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Question about a cancellation law

  1. Dec 12, 2012 #1
    1. The problem statement, all variables and given/known data
    if ba = bc and a is nonzero then b = c


    3. The attempt at a solution
    if a is nonzero I can multiply both sides by a-1 to get
    ba(a-1) = bc(a-1)
    b(1) = bc(a-1)
    b = bc(a-1)
    not sure how to proceed, is the problem statement ba = bc supposed to be ab = ac ?
     
  2. jcsd
  3. Dec 12, 2012 #2
  4. Dec 12, 2012 #3
    i'll check it out, thanks a lot.
     
  5. Dec 12, 2012 #4

    Mentallic

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    Homework Helper

    No, that's not true at all. For example, take b=2, a=4, c=4. The equality holds, the criteria are met, yet [itex]b\neq c[/itex]

    What you're looking for is

    If ba = bc and b is non-zero, then a = c.

    EDIT: Sorry, I missed this part:
    Yes, that's what it should be instead.
     
  6. Dec 12, 2012 #5

    HallsofIvy

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    Science Advisor

    You do not say what algebraic structure you are working in. In a "cancellation ring" such as the set of all integers, with ordinary addition and multiplication as operations, we do NOT have "multiplicative inverses" but the cancellation law is still true: if ab= ac and a is not the additive identity, 0, then b= c.
     
  7. Dec 12, 2012 #6
    I'm not sure if this is correct, but i'm thinking it's a field since they didn't specify. My book was using this as an example of how to show that axioms that hold in real arithmetic may not hold in matrix arithmetic: "for example, consider the following two laws of real arithmetic: If ab = bc and a is nonzero then b = c. [ Cancellation Law ] "
     
  8. Dec 12, 2012 #7
    That so-called "cancellation law" doesn't hold for matrices. Just consider ##a=\left(\begin{array}{cc} 1&0\\0&0 \end{array}\right)## (which is nonzero), ##b=\left(\begin{array}{cc} 0&1\\1&1 \end{array}\right)## and ##c=\left(\begin{array}{cc} 0&-1\\0&1 \end{array}\right)##.
     
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