Is there a standard term for $\mathbb{Q}[\sqrt{2}]$? I say it as "Q adjoin root two".
How do you pronounce $\mathbb{Q}[\sqrt{2}]$?
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abstract-algebra
notation
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3Well, I generally pronounce it as $\textbf{Q root 2}$ – 2011-09-10
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5@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. – 2011-09-10
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1@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. – 2011-09-10
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0My algebra professor always said "Q adjoin radical 2" – 2011-09-10
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2And you could say: "Q square-bracket radical two" – 2011-09-10
1 Answers
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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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1That wasn't the pronunciation mentioned in the question, which was "Q adjoin two". – 2011-09-10
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0@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 – 2011-09-10
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0There'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: http://math.stackexchange.com/revisions/63273/3. More generally, you can look at past version of questions and answers by clicking on the "edited ... ago" link underneath. – 2011-09-10
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0@Joriki:ok sir,be cool,i didnt knew that,i saw just the heading ,and i posted my answer,anyway i apologize if i behaved harshly,i thought you were mistaken,but its the OP mistake ,so what is it now,a $\sqrt 2$ or just $2$ ???,i once again apologize – 2011-09-10
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0No problem, we're cool :-) As regards $\sqrt2$ or $2$, the current version of the question makes more sense to me. – 2011-09-10
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0@joriki:ok sir,its my fault to be hasty without thinking the background – 2011-09-10
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0Now, instead of $\sqrt{2}$ suppose you have some algebraic real number $a$ so that $\mathbb Q[a] \ne \mathbb Q(a)$. Will you still say "adjoin" for one of those? – 2011-09-10
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0@GEdgar:then it becomes an Algebraic number field,if you have some algebraic number 'a',but why did you ask it,to test my level of understanding??? – 2011-09-10
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0@iyengar: I am asking about your terminology. If they are unequal, which of $\mathbb Q[a]$ and $\mathbb Q(a)$ do you call "$Q$ adjoin $a$"? – 2011-09-10
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0Remind me - what does $\mathbb{Q}(a)$ mean? – 2011-10-10
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0@GEdgar Can you give an example for such an $a$? – 2011-11-08
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2Example: $\mathbb Q[\pi]$ is all polynomials in $\pi$, while $\mathbb Q(\pi)$ is all rational functions in $\pi$. Unlike the case of $\sqrt{2}$, these are different. – 2011-11-08
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0GEdgar: In Russian the notations $f(x)$, $F[x]$, $F(x)$, $F[[x]]$, and $F((x))$ are called the same thing, essentially "eff of eks". – 2011-11-19
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1@GEdgar In your previous comment you wrote that $a$ is algebraic which was the reason for my question. – 2011-11-26