User:Aadenboy/Zerons

Experimental extension to the real and imaginary axes, introducing the null and hyperimaginary axes, notated with $q$ and $qi$ respectively.

Axioms

• $n\div 0=nq$

  $n\div mq=nmq$

  $0\div 0=0$

  Would the slope of a vertical line still be undefined?

• $n\times 0=-nq$

  $n\times mq=-nmq$

  $1\times -1q=-(1\times -1)q=-1q$

  Breaks the $n\times 0=0$ rule, but whatever.

• $q^{2}=1$

  $q^{n}$ from $0$ to $4$: 1 q 1 -q -1

• $nq\times mi=-nmqi$ (hyperimag.)
• $nq\div mi=nmqi$
• $n\times mqi=nmqi$
• $ni\times mqi=-nmq$

  $ni\div mqi=nmq$

• $nq\times mqi=nmi$

  $nq\div mqi=-nmi$

• $nqi\times mqi=-nm$

  $nqi\div nqi=nm$

• $qi^{2}=-1$
• $(a+bi+cq+dqi)$