Mingon: Huh? J numbers? You mean imaginary numbers? That's usually represented using i = +sqrt(-1) on both sides of the pond. But I have no idea what that has to do with this discussion. Where are you getting the idea that you can just replace a factor of a number with its square root? He's not squaring anyway. His example question was about cubes.
In pictorial terms, if a square has equal sides of 4 inches, then its area will be 16 inches. But in order to find the square root of a negative number, you have to imagine a negative area, with four sides of negative length. It's an impossibility, but, as Mazur argues, a mind-bendingly creative one
Originally posted by: Lynx516
Does not work you have to be able to describe teh item with a non integer number of cohordinates.
Originally posted by: Lynx516
ZeroNine you say they have to have a magnitude and directon how then do non integer dimensions work and beleive me they exist.
Does not work you have to be able to describe teh item with a non integer number of cohordinates.
Edit: I just want to clarify, I've got no problem with the math being used for negative dimensions and half dimensions, I'm just saying that these things do not exist as such (to my knowledge), rather they are only mathematical ideas with no physical counterpart. I think this was the original question, or at least my understanding of it.
Originally posted by: Pudgygiant
If it was a negative dimension, wouldn't you expect the volume of a -3 dimension cube to be negative?
Originally posted by: Chaotic42
Ok, so the area of a square is x²
The volume of a cube is x^3
The volume of a 4 dimensional hypercube is x^4
Would the volume of a -3 dimensional cube be 1/(x^3)?
If x is 3m, would the volume of the cube be (1/27) cubic meters?
Are there negative dimensions?
Um. No.Originally posted by: Mingon
The point is that the formula is the same whether the numbers are negative or positive, by substituting the - for something else (e.g. i to keep you happy) you can work it out and then just substitute - back for i afterwards. I know it sounds strange but I am pretty sure it works. Again try it, see what you get and then comment.
Originally posted by: ZeroNine8
isn't this j or i you're using the sqrt(-1)? if so, then i^2 != -1^2
and how did i get brought up anyway, like rjain said, you don't do any square roots in finding volume with normal dimensions, why is it an issue now?
Originally posted by: Mingon
You would (iirc) have to use J numbers for negative numbers you wish to square. so you substitute -3 for 3J and your answer would be 9j² But its been a while since I have done these so I could be wrong - or you might do it different stateside.
Originally posted by: ZeroNine8
Originally posted by: Lynx516
ZeroNine you say they have to have a magnitude and directon how then do non integer dimensions work and beleive me they exist.Does not work you have to be able to describe teh item with a non integer number of cohordinates.
I'm not even sure what you're trying to say here... are you talking about 1/2 dimensions?
The number of coordinates (component vectors) used to completely describe something in 3D space is 3, for 2D it is 2, it doesn't matter what coordinate system you are using as long as the dimensions you are working within are orthogonal. I don't see how integer/non integer has anything to do with it, as the component vectors are simply a magnitude and direction, typically with the direction noted as + or -. As for half dimensions, I don't know that I believe that they exist, as the only way I have encountered them are in mathematical equations that are simplified models of full dimensional systems. If you maintain that these 'non integer' dimensions truly exist, and not just as a mathematical construct in a formula, please direct me to some info on their existance.
Edit: I just want to clarify, I've got no problem with the math being used for negative dimensions and half dimensions, I'm just saying that these things do not exist as such (to my knowledge), rather they are only mathematical ideas with no physical counterpart. I think this was the original question, or at least my understanding of it.