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Can you help me to finish up this question?

by omega16
Tags: finish
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omega16
#1
Oct19-06, 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
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StatusX
#2
Oct19-06, 02:46 PM
HW Helper
P: 2,566
Well you're almost done. You need to show no such a and b will work.
Office_Shredder
#3
Oct19-06, 03:54 PM
Emeritus
Sci Advisor
PF Gold
P: 4,500
How about -1?

HallsofIvy
#4
Oct20-06, 06:45 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,352
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?
omega16
#5
Oct21-06, 03:21 AM
P: 20
Thank you very much for your opinion. I have solved this question.


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