Geometric multiplication

[Wise Young]

Here is an interesting video of a multiplication method that just requires counting and no knowledge of the multiplication table.

http://zappinternet.com/index.php?video=zebQzoBhoR

How does it work?

Wise.

[rfbdorf]

Very neat. You just count the crossings. Of course, 2 lines cross 3 lines 6 times (for example).- Richard

[In the wind]

Determinants were first discovered by Leibniz, forgotten and later worked on by Cramer and even later by Cayley.

Normally the numbers are arranged in what we call a matrix or a pattern called a determinate today

Much cleaner than the lines drawn in the video. The same system not only works with but is the basis for linear algebra

[rfbdorf]

Is this really the same thing as a determinate? There is some similarity, but I don't see the subtraction of the other terms (sorry, forgot the terminology). This looks like straightforward multiplication to me, except replacing the multiplication of single place integers with the counting of crossings.
- Richard
(gotta get to work now)

[XYNaPSE]

Determinants have to do with multiplying 2X2 and 3X3 matrices I think. What this person is doing has nothing to do with matrices. Its just straight up multiplication. Very cool.

[Wise Young]

Originally Posted by rfbdorf

Very neat. You just count the crossings. Of course, 2 lines cross 3 lines 6 times (for example).- Richard

I agree with rfbdorf. Counting the number of crossings is multiplication. By dealing with concentric layers of crossings, we are simply performing multiplication digit by digit and doing carryovers for each decimal position. It is simply a graphical depiction of the standard multiplication procedure, as illustrated below.

[In the wind]

Originally Posted by XYNaPSE

Determinants have to do with multiplying 2X2 and 3X3 matrices I think. What this person is doing has nothing to do with matrices. Its just straight up multiplication. Very cool.

Determinates (The matrices) get much larger than 3X3, there's not a limit, they tie the coefficients back to the geometry,

Penrose does an excellent job of showing the relationship (or in my mind, the “sameness”) of the number line to geometry, trigonometry, inverse, imaginary (i) and infinite numbers (and in the last case, demonstrating that there is no difference) with an abstract called the unit circle, a circle where r=1.

The video is a very neat example of this relationship, and shows that multiplication is just addition disguised, but if you really want to itch your brain ask yourself how far does this relationship between numbers and reality go.

A simple place to start, but off the subject of this thread, is to try to move from 1 to 2 without actually getting to 2 but just see how close you can get. These basic arguments (metaphysics) are 3000 years old…

[Tufelhunden]

Actually, In The Wind, you are correct in your first statement that determinants are linked to geometry. The absolute value of a determinant of an NxN matrix gives signed area and volumes of the geometric shapes in which the determinant's original matrix represents. For example, if you have a 2x2 matrix (NxN), the absolute value of the matrix equals the area of the parallelogram spanned by the row vectors. So, if you have a matrix:

51
34,

you can create a cartesian graph (the one shaped like an L with the
x-axis horizontal and the y-axis vertical, and mark the points (5,1) and (3,4) on your graph. conntect these points to the origin of the graph (the origin is the corner of the L). You now have two vectors in 2-dimensional space. You then take another vector similar in length and angle as your (3,4) vector, but instead of originating it from the origin of your cartesian graph, originate it from the point (5,1) so it looks like a "vector shooting off from another vector". You do the same thing to your original (3,4) vector, where you would replicate another (5,1) vector and "shoot it" from the point (3,4). Your two "new" vectors should now come to a single point (should be (8,5)) and the resulting geometric shape is a parallelogram. Well, the absolute value of the determinant from your original matrix is the area of this parallelogram.

If you take the determinant of a 3x3 matrix, (this matrix gives three vectors in three-dimensional space), the resulting geometric shape is a parallelopiped (a parallelogram cube, if you want a visual), and the absolute value of the determinant of the 3x3 matrix is the volume of the resulting parallelopiped.

I say absolute value because negative area or negative volume do not exist.