About quarternions. Everything I see written about the subject

>>8499029

Don't bother learning quaternions, they are useless. Litterally. Absolute garbages. The most useless field in maths.

>If i = sqrt (-1) and j = sqrt (-1) and k = sqrt (-1)

Thats your problem right there. The condition i^2 = j^2 = k^2 = ijk = -1 is strong, thats how that quarternion units are defined, similar to how the i is defined with i^2 = -1, or some equivalent form, with the complex numbers.

>>8499037

Use matrix.

http://mathworld.wolfram.com/Quaternion.html

>>8499037

Yeah, but with complex numbers, the statement,
i = sqrt (-1) is used with no contradiction.

Is this not correct?

Then if sqrt (i^2) does not equal i or sqrt (-1), what does it equal to?

>>8499029
>If i = sqrt (-1) and j = sqrt (-1) and k = sqrt (-1)
wat

learn to logic please

>>8499029
With real numbers, x^2=1 => x=1 or x=-1.
With complex numbers, x^2=-1 => x=i or x=-i.
With quaternions, x^2=-1 => x=i or x=j or x=k or x=-i or x=-j or x=-k.

It's worse with matrices. M.M=I typically has infinitely many solutions, i.e. the identity matrix has infinitely many square roots. E.g. for 3x3 matrices, a rotation of 180 degrees about any axis is a (positive-determinant) square root of the identity matrix, while a reflection in any plane is a (negative-determinant) square root of the identity matrix.

Similarly for nth roots; e.g. a rotation of 360/n degrees about any axis is an nth root of the identity matrix.

>>8499056
It doesnt equal shit because in quarternions the units are not scalars so theres no such thing as a square root and squaring one of their units is actually matrix multiplication. i = sqrt(-1) in complex numbers is just a convention adopted for solving equations. i is more formally described as a ordered pair, and "squaring it" results in another ordered pair which has many of the properties of -1 that we would like to have but is still not exactly -1. For example the set of all numbers (x,0) in R^2 is not the same thing as R. It is clearly related to R, but it is not the same, for example R is open in itself, but this described set does not even have an interior in R^2.

>>8499067

What is wrong with this?

>>8499056
>Then if sqrt (i^2) does not equal i or sqrt (-1), what does it equal to?
The problem you are having with quaternions has its root at the non-commutatitivy of the product.

Because multiplication does not commute (for example i*j=-j*i), definition of the square root is not so easy. In particular, z^2=-1 has not two (like in complex numbers), but infinitely many solutions (every number of the form b*i+c*j+d*k with b^2+c^2+d^2=1 is a solution).

>>8499073
You assumed just because j*j = -1 that j is exactly the same as i. Thats not how quarternions work. It assumes that -1 has a unique "root" in the quarternions. By construction this is emphatically not true.

>>8499073
first of all it's not provable

>>8499029
It's true that i, j and k are square roots of -1, but that doesn't mean that sqrt(-1) * sqrt(-1) != sqrt(1) = 1. It makes no sense to calculate like that.

>>8499035
They are used to program gaymen