Can you help me to finish up this question?

[omega16]

Consider Q[sqrt(2)].
Does every element of Q[sqrt(2)] have a square root in Q[sqrt(2)] ?
Prove if true, and give a counterexample if false.My solution:
sqrt(sqrt(2)) = a + bsqrt(2)

if I square both sides then I will have :

sqrt(2) = (a + b*sqrt(2))^2
= a^2 + 2ab*sqrt(2) + 2b^2

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I think the answer should be false. Am I right?

If I am right. Can you suggest me a counterexample. Thank you very much.

If I am wrong. Please correct me. Thanks

 

[StatusX]

Well you're almost done. You need to show no such a and b will work.

 

[Office_Shredder]

How about -1?

 

[HallsofIvy]

sqrt(2) = a^2 + 2ab*sqrt(2) + 2b^2
= (a^2+ b^2)+ 2ab sqrt(2),
Since every number in Q(sqrt(2)) can be written uniquely as "x+ ysqrt(2)" for rational x, y, what does that tell you about a and b?

 

[omega16]

Thank you very much for your opinion. I have solved this question.