# can you help me to finish up this question?

by omega16
Tags: finish
 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
 HW Helper P: 2,566 Well you're almost done. You need to show no such a and b will work.
 Mentor P: 4,202 How about -1?
PF Patron
Thanks
Emeritus
P: 38,424

## 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?
 P: 20 Thank you very much for your opinion. I have solved this question.

 Related Discussions Calculus & Beyond Homework 2 Academic Guidance 22 Introductory Physics Homework 2 Introductory Physics Homework 8 Calculus 7