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Show that $sqrtabc$ is irrational if $a, b, c$, and $sqrta + sqrtb + sqrtc$ are irrational. [on hold]

(How to/Can I) show irrational numbers?Prove that $sqrt 2 +sqrt 3$ is irrational.How to show that the product of two irrational numbers may be irrational?Is :$sqrtipi+sqrtipi+sqrtipi+sqrtcdots$ irrational or transcendental or real number?How to check this number $sqrt47$ is irrationalWhat are irrational real numbers?proving $ sqrt 2 + sqrt 3 $ is irrationalProof That all Positive Irrational Sqaure Roots Can be Raised to an Irrational Power to Get a Whole NumberDivision of Square Root of Primes are IrrationalIs this proof of $sqrt3$ being an irrational number correct?

0

1

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Assume $a$,$b$,$c$ are the irrational numbers, and $sqrt a + sqrt b + sqrt c$ is irrational number.
Show that $sqrtabc$ is irrational number.
Please help me this problem, thank you for watching!

irrational-numbers

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edited Mar 12 at 19:00

6005

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put on hold as off-topic by Saad, Cesareo, Gibbs, Vinyl_cape_jawa, Song Mar 13 at 12:08

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." – Saad, Cesareo, Gibbs, Vinyl_cape_jawa, Song

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• 
  
  
  

  
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  Counter Example: $sqrt2+sqrt3+sqrt6$.
  $endgroup$
  – Donald Splutterwit
  Mar 12 at 18:35
  

  
  
  
  


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  @DonaldSplutterwit No it's not, because $2,3,6$ are rational.
  $endgroup$
  – cansomeonehelpmeout
  Mar 12 at 18:45
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Assume $a$,$b$,$c$ are the irrational numbers, and $sqrt a + sqrt b + sqrt c$ is irrational number.
Show that $sqrtabc$ is irrational number.
Please help me this problem, thank you for watching!

irrational-numbers

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edited Mar 12 at 19:00

6005

37k751127

asked Mar 12 at 18:30

Chris LiChris Li

123

$endgroup$

put on hold as off-topic by Saad, Cesareo, Gibbs, Vinyl_cape_jawa, Song Mar 13 at 12:08

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." – Saad, Cesareo, Gibbs, Vinyl_cape_jawa, Song

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Counter Example: $sqrt2+sqrt3+sqrt6$.
  $endgroup$
  – Donald Splutterwit
  Mar 12 at 18:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DonaldSplutterwit No it's not, because $2,3,6$ are rational.
  $endgroup$
  – cansomeonehelpmeout
  Mar 12 at 18:45
  

  
  
  
  

add a comment | 

$begingroup$

Assume $a$,$b$,$c$ are the irrational numbers, and $sqrt a + sqrt b + sqrt c$ is irrational number.
Show that $sqrtabc$ is irrational number.
Please help me this problem, thank you for watching!

irrational-numbers

share|cite|improve this question

edited Mar 12 at 19:00

6005

37k751127

asked Mar 12 at 18:30

Chris LiChris Li

123

$endgroup$

share|cite|improve this question

edited Mar 12 at 19:00

6005

37k751127

asked Mar 12 at 18:30

Chris LiChris Li

123

share|cite|improve this question

edited Mar 12 at 19:00

6005

37k751127

asked Mar 12 at 18:30

Chris LiChris Li

123

edited Mar 12 at 19:00

6005

37k751127

edited Mar 12 at 19:00

6005

37k751127

asked Mar 12 at 18:30

Chris LiChris Li

123

asked Mar 12 at 18:30

Chris LiChris Li

123

1 Answer
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5

$begingroup$

The statement is not true. However, the counterexamples in the comments don't work. Note the requirement that $a, b,$ and $c$ (not just $sqrta$, $sqrtb$, and $sqrtc$) should be irrational.

For a correct counterexample, take
$$
a = b = c = sqrt[3]4.
$$
These are irrational (this can be proven similarly to the irrationality of the square root of $2$).
Then,
$$
sqrta + sqrtb + sqrtc = sqrt[3]2 + sqrt[3]2 + sqrt[3]2 = 3sqrt[3]2,
$$
which is also irrational -- because $sqrt[3]2$ is irrational (which can again be proven similarly to irrationality of the square root of $2$).

However,
$$
sqrtabc = sqrt4 = 2,
$$
which is rational.

share|cite|improve this answer

edited Mar 13 at 17:16

answered Mar 12 at 18:59

60056005

37k751127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hans Engler's counterexample looks good to me. $sqrt 2$ and $2^1/4$ are irrational, aren't they?
  $endgroup$
  – TonyK
  Mar 12 at 19:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
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  If $a = 2^1/2$ and $b = c = 2^1/4$, then $sqrtabc = sqrt2$, which is irrational. So that isn't a counterexample.
  $endgroup$
  – JimmyK4542
  Mar 12 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
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  @TonyK I think Hans meant rather that $a = 2$, $b = c = sqrt2$. Either way, that counterexample doesn't work.
  $endgroup$
  – 6005
  Mar 12 at 20:25
  

  
  
  
  


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  $begingroup$
  My counterexample referred to an earlier version of the post where the claim was that $sqrtabc$ is irrational. That has been fixed by the OP and I have deleted my comment.
  $endgroup$
  – Hans Engler
  Mar 13 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HansEngler Thanks for the clarification, I have removed the negative reference in my answer :)
  $endgroup$
  – 6005
  Mar 13 at 17:16
  
  

  
  
  
  

add a comment | 

5

$begingroup$

The statement is not true. However, the counterexamples in the comments don't work. Note the requirement that $a, b,$ and $c$ (not just $sqrta$, $sqrtb$, and $sqrtc$) should be irrational.

For a correct counterexample, take
$$
a = b = c = sqrt[3]4.
$$
These are irrational (this can be proven similarly to the irrationality of the square root of $2$).
Then,
$$
sqrta + sqrtb + sqrtc = sqrt[3]2 + sqrt[3]2 + sqrt[3]2 = 3sqrt[3]2,
$$
which is also irrational -- because $sqrt[3]2$ is irrational (which can again be proven similarly to irrationality of the square root of $2$).

However,
$$
sqrtabc = sqrt4 = 2,
$$
which is rational.

share|cite|improve this answer

edited Mar 13 at 17:16

answered Mar 12 at 18:59

60056005

37k751127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hans Engler's counterexample looks good to me. $sqrt 2$ and $2^1/4$ are irrational, aren't they?
  $endgroup$
  – TonyK
  Mar 12 at 19:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $a = 2^1/2$ and $b = c = 2^1/4$, then $sqrtabc = sqrt2$, which is irrational. So that isn't a counterexample.
  $endgroup$
  – JimmyK4542
  Mar 12 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK I think Hans meant rather that $a = 2$, $b = c = sqrt2$. Either way, that counterexample doesn't work.
  $endgroup$
  – 6005
  Mar 12 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My counterexample referred to an earlier version of the post where the claim was that $sqrtabc$ is irrational. That has been fixed by the OP and I have deleted my comment.
  $endgroup$
  – Hans Engler
  Mar 13 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HansEngler Thanks for the clarification, I have removed the negative reference in my answer :)
  $endgroup$
  – 6005
  Mar 13 at 17:16
  
  

  
  
  
  

add a comment | 

5

$begingroup$

The statement is not true. However, the counterexamples in the comments don't work. Note the requirement that $a, b,$ and $c$ (not just $sqrta$, $sqrtb$, and $sqrtc$) should be irrational.

For a correct counterexample, take
$$
a = b = c = sqrt[3]4.
$$
These are irrational (this can be proven similarly to the irrationality of the square root of $2$).
Then,
$$
sqrta + sqrtb + sqrtc = sqrt[3]2 + sqrt[3]2 + sqrt[3]2 = 3sqrt[3]2,
$$
which is also irrational -- because $sqrt[3]2$ is irrational (which can again be proven similarly to irrationality of the square root of $2$).

However,
$$
sqrtabc = sqrt4 = 2,
$$
which is rational.

share|cite|improve this answer

edited Mar 13 at 17:16

answered Mar 12 at 18:59

60056005

37k751127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hans Engler's counterexample looks good to me. $sqrt 2$ and $2^1/4$ are irrational, aren't they?
  $endgroup$
  – TonyK
  Mar 12 at 19:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $a = 2^1/2$ and $b = c = 2^1/4$, then $sqrtabc = sqrt2$, which is irrational. So that isn't a counterexample.
  $endgroup$
  – JimmyK4542
  Mar 12 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK I think Hans meant rather that $a = 2$, $b = c = sqrt2$. Either way, that counterexample doesn't work.
  $endgroup$
  – 6005
  Mar 12 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My counterexample referred to an earlier version of the post where the claim was that $sqrtabc$ is irrational. That has been fixed by the OP and I have deleted my comment.
  $endgroup$
  – Hans Engler
  Mar 13 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HansEngler Thanks for the clarification, I have removed the negative reference in my answer :)
  $endgroup$
  – 6005
  Mar 13 at 17:16
  
  

  
  
  
  

add a comment | 

$begingroup$

The statement is not true. However, the counterexamples in the comments don't work. Note the requirement that $a, b,$ and $c$ (not just $sqrta$, $sqrtb$, and $sqrtc$) should be irrational.

For a correct counterexample, take
$$
a = b = c = sqrt[3]4.
$$
These are irrational (this can be proven similarly to the irrationality of the square root of $2$).
Then,
$$
sqrta + sqrtb + sqrtc = sqrt[3]2 + sqrt[3]2 + sqrt[3]2 = 3sqrt[3]2,
$$
which is also irrational -- because $sqrt[3]2$ is irrational (which can again be proven similarly to irrationality of the square root of $2$).

However,
$$
sqrtabc = sqrt4 = 2,
$$
which is rational.

share|cite|improve this answer

edited Mar 13 at 17:16

answered Mar 12 at 18:59

60056005

37k751127

$endgroup$

share|cite|improve this answer

edited Mar 13 at 17:16

answered Mar 12 at 18:59

60056005

37k751127

answered Mar 12 at 18:59

60056005

37k751127

answered Mar 12 at 18:59

60056005

37k751127