dimention of R+

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

Jojo,

The dimension of over is . Indeed, note that is necessarily greater than else it would be isomorphic to , which is countable. Moreover, it's obvious that . So, we have the fact that . 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.

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?

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^ay^b) = ln(x^ay^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) = 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 = 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.

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