Re: How to define a vector without coordinate system?

In article <1147880529.606323.265350@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"andrzej" <jedrek_2004@xxxxxxxxxxx> wrote:

Thank you for your answer.

You are missing the point (and since you didn't quote any of the
message you're responding to, you're not making it any easier for
yourself to see how badly you're missing the point - please learn
how to quote in response to other posts).

The real numbers are a vector space OVER THE RATIONALS.


As such, they are not 1-dimensional, but infinite dimensional.

I believe you that it is possible make the simple problem very
complicated but
If we look for example here:
http://en.wikipedia.org/wiki/Examples_of_vector_spaces
http://en.wikipedia.org/wiki/Vector_space
http://www.cs.ut.ee/~toomas_l/linalg/lin1/node5.html
http://www.ltcconline.net/greenl/courses/203/Vectors/vectorSpaces.htm
ect.

I think in all that pages R^1=R can be treated as a particular case of
R^n which is n-dimensional vector space. Then R^1 is one dimensional
vector space.

However I agree with you that it is possible to treat R as infinite
dimensional vector space with some strange base.

You think the rationals are strange?

In your case each real number is a sum of some series of rational
numbers.

As such, for every positive integer n there are sets of real numbers
that are n-dimensional vector spaces.

What happened if n=1? We got simply R, isn't, it?

See, here's exactly where you're missing the point. If n = 1, we get
simply Q, the rationals, not R.

sqrt 2 is a vector over the
rationals, and sqrt 3 is another vector over the rationals, and these
two vectors are linearly independent over the rationals.

You are right.
In infinite dimensional spaces there are always problem with the basis
and coordinates.

I can do now some mental exercise.
Sqrt 2 is a some series of rational numbers. Then actually instead of
talking about Sqrt 2 we can talk about that series. Coefficient of such
series are simply basis vectors. So without introducing such coordinate
system your number i.e. Sqrt 2 has no meaning.
The same problem is with Sqrt 3 :).

I would say that without a coordinate system sqrt 2 has no meaning
TO YOU. It had plenty of meaning to Pythagoras & Euclid, long before
anybody ever thought of a coordinate system, and it has plenty of
meaning to me. It is possible - common, even, in some branches of
mathematics - to think of sqrt 2 as a vector in a finite-dimensional
vector space over the rationals, without ever thinking of any series
of rationals.

So if we are in the world of infinite series of rational we are in the
world of coordinates.
Sqrt 2 and Sqrt 3 are just naming convention which without appropriate
definition using appropriate series means nothing.

Nonsense. Sqrt 2 is the length of the diagonal of a square of side 1.
Sqrt 2 is the positive number whose square is 2. These are not naming
conventions, these are appropriate definitions, and no series has been
mentioned.

But I should say that infinite dimensional spaces are a little
problematic so I prefer do not talk about that.

In my opinion you can treat R as a one dimensional vector space with a
basis vector 1.

That's not an opinion, that's a fact. It's also a fact that you can
treat R as an infinite dimensional vector space over the rationals,
and you can deal with various and sundry finite-dimensional subspaces.
Moreover, this interpretation of matters is extremely useful in some
parts of mathematics. Instead of denying this, and continuing to
argue from ignorance, why not take the chance to learn something new?

That is most obvious solution. However you can use any
other number instead of 1. In that case the situation become les
trivial and your vectors Sqrt 2 and Sqrt 3 become linearly dependent.

Not over Q, they don't.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

Relevant Pages

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