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Does the set of complex numbers contain every number that exist? Around 600 years ago people thought the set of real numbers contain any number that one can think of. Slowly, with the introduction of $\sqrt{-1}$ and complex numbers that perspective changed. Now my question is will there be any other number that humans have not thought of? Or are we guaranteed that the set of complex numbers contain every number?

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    $\begingroup$ $\Bbb C$ does not contain all quaternions, nor does it contain all ordinals. Ultimately, it depends on what you call "number". $\endgroup$ – Hagen von Eitzen Mar 2 at 7:55
  • $\begingroup$ Actually I am curious to know, what is a number? $\endgroup$ – Rob Mar 2 at 7:57
  • $\begingroup$ The name of the resulting number system escapes me so early in the morning but consider $j^2=1$ but with $j\neq 1$. My point is: there are number systems independent of the complex number system (at least prima facie). $\endgroup$ – Shaun Mar 2 at 8:04
  • $\begingroup$ @Shaun then what is $j$? $\endgroup$ – Rob Mar 2 at 8:15
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    $\begingroup$ The resulting numbers are called split-complex numbers. $\endgroup$ – Shaun Mar 2 at 8:20
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you may know quaternion or p-adic number. It can be called a number.

And in algebra, people deal with a lot of what we may call number. So, I think you can answer yourself if you study algebra.

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I guess this depends on what you mean by numbers. If you are referring to closure of operations (except cases such as division by zero etc.), then the complex numbers are algebraically closed. However, we can extend complex numbers to more dimensions, such as the quaternions and so on. You could also introduce the ordinals and much more number systems.

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