User:Aadenboy/Zerons

Experimental extension to the real and imaginary axes, introducing the null and hyperimaginary axes, notated with $q$ and $h$* respectively, to "solve" the "issue" of division by zero being undefined.

Axioms

Definitions

• ${\sqrt {-1}}=i$ - Imaginary
• $1\div 0=q$ - Zeron
• $i\cdot q=h$ - Hyperimaginary
  • The symbol for hyperimaginary isn't correct, but there probably isn't any LaTeX-friendly equivalent, so $h$ will do.
• $i^{2}=-1$
• $q^{2}=1$
• $h^{2}=1$
• $q\cdot i=h$
• ${\frac {1}{i}}=-i$
• ${\frac {1}{q}}=q$
• ${\frac {1}{h}}=h$

Generalizations

• ${\sqrt {-n}}=i{\sqrt {n}}$
• $n\div 0=nq$
• $n\cdot mi=nmi$
• $-n\cdot mi=-nmi$
• $n\cdot mq=-nmq$
• $-n\cdot mi=nmq$
• $n\cdot mh=-nmh$
• $-n\cdot mh=nmh$
• $h\cdot q=i$
• $i\div q=h$
• $q\div i={\frac {1}{h}}=h$
• $h\cdot i=q$

Solving

Note: $P={\frac {\sqrt {2}}{2}}$ for ease of writing.

What is ${\sqrt {q}}$?

Taking a look at ${\sqrt {i}}=P+Pi$:

	$P$	$Pi$
$P$	$P^{2}$	$P^{2}i$
$Pi$	$P^{2}i$	$-P^{2}$
$P^{2}-P^{2}+2P^{2}i$
$2P^{2}i$
$1i$

A similar equation must be made to lead to $q$ after squaring.

$Pi+Ph$

	$Pi$	$Ph$
$Pi$	$-P^{2}$	$P^{2}q$
$Ph$	$P^{2}q$	$P^{2}$
$P^{2}-P^{2}+2P^{2}q$
$2P^{2}q$
$1q$

${\sqrt {q}}={\frac {i{\sqrt {2}}}{2}}+{\frac {h{\sqrt {2}}}{2}}$