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The natural numbers cannot provide an answer to $1-2$.

The integers cannot provide an answer to $\frac{1}{2}$.

The rational numbers cannot provide an answer to $\sqrt{2}$.

The real numbers cannot provide an answer to $\sqrt{-1}$.

The complex numbers cannot provide an answer to what (leading to quaternions)? Is this the way it works?

This question asks similar. The accepted answer points to the Cayley-Dickson construction but that doesn't seem to address an operation between complex numbers that cannot be a complex number.

KevinRethwisch
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The motivation is described in the accepted answer to the question that you mentioned. On the other hand, the field $\mathbb C$ of complex numbers is algebraically closed. This means that any nonconstant polynomial function from $\mathbb C$ into itself has a complex root. Actually, it can be written as the product of first degree polynomials. So, the need to create the quaternions was not due to something lacking in $\mathbb C$, at least from the algebraic point of view.

José Carlos Santos
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    I feel that OP was asking whether, although the historical origin of quaternions was motivated by the construction of a "vector"-like objects endowed with a product operation, there is "some" operation among complex numbers highlighting the need for an enlarged set of variables. In other words, whether or not you can cook up a crazy function $f(z)$ such that $f(z)=0$ has no solution if $z\in\mathbb C$ but it does have roots if it is extended to quaternions. – Brightsun Nov 15 '18 at 11:11
  • Thanks Jose. I'm not advanced in mathematics but your answer is restricted to polynomial functions which is not a distinction I am concerned with. As @Brightsun pointed out is there any function of complex numbers, including non-polynomial functions, whose roots require a quaternion? – KevinRethwisch Nov 16 '18 at 04:15