Consider the following function;
$$f(x,y) = \sqrt{x} + \sqrt{y}$$
If this function were to be plotted onto a 3-dimensional co-ordinate space, then the x and y axes would be orthogonal to each other.
What happens to this function however, when the values for both x and y are less than zero? For example, if we have;
$$f(x,y) = \sqrt{-1} + \sqrt{-1}$$
does this become;
$$f(x,y) = i + j$$
and if so, are i and j also considered orthogonal to each other, in the same way that the x and y axes are?
Furthermore, what if the original function is now altered so as to be defined as follows;
$$f(x,y,z) = \sqrt{x} + \sqrt{y} + \sqrt{z}$$
Similarly, what happens to this function if x, y, and z are less than zero. For example, if we have;
$$f(x,y,z) = \sqrt{-1} + \sqrt{-1} + \sqrt{-1}$$
does this become;
$$f(x,y,z) = i + j + k$$
where i, j, and k in this case are orthogonal to each other? Is what we have here, an instance of a hypercomplex number (with a real component of zero), and if it had a non-zero real component tacked onto it, would this be an example of Hamilton's quaternions?
Thanks in advance.
No(Edit: No is not completely correct. It depends should be the answer) . The solution would simply be expanded to include complex numbers. In your example this wouldnt be $ i + j $ but $2i$. Another axis in the 4th dimension would be necessary to present the solution $2i$.
Edit: You could however define that the function maps to the quaternions in which case this would be one of the infinitely many possible solutions, so it would be correct the way hou wrote it. But so would 2i, i+k, j+k,...