6
$\begingroup$

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

  • 3
    Well, I generally pronounce it as $\textbf{Q root 2}$2011-09-10
  • 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
  • 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
  • 0
    My algebra professor always said "Q adjoin radical 2"2011-09-10
  • 2
    And you could say: "Q square-bracket radical two"2011-09-10

1 Answers 1

2

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

  • 1
    That wasn't the pronunciation mentioned in the question, which was "Q adjoin two".2011-09-10
  • 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 posting2011-09-10
  • 0
    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: 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
  • 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 apologize2011-09-10
  • 0
    No problem, we're cool :-) As regards $\sqrt2$ or $2$, the current version of the question makes more sense to me.2011-09-10
  • 0
    @joriki:ok sir,its my fault to be hasty without thinking the background2011-09-10
  • 0
    Now, 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
  • 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
  • 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
  • 0
    Remind me - what does $\mathbb{Q}(a)$ mean?2011-10-10
  • 0
    @GEdgar Can you give an example for such an $a$?2011-11-08
  • 2
    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.2011-11-08
  • 0
    GEdgar: 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
  • 1
    @GEdgar In your previous comment you wrote that $a$ is algebraic which was the reason for my question.2011-11-26