
#1
Oct1906, 12:50 PM

P: 20

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 ======================= 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 



#2
Oct1906, 02:46 PM

HW Helper
P: 2,566

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




#3
Oct1906, 03:54 PM

Mentor
P: 4,499

How about 1?




#4
Oct2006, 06:45 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890

can you help me to finish up this question?
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? 



#5
Oct2106, 03:21 AM

P: 20

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



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