# dimention of R+

• Dec 9th 2012, 09:10 PM
jojo7777777
dimention of R+
What is the dimension of R+ as a vector space over R? over Q?
• Dec 10th 2012, 05:25 AM
Drexel28
Re: 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 AM
jojo7777777
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?
• Dec 11th 2012, 10:35 AM
Deveno
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.
• Dec 13th 2012, 12:18 PM
jojo7777777
Re: dimention of R+
I am grateful for your help...!!!