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/sci/ - Science & Math

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4787375 No.4787375 [Reply] [Original]

Math question, /sci/.

If we look at <span class="math">\mathbb{R}[/spoiler] (the reals) as a vector space over the field <span class="math">\mathbb{Q}[/spoiler] (the rationals), what would be a basis?

>> No.4787381

there wouldn't be any basis.
try expressing <span class="math"> \pi [/tex] in terms of a finite sum of rationals.[/spoiler]

>> No.4787384

>>4787381
Then <span class="math">\pi[/spoiler] (or a rational multiple of it) would have to be a basis vector.
We can prove by the axiom of choice that every vector space has a basis.

>> No.4787395

just define an equivalence relation, where two real numbers x,y are equivalent iff x-y=q for some rational number q.
Now you have can use the axiom of choice to pick one representant for each equivalence class.
Those are your new basis vectors

>> No.4787418

>>4787395
That was my first idea too, but how do we make sure that these vectors are independent? Like for example it could happen that the difference of two irrationals is another irrational in another equivalence class, so these wouldn't be independent.

>> No.4787439

>>4787418
yeah, you are right. I didn't think about the independency.
But that's the problem with the axiom of choice... you only know, that there exists a basis, but you can't really name one.
I'll think a little more about it

>> No.4787452

>>4787439
okay fuck it.. i have to go now.
the problem is, that by definition of this equivalence realtion, the sum of two elements from different equivalence classes is a totally different equivalence class, so that was not a clever approach from my side.

>> No.4787619

http://mathoverflow.net/questions/46063/explicit-hamel-basis-of-real-numbers