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Proving Vector Space Sets

  1. Mar 3, 2015 #1

    B18

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    1. The problem statement, all variables and given/known data
    Determine if they given set is a vector space using the indicated operations.

    2. Relevant equations


    3. The attempt at a solution
    Set {x: x E R} with operations x(+)y=xy and c(.)x=xc
    The (.) is the circle dot multiplication sign, and the (+) is the circle plus addition sign.

    I tried to use axioms 1,2,3,6,7,8,9,10 to prove it isn't a vector space and have not found anything yet. Im not entirely sure how I can define the 0 vector which is why i haven't tried axioms 4, and 5.

    Anyone able to steer me down the right path on this one?
     
  2. jcsd
  3. Mar 3, 2015 #2

    Ray Vickson

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    Nobody can say, because we do not know what book you are using and do not know exactly what is the content of Axioms 1,2,3, etc. Different books sometimes have slightly different axioms and will often number them differently.
     
  4. Mar 3, 2015 #3

    B18

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    ImageUploadedByPhysics Forums1425442787.977196.jpg
    Here is the list of axioms I'm using. I never thought of that. Thanks Ray.
     
  5. Mar 3, 2015 #4

    Dick

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    You say you did axiom 6? ##(-1)## is in ##R##. What is ##\frac{1}{2} \odot (-1)##? Are you sure you don't mean ##R## to be the positive reals? And why don't you think about trying 1 as a '0 vector'.
     
    Last edited: Mar 4, 2015
  6. Mar 4, 2015 #5

    HallsofIvy

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    The "0" vector has the property that v+ 0= v for any vector v. Here, that must be that v times "0" is equal to v. So what number has that property? Of course, every vector must have an "additive inverse" which, since "addition of vectors" is here defined as their product, means there is likely to be a problem with the number 0. That is one reason Dick asked about whether the vectors were not the positive reals rather than all real numbers.
     
  7. Mar 4, 2015 #6

    B18

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    Wow, that one flew right past me. Apparently I wasn't thinking about exponent rules enough. The set is all real numbers though, nothing indicating that it is only positives.
     
  8. Mar 4, 2015 #7

    Mark44

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    Can you post a photo of the problem? If the set involved is all real numbers, you're going to have problems with things like ##.5 \odot (-1)##, which is ##(-1)^.5##. Also not stated in this thread is the field that the scalars come from.
     
  9. Mar 4, 2015 #8

    B18

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    Sure thing!
    ImageUploadedByPhysics Forums1425500220.542684.jpg
     
  10. Mar 4, 2015 #9

    B18

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    In this case the zero vector would have to be equal to 1?
     
  11. Mar 4, 2015 #10

    HallsofIvy

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    Yes, the "zero vector", with this definition of vector addition, must be the number "1".
     
  12. Mar 4, 2015 #11

    Mark44

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    So (B18), keeping in mind that 1 in this space acts like 0, what do you have to "add" to a number x so that ##x \oplus ? = 1##? IOW, what do you have to "add" to a number x to get "zero"? Hope that doesn't boggle your mind too much!
     
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