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Linear independence

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=63039&stc=1&d=1382005383.jpg

    2. Relevant equations



    3. The attempt at a solution

    For part (a):

    a*1 + b*√2 + c*√3 = 0
    assume a, b, c not all zero
    a + b√2 = -c√3
    a2 + 2b2 + 2ab√2 = 3c2
    a2 + 2b2 - 3c2 = -2ab√2
    (a2 + 2b2 - 3c2)/(-2ab) = √2

    which is not possible since we take a, b, c to be rational, and √2 is irrational.

    thus our assumption of a, b, c not all zero was false and we must have a=b=c=0.

    For part (b), a similar argument, but easier:

    {1, 1 + √5, (1 + √5)2} = {1, 1 + √5, 1 + 2√5 + 5}

    a + b(1 + √5) + c(1 + 2√5 + 5) = 0
    assume a, b, c not all zero
    1 + b + b√5 + c + 2c√5 + 5c = 0
    1 + b + c + 5c = -b√5 - 2c√5
    1 + b + 6c = (-b - 2c)√5
    (1 + b + 6c)/(-b - 2c) = √5

    same story as before.

    Now that was super easy. And the assignment says to be careful with the structure of my argument. And I hate denominators because they can't be zero. If a = b = c = 0, then we have 0/0, which is of indeterminate form, which is ok! but i'm getting the idea that my argument is flawed because of division by zero...
     

    Attached Files:

  2. jcsd
  3. Oct 17, 2013 #2

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    Looks OK to me ... a,b,c not all zero is required for linear independence anyway.
     
  4. Oct 17, 2013 #3

    Mark44

    Staff: Mentor

    For part b, note that your set is {1, 1 + √5, 6 + 2√5}. Clearly the third "vector" is a linear combination of the first two in the list.
     
  5. Oct 17, 2013 #4
    yeah i was so caught up in things that i forgot to combine the 1 and 5, thanks everyone.
     
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