Vector division with quaternions?

[Johnny5]

 

if ij = a+bi+cj then multipliying by i on both sides means

 

-j=ai-b+cij=ai-b+a+bi+cj

 

thus c=-1' date=' a-b=0 and a+b=0 or a=b=0 so ij --j, but then i(1+j)=0 so we have zero divisors, which means it can't be a field, thus meaning it isn't that much like C.

 

Introducing k removes this problem, though we lose commutativity so strictly we only have a division algebra not a field.

 

All of these terms would pretty much have been meaningless to Hamilton, I imagine.[/quote']

 

I think you made a mistake up there. You seem to have lost a c?

 

You start off with

 

if ii=-1 & ij = a+bi+cj then

 

iij=i(a+bi+cj)

 

hence

 

-j = ai-b+cij

 

(see i think you lost that boldfaced c right there)

 

And since the assumption is that ij = a+bi+cj it would follow that:

 

-j = ai-b+cij = ai-b+c(a+bi+cj )

 

 

You say "thus c=-1" but i do not see how you drew that conclusion.

 

If I assume that c=-1, then we have:

 

-j = ai-b-ij = ai-b-(a+bi-j )

 

 

ai-ij = ai-a-bi+j

 

-ij = -a-bi+j

 

ij = a+ bi - j

 

 

Which oddly enough is true, if c=-1.

 

But how did you draw the following three conclusions?

 

1. c=-1

2. a=0

3. b=0

 

?

 

I just don't see it.

 

Regards

[matt grime]

entirely possible - i make mistakes all the time.

 

so

 

-j=ai-b+c(a+bi+cj)

 

 

is what it should read

 

so ca-b=0, a+cb=0 c^2=-1

 

well, that's even easier then, since c^2=-1 has no solution in the reals.