show that there is an infinite number of solutions to x^2= -1 in the quaternions

Question

ganeshie8 · Answer

Hey!

anonymous · Answer

hi

ganeshie8 · Answer

check the last reply http://math.stackexchange.com/questions/288938/x21-0-uncountable-many-solutions

anonymous · Answer

Let a,b,c be real numbers satisfying a2+b2+c2=1 and let x=ai+bj+ck. Then
x2=(ai+bj+ck)(ai+bj+ck)=a2i2+abij+acik+baji+b2j2+bcjk+caki+cbkj+c2k2=?−(a2+b2+c2)=−1.

where are these terms acik+baji+bcjk+caki+cbkj?

reemii · Answer

$i^2 = j^2 = k^2 = ijk = -1$. From that (or from the definition..), you see that $i^2 = ijk = -1$ $\to i = jk$. Meaning, any product of two letterrs among $i,j,k$ can be replaced by the third letter, with a + or - sign. https://en.wikipedia.org/wiki/Quaternion#Definition

kainui · Answer

I've been playing around with this, maybe you guys will be interested. Instead of representing quaternions as i,j,k with a real part, I've found the algebra to be laborious so I have been using this special factorization since ij=k

$$q=a+bi+cj+dk =( a+bi) + (c+di)j = u+vj$$
so now u and v are complex numbers and they commute with each other, however for any complex number z, we have this interesting property when you commute, it takes the complex conjugate of it.
$$zj=jz^*$$

So now I'll try to solve $q^2=-1$ with it.

$$q^2 = (u+vj)(u+vj)=u^2+uvj+vju+vjvj $$
$$q^2= u^2 + v(u+u^*)j-vv^*$$
Now I recognize that this is the real part of u $\frac{u+u^*}{2} = \Re\{u\}$ and this is the squared length of v(a real number) $vv^* = |v|^2$.

$$q^2=u^2 -|v|^2 +2 v \Re\{u\}j$$ 

So we could sorta cut this off here, and just say $\Re\{u\}=0$ to get rid of the j part, which would leave just the imaginary part left,  make $u^2 = (i\Im\{u\})^2 = - (\Im\{u\})^2 = - |u|^2$  

$$q^2 = -(|u|^2+|v|^2) = -1$$
So any 3 numbers we pick that satisfy $a^2+b^2+c^2=1$ will work (cause the real part of u was thrown out) which happens to be the same result as in @ganeshie8 's reference.

anonymous · Answer

cool