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Math Help - dimention of R+

  1. #1
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    dimention of R+

    What is the dimension of R+ as a vector space over R? over Q?
    Thank you in advance
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    Re: dimention of R+

    Jojo,

    The dimension of \mathbb{R} over \mathbb{Q} is |\mathbb{R}|=\mathfrak{c}. Indeed, note that \dim_\mathbb{Q}\mathbb{R} is necessarily greater than |\mathbb{N}|=\aleph_0 else it would be isomorphic to \mathbb{Q}[x], which is countable. Moreover, it's obvious that \dim_\mathbb{Q}\mathbb{R}\leqslant\mathfrak{c}. So, we have the fact that \aleph_0<\dim_\mathbb{Q}\mathbb{R}\leqslant \mathfrak{c}. If you're into that continuum hypothesis kind of thing you're done. If you want to see an actual proof opposed to a plausibility argument, see my blog post here.
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    Re: dimention of R+

    I appreciate your help, thank you very much!
    May I ask, is your reference to my question dealing with positive R as just R was since, they are isomorphic?
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  4. #4
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    Re: dimention of R+

    let's exhibit a linear isomorphism between the vector space U = R+, with the vector sum:

    x(+)y = xy

    and the scalar multiplication:

    c*x = xc

    and the vector space V = R, where the vector sum is ordinary real addition, and the scalar multiplication is ordinary multiplication.

    we define T:U-->V by T(u) = ln(u).

    to prove T is linear, we must show:

    T((a*x)(+)(b*y)) = aT(x) + bT(y).

    so: T((a*x)(+)(b*y)) = T((xa)(+)(yb)) = T(xayb) = ln(xayb) = ln(xa) + ln(yb) = aln(x) + bln(y)

    = aT(x) + bT(y), as desired.

    to prove that T is injective, we must show ker(T) = {1}, (since 1 is the 0-vector of R+).

    so suppose u is in ker(T). then ln(u) = 0, hence u = eln(u) = e0 = 1.

    to prove T is surjective, we need to find a u in R+, such that for any real number x, T(u) = x.

    let u = ex, which is in R+ (since e > 0).

    then T(u) = ln(ex) = x, as desired.

    since R (as V) has dimension 1 over R, (the number 1 is a basis), it follows that U = R+ also has dimension 1 over R (we can take the number e as a basis for U).

    if our base field is Q, it gets a bit more interesting: as drexel28 pointed out, R is infinite-dimensional over Q (in fact "more" than infinite, uncountably infinite!). the same mapping T will serve as an isomorphism, showing that R+ is also infinite-dimensional over Q.
    Last edited by Deveno; December 11th 2012 at 11:39 AM.
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    Re: dimention of R+

    I am grateful for your help...!!!
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