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Subsets of two-dimensional space

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data

    a) construct a subset of two-dimensional space closed under vector addition and even subtraction, but not under scalar multiplication.

    b) construct a subset of two-dimensional space (other than two opposite quadrants) closed under scalar multiplication but not under vector addition.


    3. The attempt at a solution

    a) A line going through a point on the y-axis. Not a line through the origin though. I figure since the line does not have a null-vector in it, it is not closed under scalar multiplication.

    b) I only know of two opposite quadrants :/

    Anyone? Thanks!
     
  2. jcsd
  3. Oct 30, 2008 #2

    HallsofIvy

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    Un fortunately, if it does not go through the origin, that is, does not contain 0, it is not closed under "vector addition and even subtraction" because v-v= 0.

     
  4. Nov 2, 2008 #3
    Maybe I've missunderstood what even subtraction is. Does that mean that I subtract a vector by the same vector? I thought it meant that it was closed under addition and also under subtraction..

    If even subtraction really is v-v, then the answer to a) is a line from the origin into the first quadrant, right?

    Thank you
     
  5. Nov 2, 2008 #4

    Office_Shredder

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    It means if u,v are in your subset, u+v AND u-v are in the subset. One particular example is for all v in the subset, v-v = 0 is in the subset also
     
  6. Nov 2, 2008 #5
    Damn, then I'm out of quesses.. Please help? :)
     
  7. Dec 6, 2008 #6
    I just went back to this problem and still can't find an answer.
    Could someone please help?
     
  8. Dec 6, 2008 #7
    Then for any v on this line, 0 - v must also be on it, but it's not because your line is only in the first quadrant.

    a) How about vectors with integer norm?

    b) Two non-parallel lines through the origin?
     
  9. Dec 6, 2008 #8

    HallsofIvy

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    You are looking for sets of numbers that are closed under addition and subtraction but not under multiplication by real numbers.

    Try {(n,n)} where n is an integer.
     
  10. Dec 6, 2008 #9
    Now I know that this is wrong; e.g. (1, 0) - (0, 1) does not have integer norm. See what HallsofIvy said.
     
  11. Dec 7, 2008 #10
    HallsofIvy: does {(n,n)} mean vectors in R^2 like (1,1), (2,2), (-4,-4), etc.?
     
  12. Dec 7, 2008 #11

    HallsofIvy

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    I wrote what it meant: "Try {(n,n)} where n is an integer."
    Surely you know what an integer is?
     
  13. Dec 8, 2008 #12
    Yes, I do know what they are.
    To me {(n,n)} is a line through the origin, 45 degrees on the x-axis, which is a subspace closed under addition and scalar multiplication.

    You do not have to reply to these stupid questions of mine HallsofIvy, so there is no need to be an ***.

    Thank you.
     
  14. Dec 8, 2008 #13

    HallsofIvy

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    Do you understand that insults will get you kicked of this board permanently?
    You were the one who asked if "Try {(n,n)} where n is an integer" meant "(1,1), (2,2), (-4,-4), etc." so my question was perfectly reasonable.


    In any case, do you now see what the answer to your problem is?
     
  15. Dec 8, 2008 #14
    {(n, n) : n is an integer} is not a line. If you connected the points for some reason, then it would be a line.
     
  16. Dec 9, 2008 #15
    So it is closed by addition and subtraction but not by multiplication by real numbers because then they might not be integers any more?

    I apologize for being an ***. It sometimes happens without me knowing it :p

    Thank you for your patience.
     
  17. Dec 9, 2008 #16

    HallsofIvy

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    Yes, that is correct. (n, n)+ (m, m)= (n+ m, n+ m) and (n, n)- (m, m)= (n-m, n-m), both of the form (k, k) for k integer. But (1/2)(1, 1)= (1/2, 1/2) which is NOT.
     
  18. Dec 10, 2008 #17
    I am obviously quite bad at this and am not able to figure out the second problem.
    (subset closed under multiplication but not under addition)

    Am I allowed to say that it is closed under multiplication by integers or something like that?

    Thank you.
     
  19. Dec 10, 2008 #18

    Office_Shredder

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    For part (b) you know you need a non-zero point (call it A)... then as it's closed under scalar multiplication, you need the line through the origin and A. This is a subspace, so we can't use this. So pick another point B that's not on the line. Again, you need the line passing through the origin and B. Do you need any more points?
     
  20. Dec 10, 2008 #19
    Two points in the 1st quadrant and two points in the 3. quadrant. I draw lines through the origin between the points in opposite quadrants. Then I take linear combinations of the vectors on the two lines, so I get a sector in the first quadrant, and a mirror image of the same sector in the 3. quadrant. Then it is closed under multiplication but not under addition.
    Is this right?

    I guess this would work for all "mirrored sectors".. Am I totally wrong?

    Thank you!
     
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