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

  1. #1
    Member Mollier's Avatar
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    Do sets of all pairs of real numbers form fields?

    Hi,

    problem:
    Let F be the set of all (ordered) pairs (\alpha,\beta) of real numbers.

    (a) If addition and multiplication are defined by
    (\alpha,\beta)+(\gamma,\delta)=(\alpha+\gamma,\bet  a+\delta)
    and
    (\alpha,\beta)(\gamma,\delta)=(\alpha\gamma,\beta\  delta),
    does F become a field?

    (b) If addition and multiplication are defined by
    (\alpha,\beta)+(\gamma,\delta)=(\alpha+\gamma,\bet  a+\delta)
    and
    (\alpha,\beta)(\gamma,\delta)=(\alpha\gamma-\beta\delta,\alpha\delta+\beta\gamma),
    is F a field then?

    (c) What happens (in both the preceding cases) if we consider ordered pairs of complex numbers instead?

    attempt (at (a)):
    I have gone through all the axioms for a field and am unable to find one that fails.
    For example, the multiplicative inverse:
    (\alpha,\beta)\frac{1}{\alpha,\beta} \Rightarrow (\alpha,\beta)\left(\frac{1}{\alpha},\frac{1}{\bet  a}\right)=(1,1)
    One of my friends told me that this is not a field, but I fail to see why.

    Thanks!
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  2. #2
    Senior Member Shanks's Avatar
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    for(a), (r,0) has no inverse if r is not 0!
    for(b), It is a field, because the mapping: (a,b)--> a+ib, (i is the imaginary unit) is a isomorphism. Or you can varify that all axioms for field are satisfied.
    for(c), (a) can't be a field for the same reason.
    (b) can't be a field because the inverse of some element is not unique.
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by Shanks View Post
    for(a), (r,0) has no inverse if r is not 0!
    So there is an inverse for (0,0)?

    Quote Originally Posted by Shanks View Post
    for(b), It is a field, because the mapping: (a,b)--> a+ib, (i is the imaginary unit) is a isomorphism. Or you can varify that all axioms for field are satisfied.
    I am not yet familiar with isomorphism, it's covered in a later chapter.
    As for verifying that the axioms are satisfied, I'm having some problems with it.
    Say I want to verify that there exists an multiplicative inverse:

    (\alpha,\beta)\cdot(\alpha',\beta')=(1,1)
    I see that it would be true for say,

    (1,1)\cdot(1,0)=(1-0,0+1)=(1,1)

    I am unfortunately not able to find a general expression for the inverse..

    Thank you.
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  4. #4
    Senior Member Shanks's Avatar
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    what is the inverse of (1,0)? can you answer this question?
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  5. #5
    Member Mollier's Avatar
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    If (1,0) does not have an inverse since there is no scalar \alpha such that 0\cdot\alpha=1. Correct?
    By the same reasoning, can I not say that (0,0) does not have an inverse?

    How about my second question (b) ?

    Thank you very much.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Mollier View Post
    If (1,0) does not have an inverse since there is no scalar \alpha such that 0\cdot\alpha=1. Correct?
    By the same reasoning, can I not say that (0,0) does not have an inverse?

    How about my second question (b) ?

    Thank you very much.
    What have you done to try and find this inverse so far?

    You want to find an a, b \in \mathbb{R} such that (\alpha, \beta)(a, b) = 1. This gives you 2 equations in 4 variables, but with a bit of work you should see what is going on.
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  7. #7
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    Quote Originally Posted by Mollier View Post
    If (1,0) does not have an inverse since there is no scalar \alpha such that 0\cdot\alpha=1. Correct?
    By the same reasoning, can I not say that (0,0) does not have an inverse?

    How about my second question (b) ?

    Thank you very much.
    The 0 element (additive identity), which here is (0, 0) never has an inverse. I don't see why you keep mentioning it. The crucial point is that, in a field, every member except 0 must have an inverse. (1, 0) does not.
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    The 0 element (additive identity), which here is (0, 0) never has an inverse. I don't see why you keep mentioning it. The crucial point is that, in a field, every member except 0 must have an inverse. (1, 0) does not.
    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."
    Last edited by Swlabr; January 6th 2010 at 04:26 AM. Reason: Changed what I thought the sentence should read as.
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  9. #9
    Member Mollier's Avatar
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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 if r is 0."
    Exactly. I am easily confused

    1.\;\alpha a-\beta b = 1
    2.\;\alpha b+\beta a = 1

     a_1=\frac{1+\beta b}{\alpha}
     a_2=\frac{1-\alpha b}{\beta}

    Since a_1=a_2:

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

    then

    a=\left(\frac{\alpha+\beta}{\alpha^2+\beta^2}\righ  t)

    I think that is correct.
    Thanks guys!
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  10. #10
    Senior Member Shanks's Avatar
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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 if r is 0."
    your statement does not mean the same thing as mine.
    what I want to clearify is that the element of the form (r,0) where r is not 0 has no inverse.
    (0,0) obviously has no inverse in a field.

    I appologize if I made you confused!
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  11. #11
    Kep
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    Of course, this is a justification of the "strange" multiplication law for complex numbers... clever!

    Also, the multiplicative identity for (b) is (1,0), not (1,1). Ask yourself why.
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  12. #12
    Member Mollier's Avatar
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    Quote Originally Posted by Shanks View Post
    (b) can't be a field because the inverse of some element is not unique.
    So, some element (\alpha,\beta) may have several inverses?
    Would you mind explaining this in a bit more detail?

    Thanks
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  13. #13
    Member Mollier's Avatar
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    Quote Originally Posted by Kep View Post
    Also, the multiplicative identity for (b) is (1,0), not (1,1). Ask yourself why.
    Well, I see that it is (1,0) but I don't know how to explain it without saying that it's caused by how multiplication of elements is defined...
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  14. #14
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Shanks View Post
    your statement does not mean the same thing as mine.
    what I want to clearify is that the element of the form (r,0) where r is not 0 has no inverse.
    (0,0) obviously has no inverse in a field.

    I appologize if I made you confused!
    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.
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  15. #15
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Mollier View Post
    Well, I see that it is (1,0) but I don't know how to explain it without saying that it's caused by how multiplication of elements is defined...
    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.
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