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Linear algebra - Field equations

  1. Feb 23, 2008 #1
    1. The problem statement, all variables and given/known data

    Let F be a field. For any a,b [tex]\in[/tex] F, b[tex]\neq[/tex]0, we write a/b for ab^-1. Prove the following statements for any a,a' [tex]\in[/tex]F and b,b' [tex]\in[/tex] F\{0}:

    i.) a/b = a'/b' if and only if ab' = a'b.

    ii.) a/b + a'/b' = (ab' + a'b)/bb'

    iii.) (a/b)(a'/b') = (aa')/(bb')

    iv.) ((a/b)/(a'/b')) = (ab')/(a'b) if in addition a' [tex]\neq[/tex] 0

    2. Relevant equations

    3. The attempt at a solution

    Okay im getting really confused in these questions because i dont know when it means to divide by and when it is showing that a/b = ab^-1

    i.) want to show a/b = a'/b' if and only if ab' = a'b.

    from ab' = a'b a = a'b/b' and a' = ab'/b

    So substituting in values of a and a'

    a/b = (a'b/b')/(ab'/b)

    Carrying out division:

    (a'ba')/(b'ab') and because a'b = b'a, substitute this in and cancel like terms:

    (ab'a'/ab'b') = a'/b'

    therefore a'/b' = a/b

    is this even right because its just rearranging stuff?
    Last edited: Feb 23, 2008
  2. jcsd
  3. Feb 23, 2008 #2
    Yeah, i cant be doing it right, it must mean i need to write in the inverses. Thoughts?
  4. Feb 23, 2008 #3
    So for i.)

    a/b = a'/b' becomes a(b)^-1 = a'(b')^-1

    So rearranging a'b = b'a gives a = a'b/b' and b = b'a/a', substitute this in:

    (a'b/b') (b'a/a')^-1

    Substitute in b' = a'b/a into the non inverse bracket:


    Then by the multiplicative inverse property of a field, elements and their inverses become the identity element, which is 1 in this case (do i need to show its 1?):

    So, (a'b/b)(b')^-1

    the b's cancel to give a'(b')^-1 which is written as a'/b'

    So a/b = a'/b' as asked

    Is this the right idea? Any tips to improve my setting out and explanations as i go along as i don't want to miss anything? Thanks in advance
  5. Feb 23, 2008 #4
    Should i write that i can cancel the b's by existence of multiplicative inverse as bx1/b = 1?
  6. Feb 24, 2008 #5
    I think i have done number ii.) by substituing in values for a and b, then using the inverse to rearrange into the desired form.
    Last edited: Feb 24, 2008
  7. Feb 24, 2008 #6
    I think you're really overthinking things. All of those just require you to rewrite x/y as xy^-1 then use the properties of field multiplication.
  8. Feb 24, 2008 #7
    Yea i thought so, so in my third post do you think that i have the right idea in my proof?
  9. Feb 24, 2008 #8
    Yeah, you'd just rewrite it as [itex]ab^{-1}a'b'^{-1}[/itex], then rearrange using commutativity of multiplication.

    Edit: Or you could use part 1 to show equality. In this case, it might be a better idea, because the question seems to be alluding to the kind of stuff you have to do to show that the field of fractions of an integral domain is actually a well-defined field.
    Last edited: Feb 24, 2008
  10. Feb 24, 2008 #9
    How do i show equality? What do i do differently than what i have done already?
  11. Feb 24, 2008 #10
    For part iii.) can i do it like this:

    (a/b)(a'/b') = (aa')/(bb')

    (ab^-1)(a'(b')^-1) for LHS then say that it can be rearranged to right hand side by law of associativity? Feels like one of those questions that has a suspiciously simple solution.
  12. Feb 24, 2008 #11
    Commutativity and associativity, yes. At least, that's the way it seems.
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