Complex Numbers cannot be Extended to Algebra in Three Dimensions with Real Scalars

Theorem
It is not possible to extend the complex numbers to an algebra of real-scalars dimension 3.

Proof
that just one third extraplanar basic unit
 * $j \not\in 1\R+i\R$

extended the two-dimensional plane of complex numbers
 * $1\R+i\R = (1,i)\cdot\R^2$

to an algebra of three-dimensional vector space
 * $1\R+i\R+j\R = (1,i,j)\cdot\R^3$

so that $ij$ would have to be expressable as a linear combination
 * $ij=p+qi+rj=(1,i,j)\cdot(p,q,r)\in(1,i,j)\cdot\R^3$

and $iij$ would both equal
 * $iij=i(p+qi+rj)=pi-q+rij=pi-q+r(p+qi+rj)=(1,i,j)\cdot(rp-q,rq+p,r^2)$

and
 * $iij=(-1)j=(1,i,j)\cdot(0,0,-1)$,

then the equality would require their last coordinate to equal
 * $j\cdot iij =r^2=-1<0$,

requiring an impossible negative square of a real number
 * $j\cdot ij =r\in\R$,

hence the impossibility is proven by contradiction.

Also see

 * Complex Numbers can be Extended to Algebra in Four Real Dimensions