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Is there a standard term for $\mathbb{Q}[\sqrt{2}]$? I say it as "Q adjoin root two".

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Well, I generally pronounce it as $\textbf{Q root 2}$ – user9413 Sep 10 '11 at 8:04
@MathsStudent: You've now twice edited the question such that a comment of mine no longer made sense, both times without marking the edit as such or commenting on it or notifying me. That's not good style. – joriki Sep 10 '11 at 9:02
@MathsStudent: In case you think I'm being overly pedantic, take a look at the comments under iyengar's answer for an example of what happens when you do this. – joriki Sep 10 '11 at 11:09
My algebra professor always said "Q adjoin radical 2" – Tim Seguine Sep 10 '11 at 13:06
And you could say: "Q square-bracket radical two" – GEdgar Sep 10 '11 at 13:27

We can say it as generated by $\sqrt 2$ ,but the pronunciation that you mentioned is the accurate and correct pronunciation ,i.e "$\mathbb{Q}$ adjoin $\sqrt 2$" is perfect and apt pronunciation adapted by major mathematicians

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That wasn't the pronunciation mentioned in the question, which was "Q adjoin two". – joriki Sep 10 '11 at 8:40
@joriki:I sincerely ask to read the question twice,he mentioned that $\mathbb{Q[\sqrt 2]}$,which is obtained by adjoining a $\sqrt 2$,a zero of $x^2-2$ to field of rationals $\mathbb{Q}$ ,and then we obtain $\mathbb{Q[\sqrt 2]$,then it is read as Q adjoin $\sqrt 2$,so please verify before posting – Iyengar Sep 10 '11 at 10:52
There's no need to imply that I don't verify things before posting. As I commented under the question, the OP twice changed the question without marking the edit or commenting on it. At the time I commented on your answer, the question said that the OP pronounces $\mathbb Q[\sqrt2]$ as "Q adjoin two", not as "Q adjoin root two" as it says now. You can see this in the revision logs: More generally, you can look at past version of questions and answers by clicking on the "edited ... ago" link underneath. – joriki Sep 10 '11 at 11:09
Example: $\mathbb Q[\pi]$ is all polynomials in $\pi$, while $\mathbb Q(\pi)$ is all rational functions in $\pi$. Unlike the case of $\sqrt{2}$, these are different. – GEdgar Nov 8 '11 at 20:16
@GEdgar In your previous comment you wrote that $a$ is algebraic which was the reason for my question. – Phira Nov 26 '11 at 17:55

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