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Math Help - satisfying axioms to 'prove' vector space

  1. #1
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    Question satisfying axioms to 'prove' vector space

    hello there!
    I need some clarity on the following please:

    The question asked that it be shown that the following set is not a vector space:

    'The set of all triples of real numbers with addition defined by
    (x, y, z) + (u, v, w) = (z+w, y+v, x+u)'

    I worked out that the following axiom holds:
    (k+m)u = ku + ku

    however, the answer works it out to not hold.

    It goes as follows:

    Set u = (u1, u2, u3), then
    (k+m)u
    = (k+m)(u1, u2, u3)
    = ((k+m)u1, (k+m)u2, (k+m)u3)

    and
    ku + mu
    = (ku1, ku2, ku3) + (mu1, mu2, mu3)
    = (ku3 + mu3, ku2 + mu2, ku1 + mu1)

    since these are not equal, this axiom does not hold.'

    Now, I don't understand why these two are not equal?
    Can we not further simplify:

    (ku3 + mu3, ku2 + mu2, ku1 + mu1)
    = ((k+m)u3, (k+m)u2, (k+m)u1) ?

    That way, does it not equal = ((k+m)u1, (k+m)u2, (k+m)u3)) ?

    or is it not equal because the order of the triple is different?

    I'm quite confused and any help would be much appreciated!

    Have a great day all
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  2. #2
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    Quote Originally Posted by mathchickie View Post
    hello there!
    I need some clarity on the following please:

    The question asked that it be shown that the following set is not a vector space:

    'The set of all triples of real numbers with addition defined by
    (x, y, z) + (u, v, w) = (z+w, y+v, x+u)'

    I worked out that the following axiom holds:
    (k+m)u = ku + ku
    Well, that's not what you want- you want (k+ m)u= ku+ mu. Typo?

    however, the answer works it out to not hold.

    It goes as follows:

    Set u = (u1, u2, u3), then
    (k+m)u
    = (k+m)(u1, u2, u3)
    = ((k+m)u1, (k+m)u2, (k+m)u3)

    and
    ku + mu
    = (ku1, ku2, ku3) + (mu1, mu2, mu3)
    = (ku3 + mu3, ku2 + mu2, ku1 + mu1)

    since these are not equal, this axiom does not hold.'

    Now, I don't understand why these two are not equal?
    Can we not further simplify:

    (ku3 + mu3, ku2 + mu2, ku1 + mu1)
    = ((k+m)u3, (k+m)u2, (k+m)u1) ?

    That way, does it not equal = ((k+m)u1, (k+m)u2, (k+m)u3)) ?

    or is it not equal because the order of the triple is different?

    I'm quite confused and any help would be much appreciated!

    Have a great day all
    Yes, exactly, "triples" implies "ordered triples". (1, 2, 1) is NOT the same as (2, 1, 1).
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  3. #3
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    Aha! Thank you... that makes things a lot easier.

    [quote=HallsofIvy;503590]Well, that's not what you want- you want (k+ m)u= ku+ mu. Typo?

    sorry, typo! Thanks

    Seeing then that "triples" implies ordered triples, if we look at the following axiom: with addition defined as (x, y, z) + (u, v, w) = (z+w, y+v, x+u)

    0 + u = u + 0 = u and we set
    u = (u1, u2, u3)

    Then, we get

    0 + u = (0, 0, 0) + (u1, u2, u3)
    = (0 + u3, 0 + u2, 0 + u1)
    = (u3, u2, u1) + (0, 0, 0)
    = (u1 + 0, u2 + 0, u3 + 0)
    = (u1, u2, u3)
    = u

    can this be calculated like this??

    Thanks a stack!
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  4. #4
    MHF Contributor

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    [QUOTE=mathchickie;504140]Aha! Thank you... that makes things a lot easier.

    Quote Originally Posted by HallsofIvy View Post
    Well, that's not what you want- you want (k+ m)u= ku+ mu. Typo?

    sorry, typo! Thanks

    Seeing then that "triples" implies ordered triples, if we look at the following axiom: with addition defined as (x, y, z) + (u, v, w) = (z+w, y+v, x+u)

    0 + u = u + 0 = u and we set
    u = (u1, u2, u3)

    Then, we get

    0 + u = (0, 0, 0) + (u1, u2, u3)
    = (0 + u3, 0 + u2, 0 + u1)
    = (u3, u2, u1) + (0, 0, 0)
    = (u1 + 0, u2 + 0, u3 + 0)
    = (u1, u2, u3)
    = u

    can this be calculated like this??
    No, Your first two lines are correct:
    0 + u = (0, 0, 0) + (u1, u2, u3)
    = (0 + u3, 0 + u2, 0 + u1)
    so that 0 + u= (u3, u2, u1) and so 0+ u\ne u

    Similarly u+ 0= (0, 0, 0)+ (u1, u2, u3)
    = (u3, u2, u1) \ne (u1, u2, u3) so u+ 0\ne u.

    However, it is true that u+ v= v+ u.


    Thanks a stack!
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