# Undefined and Imaginary Numbers

I found something strange with undefined and imaginary numbers.

First, please take this two mathematical definitions into consideration.

1) The square root of a negative number is undefined.

2) The square root of -1, or

*i*, is defined as an imaginary number.

This is traditionally shown using these two equations.

*i*= √ -1

*i*

^{2}= -1

So I have to ask, if √ -1 or

*i*is

**defined**as imaginary, then how is it also

**undefined**? Can something be both defined and undefined at the same time? I think not. The only possible explanation is that √ -1 or

*i*is both undefined and imaginary, and imaginary is just a mathematical representation of something undefined and not a definition in itself.

This got me thinking if undefined equations can also be imaginary too. Please take this into consideration:

If we draw the graph of:

**y = 1 / x**

As you can see, when x approaches 0, then y approaches infinity and -infinity.

Thus, y is undefined when x=0.

Alternatively, if we draw the graph of:

**y = (1 / x)**

^{2}As you can see, when x approaches 0, y approaches infinity.

Therefore, we could write:

**( 1 / 0 )**

^{2}=Thus, we can define this imaginary equation or imaginary number:

*b*= 1 / 0_{1}Then

*b*can be usable, yet still undefined and imaginary.

_{1}Note this equation above is true when the numerator is equal to any constant, except for zero (i.e. zero divided by zero). In other words,

**( c / 0 )**when c <> 0

^{2}=Therefore, if we draw the graph of:

**y = 0 / x**

As you can see, this results with a straight line or y=0, except when x=0. However when x=0, then y is approaching 0 too.

Therefore, we could write:

**0 / 0 =**

Thus, we can define this imaginary equation or imaginary number:

*b*= 0 / 0_{2}Then

*b*can be usable, yet still undefined and imaginary too.

_{2}by Phil for Humanity

on 20170901