A link between Wedge (aka cross product, bivector, etc) and one dimensional rational geometry

(Note – this is not advanced in any way, I just think it is fun and wanted to express it very plainly. Don’t wait for a big reveal at the end or anything.)

Consider two points on the one dimensional number line, both rational numbers.

3/5 and 9/12.

 ----|-------a-----b------>
     0      3/5   9/12

How do you know if one is farther from the origin (0) than the other?

The trick here is to ask a question. Are these really one dimensional values? Or, can we also think of them as two dimensional values? After all, the Rationals are defined as the ratio of two integers. That is what they are – two integers, a pair of numbers. Two numbers. In a sense, are they not two dimensional? Let’s say yes for a moment.

Imagine the number 3/5. Perhaps we can say the numerator, 3, is the x coordinate, and the denominator, 5, is the y coordinate, of a point in two dimensional space. We could even draw it as a triangle:

 |
 |
 |  /|
 | / |5
 |/__|_____________
 |  3

We could imagine a similar situation for the number 9/12, let x be 9 and y be 12. Draw the point..

 |
 |
 |       
 |       
 |     
 |         /|
 |      /   |  12
 |   /      |
 |/_________|_____________
 |    9

There is another way to think about these to points. They can be ‘vectors’ from the origin. 3,5 and 9,12.

How does this relate to the original question? Ahh…

Whichever Vector has a ‘higher slope’ in 2 dimensions…. is the vector representing the one-dimensional number closer to the origin.

Why?

Think about it this way. Say you had a billion / 2. That is a very large number on the one dimensional number line. Now say you drew the two dimensional vector (1000000000,2). That is going to have a pretty low slope. It will almost be the same as the x-axis.

Now imagine the number 3 / 100 000. In two dimensional viewpoint, this is 3,100000, which has a very high slope.

It becomes obvious after a bit of thinking that this all makes sense.

But what about the whole situation where two rational numbers represent the same number?

3/5 is the same thing as 6/10, is it not, in one dimensions? Yes, yes it is.

What happens in two dimensions, though? We get 3,5 and 6,10. What happens if we draw it?

 |
 |
 |     /|
 |    / |10
 |   /  |
 |  /|  |
 | / |5 |
 |/__|__|__________
 |  3  6

We get a line. 3,5 and 6,10 lie on the same line. Amazing!

So 3/5 and 6/10 being the same number in one dimension, corresponds to 3,5 and 6,10 being on the same line in two dimensions.

Great. But how does this help us find out which one-dimensional number is ‘bigger’ than the other? Which one is ‘smaller’, or closer to the origin?

Ahhh… this is the interesting bit. You could say “ok… take the angle between the two vectors”. You could, but you don’t need to. Angles tend to deal in irrational numbers, and we don’t need to do that.

You could say ‘compare the slopes’… yes but how do you compare slopes? “Oh… convert to decimal”…. ahh, but there is a trick. You can’t convert every rational to a decimal of finite length. For example 1/3 = 0.3333333…..

Since we are trying to actually calculate an answer, on a machine, we don’t want irratioanls. At least not in this particular way of imagining things.

Instead, we can pull out the ‘wedge’ function. What is it?

Imagine two vectors, x1 y1 and x2 y2. Wedge gives the signed area of the parallelogram formed by the two vectors.

The formula is x1 * y2 - x2 * y1

Wait, what? The parallelogram? Yes. Two vectors from the origin, form two sides of a parallelogram.

 |
 |   a
 |  /
 | /        b
_|/...---```____________
 |
 |

a and b are two vectors, and can form a parallelogram:

 |   a         a+b
 |   ...---```/
 |  /        /
 | /        /
_|/...---```b___________
 |
 |

Wait, what’s signed area? Ahh… you see the area itself is considered positive…. but if you use the formula above, it turns out that you get the opposite answer depending on which vector you plug in first (for x1,y1) and which you plug in second (for x2 y2).

If you put in x1,y1 = 3,5 and x2,y2 = 9,12 your answer is 3*12 – 9*5, or 36-45, or -9

If you put in the opposite order, x1,y1 = 9,12 and x2,y2 = 3,5 your answer is 9*5 – 3*12, or 45-36, or +9.

What does that tell you? If your answer is negative, then the first vector has a higher slope than the second. If your answer is positive, then the second vector has a higher slope than the first. You can imagine, again, using numbers like a billion over 2 and figure out which is which. (billion*billion-1*1 > 0)

Do you see? You can now determine which vector has the higher slope just by taking the signed area of the parallelogram between them… which is also called the ‘wedge’ of the vectors.

In this way, we relate the picture in two dimensions, directly to the picture in one dimension. For the higher slope 2 dimensional vector, is the ‘lowest’ number in one dimension, and is closest to the origin.

We had two one dimensional numbers, 3/5 and 9/12.
We do the wedge formula

   3*12 - 9*5 = 36-45 = -9

This answer is less than zero, so it means that 3,5 is of a higher slope, and thus that the one dimensional number 3/5 is lower than 9/12, and closer to the origin.

To check answers, think quickly of the percentages represented. 3/5 is three twenties, or 60 percent. 9/12 is 3/4, or 75 percent. Clearly 60 is less than 75. We are not crazy.

Using Wedge, we found the answer, without using a Greatest Common Divisor algorithm, without using irrational numbers or trigonometric functions like sin or cosine. We just used 2 integer multiplications and a subtraction.

That is pretty fast on a computer. Unless you are using ‘big integers’ that dont fit in a machine word… but that’s another topic for another time.

This is the type of thing that is often accomplished using ‘projection’ in Geometry. As worldly objects appear differently in their shadows on the ground, as cast by the Sun, so do geometrical objects appear differently when their shadows appear in other dimensions, cast there by the light of our imaginations. A fun example of this is the convex Hull of an 3d object and how it relates to the Delaunay triangulation of it’s shadow in two dimensions. But that is another topic for another time.

See Also

http://www.academia.edu/4162881/One_dimensional_metrical_geometry by Dr N J Wildberger

https://www.youtube.com/watch?v=q3turHmOWq4 Wild Trig 33 Projective geometry and homogeneous coordinates by Dr N J Wildberger

https://www.youtube.com/watch?v=q3turHmOWq4 Wild Trig 34 Lines and planes in projective geometry by Dr N J Wildberger

https://en.wikipedia.org/wiki/Delaunay_triangulation

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About donbright

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