How to solve $x \times 0 = n$ where $n$ does not equal $0$? So recently I was watching some videos on complex numbers. I was surprised to learn the uses of $i$ in different problems. For example, I never thought it would make sense that $x^2 + 1 = 0$ has any solutions at all. But apparently it does when you consider that this equation actually has two solutions that aren't visible by graphing this equation on your calculator.

However, my question is somewhat related yet somewhat different. What if we had the following equation:

$$x \times 0 = 1$$

Now, from my knowledge, there's no solutions to this problem because no number that I know multiplied by zero would be equal to anything but zero. But would it be possible to create your own solutions to this problem?

I noticed when I tried to solve this problem I came across two main issues:

$$x \times 0 \, / \, 0 = 1 \, / \, 0$$

First off, what is $0 \, / \, 0$? Does $0 \, / \, 0$ "cancel out?" Secondly, isn't $1 \, / \, 0$ undefined? Not to mention, what if I have another problem like this one:

$$x \times 0 = 5$$

How would the solutions, if any, differ from this problem and the first problem?

Keep in mind, I'm not a mathematics genius or anything, so I hope you won't be disappointed with me for suggesting a problem like this because I'm simply just curious. This is not meant to stump anyone, nor is it meant to waste your time. Any responses would be greatly appreciated.
You can define a new (extended) number system with the property that there is a solution to $0x=1$, and call that solution (or one of the solutions if there are more) $k$. You just have to be willing to accept the consequences. For instance, if you still want the property $0+0=0$, and the property of multiplication distributing over parentheses, then you get
$$
1=0k=(0+0)k=0k+0k=1+1=2
$$
so suddenly $1=2$ in this new number system. That's probably not desirable. Therefore you have to give up at least one of the three assumptions I've made above. I prefer to give up the existence of $k$.
The way I see it your question is this:

$x^2 + 1 = 0$ means $x^2 = -1$.  If $x \ge 0$ then $x^2 \ge 0$.  If $x &lt; 0$ then $x^2 &gt; 1$.  So $x^2 &lt; 1$ is impossible.  So there should be no answer.  But mathematicians say:  "Oh, that's okay.  It's no problem.  We just introduce a new number called $i$ where $i^2 = -1$ and we can say $i=\sqrt{-1}$.  That's easy why do you have a problem with it? You must be stupid to be stuck in your old world-view that you can't accept something new."

But if we did $x\times 0 = 1$ there is no solution.  But what if I said "Let $\mu$ be a new type of number where $\mu \times 0 =1$ and we can say $\mu = \frac 10$."  But when you say this mathematicians say:  "What?  Are you crazy?  We can't define $\frac 10$! $x\times 0 = 0$ and nothing else ever.  That's a fact.  Man, you sure are stupid, if you think you can just make up answer just by inventing numbers. "

So why is it okay to "make up" a number $i = \sqrt{-1}$ but it's not okay to "make up" a number $\mu \times 0 = 1$?  It seems like both of them should not be allowed or both should be all.  (And why are these hypothetical mathematician so dang rude?)

.....

Okay.  The idea of "making up" a new kind of number is called "extending" a number system, and it's allowed as long as the results are consistent.

For instance the first number system we have is $\mathbb N = \{1,2,3,4....\}$.  This is good.  Our basic two rules are that if you have two different numbers $a$ and $b$ they are ... different; They can't be equal to each other.  And that if you have $a+ b$ then $a + b = c$ which is also a number as well as $a \times b = d$ is also a number.  (We don't have two numbers that we can't add together.) Those are basic rules.

And there are some not so basic rules.  $1$ is the smallest number.  Every number has a number that comes immediately after it.  Etc.  Those are interesting rules but they are not fundamental.

If I asked you to solve $x + a = a$... there is not solution.  So we say, let's make up a new type of number called $0$ where $0 + a = a$ for all $a$.

This is consistent with our idea that all numbers are distinct and that we can add any two numbers together.  

To make it consistent if $a \times b = b + b+ b....$ then $0\times a = nothing$ and $a*0 = 0+ 0+0+.... = 0$ so that's a new rule.  $0*a = a*0 = 0$.

Some of a rules change.  $1$ isn't the smallest number; $0$ is.

The next  extension: Can we solve $x + a = 0$.  We "make up" negative numbers.  They are consistent with our most important rules. But we no longer have the rule that there is a smallest number.

The next extension: Can we solve $nk = m$.  We have to make up the rationals.  We no longer have the rules that every number has a number that comes immediately after it.  But the must important one is that numbers are distinct.  We have to make an exception that $0k = m$ has not solution. As that would mean $m=0*k = 0+0+0... = 0$ and we have two different numbers equaling the same.

We also extended and discovered that $\frac km*\frac jn$ doesn't mean adding $\frac jn$ a total of $\frac km$ times.  We are no longer doing whole numbers but that's okay.  $\frac km *\frac jn = \frac {kj}{mn}$ works and is consistent.

So.....

Upshot.  Allowing $x^2 + 1 =0$ by "making up" $i^2 = -1$ makes things weird BUT consistent.  We still have multiplication and addition are close operations.  (We always add or multiply any two numbers together) and numbers are distinct.

We lose some basic rules.  The most strange is we can no longer say numbers are bigger or smaller than each other!  $i \not &lt; 0; i\ne 0$ and $i \not  &gt; 0$.  $i$ simply... goes off at a right a right angle.  We just can't say that one complex number is bigger than another!

But otherwise things are consistent.

Not so if we try to "make up"  $\mu\times 0 = 1$ and $\mu = \frac 10$.  We can simply get $(a*\mu)*0 = a*(\frac 10\times 0) = a*1 = a$ but $a*0= 0$ so $a*0^2 = 0$ so $0 = a*0^2\mu = 0*\mu = 1$.  .... we just can not make consistent sense of this.

.....

So we have:  $\mathbb N\subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset\mathbb C$ are all consistent exstensions.

$x + a = a$ means $x \in \mathbb N$ is not possible but $x \in \mathbb Z$ is.  $\frac x7 = 4$ means $x \not \in \mathbb N$ but $x \in \mathbb Q$ is possible.  And $x^2 = 2$ means $x \not \in \mathbb Q$ but $x \in \mathbb R$ is possible. And $x^2 + 1=0$ means $x\not \in \mathbb R$ but $x \in \mathbb C$ is possible.  All that is consistent.

But $x\time 0 = 1$ is simply not consistent and can't happen in any extension, on which we have basic rules of arithmetic.