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We define $i$ to be $i^2 = -1$. Can we also define $\omega$ such that if it is multiplied by $0$, we get $1$? Could we perform operations on it just like we do on imaginary numbers? If so, does $\omega \in \mathbb{C}$ or not?

I read Why can one multiply by zero and Max's answer suggests that the answer to my last question is no.

I am not sure about the tags I have given to the question. Please edit them if needed.

sato
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    $$1 = \omega \times 0 = \omega \times (0 \times 0) = (\omega \times 0)\times 0 = 1 \times 0 = 0$$ – Calvin Khor Sep 23 '21 at 12:31
  • We like certain properties of the standard number systems. Suppose you add $\omega$ so that $\omega\cdot 0 = 1$. Can you use the ordinary properties of a field to obtain a contradiction? For example, @CalvinKhor comment shows multiplication cannot be associative. From that conclude that the new system is not contained in$\mathbb C$. – GEdgar Sep 23 '21 at 12:31
  • In general, we can add a lot of things to number systems. Any time you add something, you lose some properties. Generally speaking, we like adding things that remove as few desirable properties and add more desirable properties as possible. – Rushabh Mehta Sep 23 '21 at 13:05

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This might not be an answer but it is too long to be posted as a comment.

So if I read you right you want to add an element $\omega$ to $\mathbb{C}$ so that $\omega \times 0 = 1$. You can do so! Take any element not in $\mathbb{C}$, denote it $\omega$ and fix $\mathbb{C}' = \mathbb{C} \cup \{\omega\}$. As you did, denote by $*$ your new "multiplication", $x*y$ is defined as the complex product if $x$ and $y$ are both complex ; as $0$ if $\{x,y\} = \{0,\omega\}$ ; and whatever you want in the remaining case.

Now $\mathbb{C}'$ is a set endowed of a "multiplication", which turns to merely be an application from $\mathbb{C}' \times \mathbb{C}'$ to $\mathbb{C}'$. But what axiom does it satifsfy? Clearly nothing interesting. As mentioned in the first comment the assiociativity is not granted.

The set of complex numbers is (at least, I am not an algebrist) a field. The set $\mathbb{C}'$ endowed of $*$ is not a group so you cannot perform operations as you do in $\mathbb{C}$.

Moral of the story : the operation you want to add can be supported but destroys the algebaic structure which allows you to perform nice computations.

blamethelag
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