I'm back with new idea!

[Muzozavr]

Bob can see/imagine:

3D
2D
1D
0D (point)

It means, Bob can see "three" dimensions behind him. (with 0D)
BUT can Fred see only two dimensions behind him? Can Emily see all four behind her?

I'm starting to think that Emily can see/imagine:

4D
3D
2D
1D (line) STOP! No point for her.

Again, Fred can see/imagine:

2D
1D
0D
-1D (nulloid)

So, the dimensional line have no end in both sides, in both sides it just goes on and on... :?

[jinydu]

That doesn't make any sense. What basis do you have for believing that an n-dimensional being can only see/imagine dimensions between n-3 and n?

[Muzozavr]

You are in the dark and your pupils are big, you are in the light and your pupils are small. But if someone lives in other (more dark) diapasone of light he will catch the light in darkness much better. For US it's no light. For THEM there is still quite many light.
For Fred, who lives in a smaller dimension, than Bob, there are another DIAPASONE of dimensions. He can imagine nulloid, for HIM point still has some directions.
For Emily (she lives in 4D) 3D feels FLAT. That's the key. Line will feel for her like point for us. So, she can't imagine point like we can't imagine nulloid.

[jinydu]

A point, by definition, cannot have any directions (not counting position vectors, of course).

Using mathematics, it is possible to study spaces with any number of dimensions, regardless of one's sight. Of course a four-dimensional being could imagine a point: A point is just the limiting case where all four dimensions of an object approach zero.

[Muzozavr]

Are you sure? For US the point doesn't have any directions. And for flatlanders?

For Emily 3D feels flat -- someone already said it. For her: 2D feels like 1D. 1D feels like 0D. Continue, stretch your imgaination, 0D feels like... -1D.

Can Bob imagine -1D?
Can Emily imagine 0D?
Can Fred imagine -2D?


EDIT: May be Emilly will be able to define what is point, but NOT imagine it.

[pat]

Let's reframe your argument.

Pentagons have five sides.
Squares have four sides.
Triangles have three sides.

Assume that to squares, our triangles are triangles.

Then, to pentagons, our squares are triangles.

Furthermore, to septagons (7-side polygons), our hexagons are triangles and our triangles have -1 sides!!

Notice a problem?

[Muzozavr]

No problem here. That's quite possible. If something feels one way, it will be imagined quite the same way.

[jinydu]

Muzozavr, you have stretched analogy way beyond its logical limits and seem to have no understanding of mathematical reasoning.

I recommend you study some geometry. Consult a textbook, or better yet, read a few pages from Euclid's Elements.

[Muzozavr]

1. I will.
2. And mathematical reasoning of different dimensdions may be different...

[bo198214]

And mathematical reasoning of different dimensdions may be different...

For me mathematical reasoning is time and spaceless. Even 0-d beeings can reason about euclidean space (though probably wouldnt get a feeling of it).

[jinydu]

1. I will.
2. And mathematical reasoning of different dimensdions may be different...

1. Thank you

2. Not if you use an agreed upon logical framework.

In the linear algebra class I'm taking at university right now, there are many theorems that can be proven in any number of dimensions. Here's an example:

Any n linearly independent vectors in R^n span R^n.

Thus, for instance, if I have four vectors in R^4 (roughly speaking, four-dimensional space), and none of them can be written as linear combinations of the others, then any vector in R^4 can be written as a linear combination of those four vectors.

[thigle]

i don't know but has that theorem anything to do with root-vectors, i mean coordinate bases ? does it mean that n vectors (if linearly independent) are enough to serve as root vectors for nD space ? (roughly speaking)

what does it mean that n vectors are linearly dependent ? for exemple 3 vectors (010, 100, 111) are base for some affine geometry. how do i find if they are linearly in/dependent ?

[jinydu]

i don't know but has that theorem anything to do with root-vectors, i mean coordinate basis ? does it mean that n vectors (if lineary independent) are enough to serve as root vectors for nD space ? (roughly speaking)

n linearly independent vectors in R^n form a basis for all of R^n; but they do not necessarily form an orthonormal basis.

[PWrong]

what does it mean that n vectors are linearly dependent ? for exemple 3 vectors (010, 100, 111) are base for some affine geometry. how do i find if they are linearly in/dependent ?

Good question. If n vectors are linearly dependent, that means one of them is just a simple combination of the others. If they are linearly independent, then they make a basis for R^n. That means you can express any point as a combination of these vectors.

For instance, (010, 100, 210) is linearly dependent, because the third vector is just the sum of the first vector, and two of the third. That is:

u₃ = u₁ + 2 u₂

Your example is linearly independent. I'll show you how to prove this.

u₃ = (1,1,1)

All we have to do is prove that u_3 is not a combination of u_1 and u_2.
In this case, it's easy.

u₁ = (0,1,0)
u₂ = (1,0,0)
u₃ - u₂ - u₁ = (0,0,1)

These are the three standard basis vectors for R^3, which we already know these are linearly independent.

If you tried this with my linearly dependent set, you'd eventually end up with:
(1,0,0)
(0,1,0)
(0,0,0)

For trickier sets of vectors, you use a matrix to represent them, and try to reduce it to either an Identity matrix:
1 0 0
0 1 0
0 0 1

or some ugly kind of matrix:
1 0 0
0 1 0
0 0 0

Incidentally, not all maths is as boring as linear algebra. I learnt most of this information in the two weeks before my maths exam last year. I think I went to 5 or 6 linear algebra lectures that semester.

[thigle]

:
For trickier sets of vectors, you use a matrix to represent them, and try to reduce it to either an Identity matrix:
1 0 0
0 1 0
0 0 1

or some ugly kind of matrix:
1 0 0
0 1 0
0 0 0

this i don't understand, the first part of your reply i do and tried for that affine base.

you also state: "...not all maths is as boring as linear algebra." what math-label does the above quote fit in ? i mean what subdomains of math discipline are those concepts from ? and what is the most provoking/exciting to you in geometry ?

(am i becoming a notoric off-topicER ? what's off an what's not in a thread "...new idea" ? anything new ? anything ?:shock:)

[jinydu]

The study of matrices is definitely part of linear algebra.