What is the dimension of R^{+}as a vector space over R? over Q?

Thank you in advance

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- Dec 9th 2012, 09:10 PMjojo7777777dimention of R+
What is the dimension of R

^{+}as a vector space over R? over Q?

Thank you in advance - Dec 10th 2012, 05:25 AMDrexel28Re: dimention of R+
Jojo,

The dimension of $\displaystyle \mathbb{R}$ over $\displaystyle \mathbb{Q}$ is $\displaystyle |\mathbb{R}|=\mathfrak{c}$. Indeed, note that $\displaystyle \dim_\mathbb{Q}\mathbb{R}$ is necessarily greater than $\displaystyle |\mathbb{N}|=\aleph_0$ else it would be isomorphic to $\displaystyle \mathbb{Q}[x]$, which is countable. Moreover, it's obvious that $\displaystyle \dim_\mathbb{Q}\mathbb{R}\leqslant\mathfrak{c}$. So, we have the fact that $\displaystyle \aleph_0<\dim_\mathbb{Q}\mathbb{R}\leqslant \mathfrak{c}$. 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. - Dec 11th 2012, 08:06 AMjojo7777777Re: 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? - Dec 11th 2012, 10:35 AMDevenoRe: 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 = x^{c}

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((x^{a})(+)(y^{b})) = T(x^{a}y^{b}) = ln(x^{a}y^{b}) = ln(x^{a}) + ln(y^{b}) = 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 = e^{ln(u)}= e^{0}= 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 = e^{x}, which is in R^{+}(since e > 0).

then T(u) = ln(e^{x}) = 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. - Dec 13th 2012, 12:18 PMjojo7777777Re: dimention of R+
I am grateful for your help...!!!