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X^(n-1) = 0, what does x equal ?

  1. May 11, 2006 #1
    tnx

    tnxtnxtnxtnx
     
    Last edited: May 11, 2006
  2. jcsd
  3. May 11, 2006 #2

    matt grime

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    In the reals ab=0 if and only if one of a or b is zero.
     
  4. May 11, 2006 #3
    but its a^b = 0
    soz, n is a natural number
     
  5. May 11, 2006 #4
    matt wouldn't have said that unless it applied to your problem.

    is there any way of writing your problem in the form a.b=0?
     
  6. May 11, 2006 #5
    not that i know of? pls explian
     
    Last edited: May 11, 2006
  7. May 11, 2006 #6

    matt grime

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    What does x^m mean? (For m a natural number)
     
  8. May 11, 2006 #7
    x*x m times

    x^2 = xx
    x^3 = xxx
    x^m = x.....?
     
  9. May 11, 2006 #8

    matt grime

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    So you do know how to make a power of x into a product of two (or more) numbers, all of which are x. Now, apply the first thing I posted.
     
  10. May 11, 2006 #9
    x^(n-1) = 0

    x*x*x*x = 0?
     
  11. May 11, 2006 #10

    matt grime

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    If you multiply any collection of (real) numbers together and get zero one of them must be zero, that is an underlying and very important fact about them, it is how you factor polynomials, remember?
     
  12. May 11, 2006 #11
    oh yeah... thanks :)
     
  13. May 11, 2006 #12

    HallsofIvy

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    There are actually two answers to your question. For some values of n,
    x= ___ . For other values of n, there is no such x.
     
  14. May 11, 2006 #13
    Somewhat unrelated, but that's because the Reals are a principle ideal domain isn't it? I'm vaguely trying to remember my 'Groups, Rings and Modules' course from 2 years ago.
     
  15. May 11, 2006 #14

    matt grime

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    No. A principal ideal is one in which any ideal is generated by a single element. The reals, being a field, only have trivial ideals anyway (ie 0 and R). This has nothing to do with zero divisors. You're thinking of an integral domain. There are non principle ideal domains that are integral (eg Z[x,y]: the ideal (x,y) is not principle), and there are principal ideal domains that are not integral like Z/4Z.

    As far as I'm concerned, the fact that there are no zero divisors in R comes first, therefore it implies they are an integral domain, rather than they are an integral domain therefore there are no zero divisors. Small point, and wholly semantic.
     
    Last edited: May 11, 2006
  16. May 12, 2006 #15

    matt grime

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    Now I'm having second thoughts. A PID has no zero divisors because it is required to be an integral domain as well. Domain being, apparently, synonymous with integral domain, though that isn't necessarily universal: are domains presumed commutative?
     
  17. May 12, 2006 #16
    In my algebra class we assumed domains to be commutative. But we didn't assume rings to have unity. Strange, in my opinion.
     
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