Solving $f(x)=a,a \in A$?

Asked Jun 05 '18 at 17:15

Active Jun 05 '18 at 17:15

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In the beginning of mathematics, Number system was restricted to only integers.

then solutions of equations like $2x=5$ Needed us to extend it to rationals.

then solution of $x^2=\frac{1}{2}$ needed us to extend it to irrational.

Then solutions of $x^2=-1$ needed us to extend it too complex as well.

So considering this analogy solution of $f(x)=a,a \in A$ may not belong to A necessarily, Thus can we not say that numbers beyond complex numbers may exist.

ie. If $f(x)=z,z \in C$ but $x \notin C$

Is there any fundamental theorem which restricts this extension, Are there any clues of this extension, Please discuss and help.

asked Jun 05 '18 at 17:15

Maths textbook

If you mean numbers that are the zero of a polynomial in $\Bbb C$ then no because $\Bbb C$ is algebraically closed, meaning any polynomial with coefficients in it has a root in it. – John Cataldo Jun 05 '18 at 17:18
Well, Quaternions exist. – Rhys Hughes Jun 05 '18 at 17:18
@RhysHughes Thank you and it would be of great help if you let me know what king of equation can be solved only using quaternions – Maths textbook Jun 05 '18 at 17:21
Quaternions exist indeed but as John states, $\mathbb{C}$ is algebraically closed. Hence, there is no such extension from $\mathbb{C}$ to the quaternions. – Stan Tendijck Jun 05 '18 at 17:32

Solving $f(x)=a,a \in A$?

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