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Math Help - Do sets of all pairs of real numbers form fields?

  1. #16
    Member Mollier's Avatar
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    Quote Originally Posted by Swlabr View Post
    Yes, but this implies that if r=0 there is an inverse.

    As to an element having more than one inverse, this is impossible for the definition of inverse here. There is a short proof of this, which I will challenge whoever believes it to not hold to do.
    So then the answers to (c) is the same as for (a) and (b)?

    Quote Originally Posted by Swlabr View Post
    Identities are unique. So, simply show that (\alpha, \beta)(1, 0) = (\alpha, \beta) to get that this is your identity.

    In general you would also want to show that (1, 0)(\alpha, \beta) = (\alpha, \beta), however your multiplication here is commutative so this doesn't need to be shown.
    That's the thing, I can show it, but I would not have noticed it myself..

    Again, Thanks!
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  2. #17
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Mollier View Post
    So then the answers to (c) is the same as for (a) and (b)?
    ...but the answers to (a) and (b) are different...

    Quote Originally Posted by Mollier View Post
    That's the thing, I can show it, but I would not have noticed it myself..

    Again, Thanks!
    These types of questions (you are given a strange multiplication in this way) are relatively easy to do once you know how, and are a gift in an exam!

    Firstly, look to see if the multiplication is commutative. This is usually quite easy to spot if it is.

    Secondly, you need to find an identity element. So, let (\alpha, \beta) be an arbitrary element. We want to find an a and a b such that (\alpha, \beta)(a, b) = (\alpha, \beta). So, just use the multiplication given to work out what (a, b) is.

    This is your identity element. You know it is an identity on the right, but unless the multiplication is commutative you do not know if it is an identity on the left. So, knowing what (a, b) is check to see if (a, b)(\alpha, \beta) = (\alpha, \beta). Usually (always if you have been asked to find an identity or an inverse not a right identity or a right inverse (alternatively, left identity etc.)) this will be so. But do it all the same as it is part of the proof. Once you have shown this you have found your identity element.

    Now you need to find your inverse. You want to find an element (\alpha^{\prime}, \beta^{\prime}) such that (\alpha, \beta)(\alpha^{\prime}, \beta^{\prime}) = (a, b). Again, just use the fact that you know what the multiplication is to solve for (\alpha, \beta).

    You will usually find a generic inverse, but this is not always an inverse for all elements, as with question (a) here. Usually if this is the case you will be dividing by zero.
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  3. #18
    Member Mollier's Avatar
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    Quote Originally Posted by Swlabr View Post
    ...but the answers to (a) and (b) are different...
    Aha,
    \alpha^2+\beta^2=(a_1+b_1i)^2+(a_2+b_2i)^2
    So for say a_1=b_1=1,a_2=-1\;and\;b_2=1 I would be dividing by zero.
    In other words, for some elements there is no multiplicative inverse.

    Quote Originally Posted by Swlabr View Post
    These types of questions (you are given a strange multiplication in this way) are relatively easy to do once you know how, and are a gift in an exam!
    Luckily, I am doing this just for fun so no exams are involved
    I do appreciate the detailed explanation though, thanks!
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  4. #19
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    Quote Originally Posted by Swlabr View Post
    Shanks added confusion:

    His claim "for(a), (r,0) has no inverse if r is not 0!" should read "for(a), (r,0) has no inverse. Ever."
    But his point was that it is "(r, 0) if r is not 0" not having an inverse that says this is not a field!
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  5. #20
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    But his point was that it is "(r, 0) if r is not 0" not having an inverse that says this is not a field!
    Yeah, but that is what was causing the confusion. Although a correct statement, it made Mollier think that this was the only time there was not an inverse.
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  6. #21
    Member Mollier's Avatar
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    It's all good now
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