What are the group axioms?

[matt grime]

why are you hatting your i's etc? THere's no need to do that.

[Johnny5]

why are you hatting your i's etc? THere's no need to do that.

 

 

They are unit vectors, einheitsvektors, versors, vectors with magnitude one. Why not do it?

 

As in here is the following standard orthonormal set of basis vectors in R^3:

 

[math] \mathcal{f} \hat e_1, \hat e_2, \hat e_3 \mathcal{g} [/math]

[matt grime]

Johnny, the multiplication of two quaternions is simply the algebra multiplication according to the rules that ij=-ji etc. It is NOT the cross product as you assume it is. It can be written in terms of the cross product and the scalar product, but in your first line you say:

 

"interpret the multiplication as vector cross product"

 

if you do that then it isn't the product of the elements as quaternions.

 

There is no need to hat or underline vectors. This is maths, we dont' do that, physicists might but we don't. It is completely unnecessary and wastes time, and we know from the definition of the quaternion norm that i, j and k have magnitude 1.

[Johnny5]

we know from the definition of the quaternion norm that i, j and k have magnitude 1.

 

What is the definition of the quaternion norm?

[matt grime]

if you don't know the definition of quaternion norm how were you able to state that i had length one? Ie how did you decide it was a unit vector if you don't konw what the norm is?

 

If you wre writing complex numbers would you hat the i in a+ib? No, of course not.

[Johnny5]

if you don't know the definition of quaternion norm how were you able to state that i had length one?Ie how did you decide it was a unit vector if you don't konw what the norm is?

 

I treated i,j,k as a unit vectors. No reasoning went into that.

 

 

If you wre writing complex numbers would you hat the i in a+ib? No' date=' of course not.[/quote']

 

No of course not. Are you suggesting that i treat i,j,k as imaginary numbers, instead of unit vectors? Or i should say, imaginary units, that is:

 

[math] i=j=k=\sqrt{-1} [/math]

 

?

[matt grime]

Seeing as i^2=-1=j^2=k^2 then treating them as "imaginary units" would seem better than treating them as vector, though why you can't simply call them quaternions is a mystery. We rarely point out something is a vector anyway, since the fact it is in a vector space implies that already.

 

You assumed i, j and k had length 1 when you called them unit vectors. That is what a unit vector is for heaven's sake: a vector of length 1.

[Johnny5]

Seeing as i^2=-1=j^2=k^2 then treating them as "imaginary units" would seem better than treating them as vector' date=' though why you can't simply call them quaternions is a mystery. We rarely point out something is a vector anyway, since the fact it is in a vector space implies that already.

 

You assumed i, j and k had length 1 when you called them unit vectors. That is what a unit vector is for heaven's sake: a vector of length 1.[/quote']

 

Well here is my next question.

 

What kind of multiplication allows you to write:

 

ii=-1

 

?

 

Not the ordinary multiplication of two real numbers that we learn in grade school.

 

So what?

 

As I previously showed, in this thread, it cannot be cross product, because then we do not get Q1Q2 = -(dot product Q1,Q2) + (cross product Q1,Q2), like many sources suggest you should.

 

In the case where

 

Q1=(0,x1,y1,z1)

Q2=(0,x2,y2,z2)

 

So where was my error?

 

Treating the unknown multiplcation as vector cross product, i did get part of the answer but then

 

ii=0 rather than -1.

 

So??

[matt grime]

The error was you never bothered to work out the dot product part and include it, you only worked out the cross prodcut part. The fact that there is no dot product part in the method you used exactly accounts for the reason why it fails to appear in your answer. It really is no simpler, nor more complicated than that.

 

 

Quote:

"What kind of multiplication allows you to write:

 

ii=-1

 

?

 

Not the ordinary multiplication of two real numbers that we learn in grade school."

 

"What kind"???? Never mind, Johnny.

 

You do understand that I can declare the composition of any two elements of any ring to be anything I care as long as they satisfy the rules. What has "grade school" maths got to do with this? You seem happy enough with the complex numbers as a+ib. They're just symbols to manipulate.

[Johnny5]

"What kind of multiplication allows you to write:

 

ii=-1

 

?

 

Not the ordinary multiplication of two real numbers that we learn in grade school.

 

I was thinking about this last night.

 

The angle between i and itself is 2pi.