(mathbbQ, + , times)$ is a field.)

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answered Jul 22 at 14:01

Siong Thye Goh

77.6k134795

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Since $(2+sqrt2)(2-sqrt2)=2$, $a,,binmathbbZ$ contains $2+sqrt2$ but no inverse for it.

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answered Jul 22 at 14:01

J.G.

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Does the number $2$ have inverse? No.

Suppose it has an inverse $a+bsqrt 2$, with $a,bin mathbb Z$. Then we have $2(a+bsqrt 2)=1$, which means $2a=1$ and $2bsqrt 2 =0$.

Now, there is no integer $a$ such that $2a=1$, which is a contradiction.

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answered Jul 22 at 14:01

Hugocito

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The problem is in finding the inverse for nonzero elements.

The inverse of $$a+bsqrt 2$$ is $$frac 1a+bsqrt 2= frac a-bsqrt 2a^2-2b^2$$

What if $$a^2-2b^2 =0$$ in which case the inverse is undefined ?

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answered Jul 22 at 14:48

Mohammad Riazi-Kermani

27.5k41852

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

To show that $(a,b in mathbbZ, + , times)$ is not a field, it suffices to find a particular non-zero element from the set that does not have an inverse.

$2 in a,b in mathbbZ$, hence, if it doesn't have a multiplicative inverse, then $(a,b in mathbbZ, + , times)$ is not a field.

$pi notin mathbbQ$, it can't be used to show that $mathbbQ$ is not a field. (and actually $(mathbbQ, + , times)$ is a field.)

share|cite|improve this answer

answered Jul 22 at 14:01

Siong Thye Goh

77.6k134795

add a comment | 

up vote
3
down vote

To show that $(a,b in mathbbZ, + , times)$ is not a field, it suffices to find a particular non-zero element from the set that does not have an inverse.

$2 in a,b in mathbbZ$, hence, if it doesn't have a multiplicative inverse, then $(a,b in mathbbZ, + , times)$ is not a field.

$pi notin mathbbQ$, it can't be used to show that $mathbbQ$ is not a field. (and actually $(mathbbQ, + , times)$ is a field.)

share|cite|improve this answer

answered Jul 22 at 14:01

Siong Thye Goh

77.6k134795

add a comment | 

up vote
3
down vote

up vote
3
down vote

To show that $(a,b in mathbbZ, + , times)$ is not a field, it suffices to find a particular non-zero element from the set that does not have an inverse.

$2 in a,b in mathbbZ$, hence, if it doesn't have a multiplicative inverse, then $(a,b in mathbbZ, + , times)$ is not a field.

$pi notin mathbbQ$, it can't be used to show that $mathbbQ$ is not a field. (and actually $(mathbbQ, + , times)$ is a field.)

share|cite|improve this answer

answered Jul 22 at 14:01

Siong Thye Goh

77.6k134795

To show that $(a,b in mathbbZ, + , times)$ is not a field, it suffices to find a particular non-zero element from the set that does not have an inverse.

$2 in a,b in mathbbZ$, hence, if it doesn't have a multiplicative inverse, then $(a,b in mathbbZ, + , times)$ is not a field.

$pi notin mathbbQ$, it can't be used to show that $mathbbQ$ is not a field. (and actually $(mathbbQ, + , times)$ is a field.)

share|cite|improve this answer

answered Jul 22 at 14:01

Siong Thye Goh

77.6k134795

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 14:01

Siong Thye Goh

77.6k134795

answered Jul 22 at 14:01

Siong Thye Goh

77.6k134795

answered Jul 22 at 14:01

Siong Thye Goh

77.6k134795

77.6k134795

add a comment | 

add a comment | 

up vote
0
down vote

Since $(2+sqrt2)(2-sqrt2)=2$, $a,,binmathbbZ$ contains $2+sqrt2$ but no inverse for it.

share|cite|improve this answer

answered Jul 22 at 14:01

J.G.

13.2k11424

add a comment | 

up vote
0
down vote

Since $(2+sqrt2)(2-sqrt2)=2$, $a,,binmathbbZ$ contains $2+sqrt2$ but no inverse for it.

share|cite|improve this answer

answered Jul 22 at 14:01

J.G.

13.2k11424

add a comment | 

up vote
0
down vote

up vote
0
down vote

Since $(2+sqrt2)(2-sqrt2)=2$, $a,,binmathbbZ$ contains $2+sqrt2$ but no inverse for it.

share|cite|improve this answer

answered Jul 22 at 14:01

J.G.

13.2k11424

Since $(2+sqrt2)(2-sqrt2)=2$, $a,,binmathbbZ$ contains $2+sqrt2$ but no inverse for it.

share|cite|improve this answer

answered Jul 22 at 14:01

J.G.

13.2k11424

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 14:01

J.G.

13.2k11424

answered Jul 22 at 14:01

J.G.

13.2k11424

answered Jul 22 at 14:01

J.G.

13.2k11424

13.2k11424

add a comment | 

add a comment | 

up vote
0
down vote

Does the number $2$ have inverse? No.

Suppose it has an inverse $a+bsqrt 2$, with $a,bin mathbb Z$. Then we have $2(a+bsqrt 2)=1$, which means $2a=1$ and $2bsqrt 2 =0$.

Now, there is no integer $a$ such that $2a=1$, which is a contradiction.

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answered Jul 22 at 14:01

Hugocito

1,6451019

add a comment | 

up vote
0
down vote

Does the number $2$ have inverse? No.

Suppose it has an inverse $a+bsqrt 2$, with $a,bin mathbb Z$. Then we have $2(a+bsqrt 2)=1$, which means $2a=1$ and $2bsqrt 2 =0$.

Now, there is no integer $a$ such that $2a=1$, which is a contradiction.

share|cite|improve this answer

answered Jul 22 at 14:01

Hugocito

1,6451019

add a comment | 

up vote
0
down vote

up vote
0
down vote

Does the number $2$ have inverse? No.

Suppose it has an inverse $a+bsqrt 2$, with $a,bin mathbb Z$. Then we have $2(a+bsqrt 2)=1$, which means $2a=1$ and $2bsqrt 2 =0$.

Now, there is no integer $a$ such that $2a=1$, which is a contradiction.

share|cite|improve this answer

answered Jul 22 at 14:01

Hugocito

1,6451019

Does the number $2$ have inverse? No.

Suppose it has an inverse $a+bsqrt 2$, with $a,bin mathbb Z$. Then we have $2(a+bsqrt 2)=1$, which means $2a=1$ and $2bsqrt 2 =0$.

Now, there is no integer $a$ such that $2a=1$, which is a contradiction.

share|cite|improve this answer

answered Jul 22 at 14:01

Hugocito

1,6451019

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 14:01

Hugocito

1,6451019

answered Jul 22 at 14:01

Hugocito

1,6451019

answered Jul 22 at 14:01

Hugocito

1,6451019

1,6451019

add a comment | 

add a comment | 

up vote
0
down vote

The problem is in finding the inverse for nonzero elements.

The inverse of $$a+bsqrt 2$$ is $$frac 1a+bsqrt 2= frac a-bsqrt 2a^2-2b^2$$

What if $$a^2-2b^2 =0$$ in which case the inverse is undefined ?

share|cite|improve this answer

answered Jul 22 at 14:48

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

The problem is in finding the inverse for nonzero elements.

The inverse of $$a+bsqrt 2$$ is $$frac 1a+bsqrt 2= frac a-bsqrt 2a^2-2b^2$$

What if $$a^2-2b^2 =0$$ in which case the inverse is undefined ?

share|cite|improve this answer

answered Jul 22 at 14:48

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

up vote
0
down vote

The problem is in finding the inverse for nonzero elements.

The inverse of $$a+bsqrt 2$$ is $$frac 1a+bsqrt 2= frac a-bsqrt 2a^2-2b^2$$

What if $$a^2-2b^2 =0$$ in which case the inverse is undefined ?

share|cite|improve this answer

answered Jul 22 at 14:48

Mohammad Riazi-Kermani

27.5k41852

The problem is in finding the inverse for nonzero elements.

The inverse of $$a+bsqrt 2$$ is $$frac 1a+bsqrt 2= frac a-bsqrt 2a^2-2b^2$$

What if $$a^2-2b^2 =0$$ in which case the inverse is undefined ?

share|cite|improve this answer

answered Jul 22 at 14:48

Mohammad Riazi-Kermani

27.5k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 14:48

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 14:48

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 14:48

Mohammad Riazi-Kermani

27.5k41852

27.5k41852

add a comment | 

add a comment |