# Representing vector by a number.

1. Jun 7, 2009

### dE_logics

Since vector has a direction and a number alone does not, I think a vector cannot be represented using just a number (except in a case of 1d system).

Am I right?

In a 1d system, it can have only 2 directions...how do we represent that?

2. Jun 7, 2009

### Simon-M

The "sign" of a number could act as a "direction" (using your definition of a vector)

3. Jun 7, 2009

### maze

Well, since there exists a bijection from Rn -> R, it should be theoretically possible, though it would be pretty crazy.

4. Jun 7, 2009

### disregardthat

If you allow complex numbers it's pretty easy

5. Jun 7, 2009

### Tibarn

Depends on how useful you want the representation to be. There is a one-to-one correspondence between, say, n-dimensional vectors and numbers. So, each vector gets mapped to a unique number, and each number corresponds to a unique vector. Unfortunately, there's no quick way of finding out which vector a given number represents. Even if we limit ourselves to vectors with rational coordinates and represent them by rational numbers, you'd have to look through a very large table to figure out which number corresponds to which vector.

In conclusion, you can always represent vectors by numbers, but it's almost useless for all practical purposes. The only times it makes sense is if you have one-dimensional vectors mapped to the reals or two-dimensional vectors corresponding to complex numbers.

6. Jun 8, 2009

### dE_logics

But in 1-d only right?

In 2-d signs wont persist.

Ok...integers.

So with complex numbers...you represent the real part as x and imaginary as y?

So we actually have to 'devlope' such a coordinate system.

Lets just talk integers, it can't be represented by integers right? (in cases of above 1-d)

7. Jun 8, 2009

### Simon-M

Only in 1D, yes. Assuming your definitions are:

"A vector is a quantity with 'magnitude' and 'direction'"

"A number is a real number"

When working in 1D we can show 'direction' with the sign of the number. That's all I was saying.

Going along the lines of a bijection between the reals and the n dimensional vectors, one would also need to prove that it respects the vector space axioms. This cannot be the case because we're "forming" a bijection between a vector space of dimension n and a vector space of dimension 1. (Which obviously can't exist)

8. Jun 8, 2009

### Klockan3

You are talking out of your ***, you can easily create a bijection between R^n and R, just follow this:
The first number in both directions;10^0 and 10^-1, denotes the same numbers in the first dimension in R^n, the second in both direction;10^1, 10^-1, denotes the first numbers in the second dimension...... then the n+1th number in both direction; 10^(n+1) and 10^(-2-n), denotes the second numbers in the first dimension.

Reiterate to infinity. Also this is basically the same thing as just denoting them with (a,b,c) since you are just splitting the numbers up.

9. Jun 8, 2009

### Simon-M

There is a difference between creating a bijection between a set and a bijection between vector spaces.

Sure, we could "number" them all, but it wouldn't be a great "representation" as it wouldn't even respect addition

10. Jun 8, 2009

### Klockan3

Then you are not talking about a bijection, you are talking about an isomorphism.

Edit: And you could create an algebra on R which would be isomorph with the vector algebras in R^n. You would just get some strange operations.

Also you could make a total order which respects the normal vector addition laws, just put x>>y>>z.

11. Jun 8, 2009

### D H

Staff Emeritus
You are playing semantic games, Klockan3. As all Rn have the same cardinality, there obviously exists bijections between Rn and Rm for all n and m. That the bijection has essentially zero value is equally obvious: No such bijection is homeomorphic if n is not equal to m.

12. Jun 8, 2009

### Klockan3

But you can still put signs on the vectors, with the ordering x>>y>>z, put those larger than 0 as positive and smaller than 0 as negative.

The deal is that "signs" is a dumb concept in itself and shouldn't really be discussed outside of the first definition. We define sign for R as the numbers smaller than 0 have negative and larger have positive, that can of course not be 100% transferred to the higher dimensions but that doesn't mean that you can't put signs on the higher either.

Do bijections between finite discrete objects have signs?

13. Jun 8, 2009

### dE_logics

We have an on-going online war here....

14. Jun 8, 2009

### dE_logics

Yeah, 1 more question...if this same 1-d vector rotates (i.e exhibits more than 1-d), it will loose on its sign right?

15. Jun 8, 2009

### Klockan3

Eliminate your whole notion of "sign" as something special. The one dimensional "Sign" just means "The vector points in positive/negative direction". That version of "sign" is not possible to apply to anything else.

Edit: But there are a lot of "sign" in mathematics, like matrix determinants denotes the matrix's "sign", or the number of 2-cycles needed to create a permutation is denoted as "sign", with odd numbers being negative and even positive. The word "sign" itself doesn't really mean anything in mathematics except to give you some rules to make computations easier.

16. Jun 8, 2009

### Tibarn

Assuming, of course, the standard topology on R. However, you can use your bijection to induce a quotient topology on R, in which case they would be homeomorphic. If you suitably define new operations on R, you can even get your function to be an isomorphism. These will most likely bear little or no resemblance to the familiar ones. It's useless in practice because there may not be any simple rules governing addition or which vector a given number represents, but it's certainly possible in theory.

17. Jun 8, 2009

Ok...thanks!