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Math Help - Is it a field? set of real numbers in the form...

  1. #1
    Member Mollier's Avatar
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    Is it a field? set of real numbers in the form...

    Hi,

    problem:
    Let Q(\sqrt{2}) be the set of all real numbers of the form
    \alpha+\beta\sqrt{2}, where \alpha\;and\;\beta are rational.

    (a) Is Q(\sqrt{2}) a field?
    (b) What if \alpha\;and\;\beta are required to be integers?

    attempt:


    (a)
    I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
    Is it ok to say that I can express any real number in the form \alpha+\beta\sqrt{2} as long as \alpha\;and\;\beta are rational?

    (b)
    I am unable to find an axiom that "fails", but say I want to write the real number \frac{\sqrt{2}}{2}. I don't think I can do this if \alpha\;and\;\beta are integers..

    Any comments are very welcome!
    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mollier View Post
    Hi,

    problem:
    Let Q(\sqrt{2}) be the set of all real numbers of the form
    \alpha+\beta\sqrt{2}, where \alpha\;and\;\beta are rational.

    (a) Is Q(\sqrt{2}) a field?
    (b) What if \alpha\;and\;\beta are required to be integers?

    attempt:


    (a)
    I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
    Is it ok to say that I can express any real number in the form \alpha+\beta\sqrt{2} as long as \alpha\;and\;\beta are rational?

    (b)
    I am unable to find an axiom that "fails", but say I want to write the real number \frac{\sqrt{2}}{2}. I don't think I can do this if \alpha\;and\;\beta are integers..

    Any comments are very welcome!
    Thanks
    just go through each property for a field and make sure that they all hold. you have to verify these explicitly.

    and no, you cannot say any real number can be expressed in that form. how would you express \pi that way, for instance?
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Mollier View Post
    Hi,

    problem:
    Let Q(\sqrt{2}) be the set of all real numbers of the form
    \alpha+\beta\sqrt{2}, where \alpha\;and\;\beta are rational.

    (a) Is Q(\sqrt{2}) a field?
    (b) What if \alpha\;and\;\beta are required to be integers?

    attempt:


    (a)
    I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
    Is it ok to say that I can express any real number in the form \alpha+\beta\sqrt{2} as long as \alpha\;and\;\beta are rational?

    (b)
    I am unable to find an axiom that "fails", but say I want to write the real number \frac{\sqrt{2}}{2}. I don't think I can do this if \alpha\;and\;\beta are integers..

    Any comments are very welcome!
    Thanks
    Firstly, if every real number was of the form \alpha+\beta \sqrt{2} with \alpha, \beta \in \mathbb{Q} then we can easily define a bijection \mathbb{R} \rightarrow \mathbb{Q} \times \mathbb{Q}, \alpha + \beta \sqrt{2} \mapsto (\alpha, \beta). As there exists a bijection between \mathbb{Q} and \mathbb{Q} \times \mathbb{Q} we have that |\mathbb{R}| = |\mathbb{Q}|, a contradiction.

    Secondly, in both of your problems you know that "most" of the axioms for fields holds (both are clearly rings, and multiplication in both cases is commutative). The one to concentrate on will be multiplicative inverses.

    What is the inverse of \alpha+\beta \sqrt{2} with \alpha and \beta rationals?

    What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take \beta = 0. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form \alpha + \beta \sqrt{2}. That is, every rational number of the form \frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\} can be written as \alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}. As we are, allegedly, in a field we have that \frac{n}{\gamma} is also of this form, with n \in \mathbb{N}. Thus, every rational number is of the form \alpha + \beta \sqrt{2}. This is silly!)
    Last edited by Swlabr; January 5th 2010 at 12:52 AM.
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  4. #4
    Member Mollier's Avatar
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    Quote Originally Posted by Jhevon View Post
    and no, you cannot say any real number can be expressed in that form. how would you express \pi that way, for instance?
    Thanks

    Quote Originally Posted by Swlabr View Post
    The one to concentrate on will be multiplicative inverses.

    What is the inverse of \alpha+\beta \sqrt{2} with \alpha and \beta rationals?

    What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take \beta = 0. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form \alpha + \beta \sqrt{2}. That is, every rational number of the form \frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\} can be written as \alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}. As we are, allegedly, in a field we have that \frac{n}{\gamma} is also of this form, with n \in \mathbb{N}. Thus, every rational number is of the form \alpha + \beta \sqrt{2}. This is silly!)
    It would be \frac{1}{\alpha+\beta\sqrt{2}} .
    Now, the inverse would have to be in Q(\sqrt{2}), so it would have to be in the form \alpha+\beta\sqrt{2}.
    Since \frac{1}{\alpha+\beta\sqrt{2}} is not in that form, Q(\sqrt{2}) is not a field.
    Is this totally off? I think I need to get (a) right, before I move on to (b).

    Thanks guys!
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Mollier View Post
    Thanks



    It would be \frac{1}{\alpha+\beta\sqrt{2}} .
    Now, the inverse would have to be in Q(\sqrt{2}), so it would have to be in the form \alpha+\beta\sqrt{2}.
    Since \frac{1}{\alpha+\beta\sqrt{2}} is not in that form, Q(\sqrt{2}) is not a field.
    Is this totally off? I think I need to get (a) right, before I move on to (b).

    Thanks guys!
    You want to try and write \frac{1}{\alpha + \beta \sqrt{2}} as \alpha^{\prime} + \beta^{\prime}\sqrt{2}. I mean, \sqrt{2} = \frac{2}{\sqrt{2}}...numbers are strange things...just because something isn't obviously in a certain form does not mean you cannot write it in that form.

    I shall give you a tantalising hint: It is a field.
    Last edited by Swlabr; January 5th 2010 at 01:49 AM.
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  6. #6
    Member Mollier's Avatar
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    <br />
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}<br />

    \alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)

    \beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)

    Since both \alpha'\;and\;\beta' are rationals, Q(\sqrt{2}) is a field?

    Thanks mate!
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Mollier View Post
    <br />
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}<br />

    \alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)

    \beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)

    Since both \alpha'\;and\;\beta' are rationals, Q(\sqrt{2}) is a field?

    Thanks mate!
    Yeah, that's it! Although to be thorough you would need to make sure there is no division by 0 going on.
    Last edited by Swlabr; January 5th 2010 at 04:59 AM.
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  8. #8
    Member Mollier's Avatar
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    So, \alpha^2-2\beta^2\neq 0,
    then  \beta \neq \frac{1}{\sqrt{2}}\alpha , but \beta is then not a rational number anyways..

    As for (b), can I use a similar argument by saying that since \alpha' and \beta' are rational numbers, Q(\sqrt{2}) is not a field when \alpha,\beta\in Z?

    Thank you very much.
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  9. #9
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Mollier View Post
    So, \alpha^2-2\beta^2\neq 0,
    then  \beta \neq \frac{1}{\sqrt{2}}\alpha , but \beta is then not a rational number anyways..

    As for (b), can I use a similar argument by saying that since \alpha' and \beta' are rational numbers, Q(\sqrt{2}) is not a field when \alpha,\beta\in Z?

    Thank you very much.
    For part (b) you need to find a counter-example - \alpha^{\prime} and \beta^{\prime} may look like non-integers, but this may not always be the case.

    So, to complete the problem simply take an integer value of \alpha and an integer value of \beta such that \alpha^{\prime} \notin \mathbb{Z} or \beta^{\prime} \notin \mathbb{Z}. For simplicity, let one of them be zero.
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Mollier View Post
    So, \alpha^2-2\beta^2\neq 0,
    then  \beta \neq \frac{1}{\sqrt{2}}\alpha , but \beta is then not a rational number anyways..
    No, this is not a proof. You want to assume that \alpha^2 -2\beta^2=0 and look for a contradiction.
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  11. #11
    Senior Member Shanks's Avatar
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    \alpha^2-2\beta^2\neq 0 because of the irrationality of \sqrt{2}, it can be easily proved in number theory.
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  12. #12
    Member Mollier's Avatar
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    Quote Originally Posted by Swlabr View Post
    No, this is not a proof. You want to assume that \alpha^2 -2\beta^2=0 and look for a contradiction.
    \alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta
    Contradiction since \alpha\in Q
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  13. #13
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Mollier View Post
    \alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta
    Contradiction since \alpha\in Q
    Precisely!

    Sorry to be pedantic about it, but in maths the proof is the king...
    Last edited by Jhevon; January 5th 2010 at 09:40 AM.
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  14. #14
    Member Mollier's Avatar
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    Please don't apologize. I have a BS in mechanical engineering from a horrible school, so I have now decided to actually learn some math. You being "pedantic" is exactly what I need! Thanks!
    Last edited by Jhevon; January 5th 2010 at 09:34 AM.
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